Metallica Batman Statics

PROBLEM 4.1 The boom on a 4300-kg truck is used to unload a pallet of shingles of mass 1600 kg. Determine the reaction at each of the two (a) rear wheels B, (b) front wheels C.

SOLUTION

(

)

(

)

WA = mA g = (1600 kg ) 9.81 m/s 2 = 15696 N

WA = 15.696 kN

or

WG = mG g = ( 4300 kg ) 9.81 m/s 2

= 42 183 N WG = 42.183 kN

or (a) From f.b.d. of truck with boom ΣM C = 0:

(15.696 kN ) ( 0.5 + 0.4 + 6 cos15° ) m  − 2FB ( 0.5 + 0.4 + 4.3) m  + ( 42.183 kN )( 0.5 m ) = 0 ∴ 2 FB =

126.185 = 24.266 kN 5.2 or FB = 12.13 kN

(b) From f.b.d. of truck with boom ΣM B = 0:

(15.696 kN ) ( 6 cos15° − 4.3) m  − ( 42.183 kN ) ( 4.3 + 0.4 ) m  + 2 FC ( 4.3 + 0.9 ) m  = 0

∴ 2 FC =

174.786 = 33.613 kN 5.2 or FC = 16.81 kN

Check:

ΣFy = 0:

( 33.613 − 42.183 + 24.266 − 15.696 ) kN = 0? ( 57.879 − 57.879 ) kN = 0

ok

PROBLEM 4.2 Two children are standing on a diving board of mass 65 kg. Knowing that the masses of the children at C and D are 28 kg and 40 kg, respectively, determine (a) the reaction at A, (b) the reaction at B.

SOLUTION

(

)

(

)

(

)

WG = mG g = ( 65 kg ) 9.81 m/s 2 = 637.65 N WC = mC g = ( 28 kg ) 9.81 m/s 2 = 274.68 N WD = mD g = ( 40 kg ) 9.81 m/s 2 = 392.4 N (a) From f.b.d. of diving board ΣM B = 0: − Ay (1.2 m ) − ( 637.65 N )( 0.48 m ) − ( 274.68 N )(1.08 m ) − ( 392.4 N )( 2.08 m ) = 0 ∴ Ay = −

1418.92 = −1182.43 N 1.2 or A y = 1.182 kN

(b) From f.b.d. of diving board ΣM A = 0: By (1.2 m ) − 637.65 N (1.68 m ) − 274.68 N ( 2.28 m ) − 392.4 N ( 3.28 m ) = 0 ∴ By =

2984.6 = 2487.2 N 1.2 or B y = 2.49 kN

Check:

ΣFy = 0:

( −1182.43 + 2487.2 − 637.65 − 274.68 − 392.4 ) N = 0? ( 2487.2 − 2487.2 ) N = 0

ok

PROBLEM 4.3 Two crates, each weighing 250 lb, are placed as shown in the bed of a 3000-lb pickup truck. Determine the reactions at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

(a) From f.b.d. of truck ΣM B = 0:

( 250 lb )(12.1 ft ) + ( 250 lb )( 6.5 ft ) + ( 3000 lb )( 3.9 ft ) − ( 2FA )( 9.8 ft ) = 0 ∴ 2 FA =

16350 = 1668.37 lb 9.8 ∴ FA = 834 lb

(b) From f.b.d. of truck ΣM A = 0:

( 2FB )( 9.8 ft ) − ( 3000 lb )( 5.9 ft ) − ( 250 lb )( 3.3 ft ) + ( 250 lb )( 2.3 ft ) = 0 ∴ 2 FB =

17950 = 1831.63 lb 9.8 ∴ FB = 916 lb

Check:

ΣFy = 0:

( −250 + 1668.37 − 250 − 3000 + 1831.63) lb = 0? ( 3500 − 3500 ) lb = 0

ok

PROBLEM 4.4 Solve Problem 4.3 assuming that crate D is removed and that the position of crate C is unchanged.

P4.3 The boom on a 4300-kg truck is used to unload a pallet of shingles of mass 1600 kg. Determine the reaction at each of the two (a) rear wheels B, (b) front wheels C

SOLUTION

(a) From f.b.d. of truck ΣM B = 0:

( 3000 lb )( 3.9 ft ) − ( 2FA )( 9.8 ft ) + ( 250 lb )(12.1 ft ) = 0 ∴ 2 FA =

14725 = 1502.55 lb 9.8 or FA = 751 lb

(b) From f.b.d. of truck ΣM A = 0:

( 2FB )( 9.8 ft ) − ( 3000 lb )( 5.9 ft ) + ( 250 lb )( 2.3 ft ) = 0 ∴ 2 FB =

17125 = 1747.45 lb 9.8 or FB = 874 lb

Check:

ΣFy = 0:  2 ( 751 + 874 ) − 3000 − 250  lb = 0?

( 3250 − 3250 ) lb = 0

ok

PROBLEM 4.5 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B if (a) a = 100 mm, (b) a = 70 mm.

SOLUTION (a)

From f.b.d. of bracket ΣM B = 0: − (10 N )( 0.18 m ) − ( 30 N )( 0.1 m ) + ( 40 N )( 0.06 m ) + A ( 0.12 m ) = 0 ∴ A=

2.400 = 20 N 0.12

or A = 20.0 N

ΣM A = 0: B ( 0.12 m ) − ( 40 N )( 0.06 m ) − ( 50 N )( 0.12 m ) − ( 30 N )( 0.22 m ) − (10 N )( 0.3 m ) = 0 ∴ B=

18.000 = 150 N 0.12

or B = 150.0 N

(b)

From f.b.d. of bracket ΣM B = 0: − (10 N )( 0.15 m ) − ( 30 N )( 0.07 m ) + ( 40 N )( 0.06 m ) + A ( 0.12 m ) = 0 ∴ A=

1.200 = 10 N 0.12

or A = 10.00 N

ΣM A = 0: B ( 0.12 m ) − ( 40 N )( 0.06 m ) − ( 50 N )( 0.12 m ) − ( 30 N )( 0.19 m ) − (10 N )( 0.27 m ) = 0 ∴ B=

16.800 = 140 N 0.12

or B = 140.0 N

PROBLEM 4.6 For the bracket and loading of Problem 4.5, determine the smallest distance a if the bracket is not to move. P4.5 A T-shaped bracket supports the four loads shown. Determine the reactions at A and B if (a) a = 100 mm, (b) a = 70 mm.

SOLUTION The amin value will be based on A = 0

From f.b.d. of bracket ΣM B = 0:

( 40 N )( 60 mm ) − ( 30 N )( a ) − (10 N )( a + 80 mm ) = 0 ∴a=

1600 = 40 mm 40 or amin = 40.0 mm

PROBLEM 4.7 A hand truck is used to move two barrels, each weighing 80 lb. Neglecting the weight of the hand truck, determine (a) the vertical force P which should be applied to the handle to maintain equilibrium when α = 35o , (b) the corresponding reaction at each of the two wheels.

SOLUTION a1 = ( 20 in.) sin α − ( 8 in.) cos α a2 = ( 32 in.) cos α − ( 20 in.) sin α b = ( 64 in.) cos α

From f.b.d. of hand truck ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0 ΣFy = 0: P − 2w + 2 B = 0

(2)

α = 35°

For

a1 = 20sin 35° − 8cos 35° = 4.9183 in. a2 = 32cos 35° − 20sin 35° = 14.7413 in. b = 64cos 35° = 52.426 in.

(a)

From Equation (1) P ( 52.426 in.) − 80 lb (14.7413 in.) + 80 lb ( 4.9183 in.) = 0

∴ P = 14.9896 lb (b)

or P = 14.99 lb

From Equation (2) 14.9896 lb − 2 ( 80 lb ) + 2 B = 0 ∴ B = 72.505 lb

(1)

or B = 72.5 lb

PROBLEM 4.8 Solve Problem 4.7 when α = 40o. P4.7 A hand truck is used to move two barrels, each weighing 80 lb. Neglecting the weight of the hand truck, determine (a) the vertical force P which should be applied to the handle to maintain equilibrium when α = 35o , (b) the corresponding reaction at each of the two wheels.

SOLUTION a1 = ( 20 in.) sin α − ( 8 in.) cos α a2 = ( 32 in.) cos α − ( 20 in.) sin α b = ( 64 in.) cos α

From f.b.d. of hand truck ΣM B = 0: P ( b ) − W ( a2 ) + W ( a1 ) = 0 ΣFy = 0: P − 2w + 2 B = 0

(2)

α = 40°

For

a1 = 20sin 40° − 8cos 40° = 6.7274 in. a2 = 32cos 40° − 20sin 40° = 11.6577 in. b = 64cos 40° = 49.027 in.

(a)

From Equation (1) P ( 49.027 in.) − 80 lb (11.6577 in.) + 80 lb ( 6.7274 in.) = 0

∴ P = 8.0450 lb or P = 8.05 lb (b)

(1)

From Equation (2) 8.0450 lb − 2 ( 80 lb ) + 2 B = 0 ∴ B = 75.9775 lb or B = 76.0 lb

PROBLEM 4.9 Four boxes are placed on a uniform 14-kg wooden plank which rests on two sawhorses. Knowing that the masses of boxes B and D are 4.5 kg and 45 kg, respectively, determine the range of values of the mass of box A so that the plank remains in equilibrium when box C is removed.

SOLUTION WA = m A g WB = mB g = 4.5 g

WD = mD g = 45 g WG = mG g = 14 g

For ( m A )min, E = 0 ΣM F = 0:

( mA g )( 2.5 m ) + ( 4.5g )(1.6 m ) + (14 g )(1 m ) − ( 45g )( 0.6 m ) = 0 ∴ m A = 2.32 kg

For ( m A )max, F = 0: ΣM E = 0: mA g ( 0.5 m ) − ( 4.5g )( 0.4 m ) − (14 g )(1 m ) − ( 45g )( 2.6 m ) = 0 ∴ mA = 265.6 kg or 2.32 kg ≤ mA ≤ 266 kg

PROBLEM 4.10 A control rod is attached to a crank at A and cords are attached at B and C. For the given force in the rod, determine the range of values of the tension in the cord at C knowing that the cords must remain taut and that the maximum allowed tension in a cord is 180 N.

SOLUTION

(TC )max,

For ΣM O = 0:

TB = 0

(TC )max ( 0.120 m ) − ( 400 N )( 0.060 m ) = 0 (TC )max

= 200 N > Tmax = 180 N

∴ (TC )max = 180.0 N

(TC )min ,

For ΣM O = 0:

TB = Tmax = 180 N

(TC )min ( 0.120 m ) + (180 N )( 0.040 m ) − ( 400 N )( 0.060 m ) = 0 ∴ (TC )min = 140.0 N

Therefore,

140.0 N ≤ TC ≤ 180.0 N

PROBLEM 4.11 The maximum allowable value of each of the reactions is 360 N. Neglecting the weight of the beam, determine the range of values of the distance d for which the beam is safe.

SOLUTION From f.b.d. of beam ΣFx = 0: Bx = 0

so that

B = By

ΣFy = 0: A + B − (100 + 200 + 300 ) N = 0

A + B = 600 N

or

Therefore, if either A or B has a magnitude of the maximum of 360 N, the other support reaction will be < 360 N ( 600 N − 360 N = 240 N ) . ΣM A = 0:

(100 N )( d ) − ( 200 N )( 0.9 − d ) − ( 300 N )(1.8 − d ) + B (1.8 − d ) = 0 d =

or

720 − 1.8B 600 − B

Since B ≤ 360 N, d =

720 − 1.8 ( 360 ) = 0.300 m 600 − 360

ΣM B = 0:

d ≥ 300 mm

(100 N )(1.8) − A (1.8 − d ) + ( 200 N )( 0.9 ) = 0 d =

or

or

1.8 A − 360 A

Since A ≤ 360 N, d =

1.8 ( 360 ) − 360 = 0.800 m 360

or

d ≤ 800 mm

or 300 mm ≤ d ≤ 800 mm

PROBLEM 4.12 Solve Problem 4.11 assuming that the 100-N load is replaced by a 160-N load. P4.11 The maximum allowable value of each of the reactions is 360 N. Neglecting the weight of the beam, determine the range of values of the distance d for which the beam is safe.

SOLUTION From f.b.d of beam ΣFx = 0: Bx = 0

so that

B = By

ΣFy = 0: A + B − (160 + 200 + 300 ) N = 0 A + B = 660 N

or

Therefore, if either A or B has a magnitude of the maximum of 360 N, the other support reaction will be < 360 N ( 660 − 360 = 300 N ) . ΣM A = 0: 160 N ( d ) − 200 N ( 0.9 − d ) − 300 N (1.8 − d ) + B (1.8 − d ) = 0 d =

or

720 − 1.8B 660 − B

Since B ≤ 360 N, d =

720 − 1.8 ( 360 ) = 0.240 m 660 − 360

or

d ≥ 240 mm

ΣM B = 0: 160 N (1.8 ) − A (1.8 − d ) + 200 N ( 0.9 ) = 0 d =

or

1.8 A − 468 A

Since A ≤ 360 N, d =

1.8 ( 360 ) − 468 = 0.500 m 360

or

d ≥ 500 mm

or 240 mm ≤ d ≤ 500 mm

PROBLEM 4.13 For the beam of Sample Problem 4.2, determine the range of values of P for which the beam will be safe knowing that the maximum allowable value of each of the reactions is 45 kips and that the reaction at A must be directed upward. SOLUTION

For the force of P to be a minimum, A = 0. With A = 0, ΣM B = 0: Pmin ( 6 ft ) − ( 6 kips )( 2 ft ) − ( 6 kips )( 4 ft ) = 0

∴ Pmin = 6.00 kips For the force P to be a maximum, A = A max = 45 kips With A = 45 kips, ΣM B = 0: − ( 45 kips )( 9 ft ) + Pmax ( 6 ft ) − ( 6 kips )( 2 ft ) − ( 6 kips )( 4 ft ) = 0 ∴ Pmax = 73.5 kips A check must be made to verify the assumption that the maximum value of P is based on the reaction force at A. This is done by making sure the corresponding value of B is < 45 kips. ΣFy = 0: 45 kips − 73.5 kips + B − 6 kips − 6 kips = 0 ∴ B = 40.5 kips < 45 kips

∴ ok

or Pmax = 73.5 kips and 6.00 kips ≤ P ≤ 73.5 kips

PROBLEM 4.14 For the beam and loading shown, determine the range of values of the distance a for which the reaction at B does not exceed 50 lb downward or 100 lb upward.

SOLUTION To determine amax the two 150-lb forces need to be as close to B without having the vertical upward force at B exceed 100 lb. From f.b.d. of beam with B = 100 lb ΣM D = 0: − (150 lb ) ( amax − 4 in.) − (150 lb ) ( amax − 1 in.) − ( 25 lb )( 2 in.) + (100 lb )( 8 in.) = 0 amax = 5.00 in.

or

To determine amin the two 150-lb forces need to be as close to A without having the vertical downward force at B exceed 50 lb. From f.b.d. of beam with B = 50 lb ΣM D = 0:

(150 lb )( 4 in. − amin ) − (150 lb )( amin

− 1 in.)

− ( 25 lb )( 2 in.) − ( 50 lb )( 8 in.) = 0

or Therefore,

amin = 1.00 in.

or 1.00 in. ≤ a ≤ 5.00 in.

PROBLEM 4.15 A follower ABCD is held against a circular cam by a stretched spring, which exerts a force of 21 N for the position shown. Knowing that the tension in rod BE is 14 N, determine (a) the force exerted on the roller at A, (b) the reaction at bearing C.

SOLUTION Note: From f.b.d. of ABCD Ax = A cos 60° = Ay = A sin 60° = A

A 2

3 2

(a) From f.b.d. of ABCD  A ΣM C = 0:   ( 40 mm ) − 21 N ( 40 mm ) 2 + 14 N ( 20 mm ) = 0 ∴ A = 28 N or A = 28.0 N

60°

(b) From f.b.d. of ABCD ΣFx = 0: C x + 14 N + ( 28 N ) cos 60° = 0 ∴ C x = −28 N ΣFy = 0:

C y − 21 N + ( 28 N ) sin 60° = 0

∴ C y = −3.2487 N Then and

C =

C x = 28.0 N

or

C x2 + C y2 =

or

C y = 3.25 N

( 28)2 + ( 3.2487 )2

= 28.188 N

 Cy  −1  −3.2487   = tan   = 6.6182°  −28   Cx 

θ = tan −1 

or C = 28.2 N

6.62°

PROBLEM 4.16 A 6-m-long pole AB is placed in a hole and is guyed by three cables. Knowing that the tensions in cables BD and BE are 442 N and 322 N, respectively, determine (a) the tension in cable CD, (b) the reaction at A.

SOLUTION Note: DB =

( 2.8)2 + ( 5.25)2

= 5.95 m

DC =

( 2.8)2 + ( 2.10 )2

= 3.50 m

(a) From f.b.d. of pole  2.8 m   ΣM A = 0: − ( 322 N )( 6 m ) +   ( 442 N )  ( 6 m ) 5.95 m     2.8 m   +   TCD  ( 2.85 m ) = 0  3.50 m   ∴ TCD = 300 N or TCD = 300 N (b) From f.b.d. of pole  2.8 m  ΣFx = 0: 322 N −   ( 442 N )  5.95 m   2.8 m  −  ( 300 N ) + Ax = 0  3.50 m  ∴ Ax = 126 N

or

A x = 126 N

 5.25 m  ΣFy = 0: Ay −   ( 442 N )  5.95 m  ∴ Ay = 570 N Then and

A=

Ax2 + Ay2 =

or

 2.10 m  −  ( 300 N ) = 0  3.50 m 

A y = 570 N

(126 )2 + ( 570 )2

= 583.76 N

 570 N 

θ = tan −1   = 77.535°  126 N  or A = 584 N

77.5°

PROBLEM 4.17 Determine the reactions at A and C when (a) α = 0, (b) α = 30o.

SOLUTION (a)

(a) α = 0° From f.b.d. of member ABC

(80 lb )(10 in.) + (80 lb )( 20 in.) − A ( 40 in.) = 0

ΣM C = 0:

∴ A = 60 lb or A = 60.0 lb ΣFy = 0: C y + 60 lb = 0 ∴ C y = −60 lb

or

C y = 60 lb

ΣFx = 0: 80 lb + 80 lb + Cx = 0 ∴ C x = −160 lb C =

Then

C x2 + C y2 =

or

C x = 160 lb

(160 )2 + ( 60 )2

= 170.880 lb

 Cy  −1  −60   = tan   = 20.556°  −160   Cx 

θ = tan −1 

and

or C = 170.9 lb (b)

20.6°

(b) α = 30° From f.b.d. of member ABC ΣM C = 0:

(80 lb )(10 in.) + (80 lb )( 20 in.) − ( A cos 30° )( 40 in.) + ( A sin 30° )( 20 in.) = 0 ∴ A = 97.399 lb or A = 97.4 lb

60°

PROBLEM 4.17 CONTINUED ΣFx = 0: 80 lb + 80 lb + ( 97.399 lb ) sin 30° + Cx = 0 ∴ C x = −208.70 lb

C x = 209 lb

or

ΣFy = 0: C y + ( 97.399 lb ) cos 30° = 0 ∴ C y = −84.350 lb Then and

C =

C x2 + C y2 =

or

C y = 84.4 lb

( 208.70 )2 + (84.350 )2

= 225.10 lb

 Cy  −1  −84.350   = tan   = 22.007° C  −208.70   x

θ = tan −1 

or C = 225 lb

22.0°

PROBLEM 4.18 Determine the reactions at A and B when (a) h = 0, (b) h = 8 in.

SOLUTION (a)

(a) h = 0 From f.b.d. of plate ΣM A = 0:

( B sin 30° )( 20 in.) − ( 40 lb )(10 in.) = 0 ∴ B = 40 lb or B = 40.0 lb

30°

ΣFx = 0: Ax − ( 40 lb ) cos 30° = 0 ∴ Ax = 34.641 lb

or

A x = 34.6 lb

ΣFy = 0: Ay − 40 lb + ( 40 lb ) sin 30° = 0 ∴ Ay = 20 lb A=

Then

Ax2 + Ay2 =

A y = 20.0 lb

or

( 34.641)2 + ( 20 )2

= 39.999 lb

 Ay  20  −1   = tan   = 30.001°  34.641   Ax 

θ = tan −1 

and

or A = 40.0 lb (b)

30°

(b) h = 8 in. From f.b.d. of plate ΣM A = 0:

( B sin 30° )( 20 in.) − ( B cos 30° )(8 in.) − ( 40 lb )(10 in.) = 0 ∴ B = 130.217 lb or B = 130.2 lb

30.0°

PROBLEM 4.18 CONTINUED ΣFx = 0: Ax − (130.217 lb ) cos30° = 0 ∴ Ax = 112.771 lb

or

A x = 112.8 lb

ΣFy = 0: Ay − 40 lb + (130.217 lb ) sin 30° = 0 ∴ Ay = −25.108 lb Then and

A=

Ax2 + Ay2 =

or

A y = 25.1 lb

(112.771)2 + ( 25.108)2

= 115.532 lb

 Ay  −1  −25.108   = tan   = −12.5519° A  112.771   x

θ = tan −1 

or A = 115.5 lb

12.55°

PROBLEM 4.19 The lever BCD is hinged at C and is attached to a control rod at B. If P = 200 N, determine (a) the tension in rod AB, (b) the reaction at C.

SOLUTION (a) From f.b.d. of lever BCD ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0 ∴ TAB = 300 N (b) From f.b.d. of lever BCD ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0 ∴ C x = −380 N

C x = 380 N

or

ΣFy = 0: C y + 0.8 ( 300 N ) = 0 ∴ C y = −240 N Then and

C =

C x2 + C y2 =

C y = 240 N

or

( 380 )2 + ( 240 )2

= 449.44 N

 Cy  −1  −240   = tan   = 32.276°  −380   Cx 

θ = tan −1 

or C = 449 N

32.3°

PROBLEM 4.20 The lever BCD is hinged at C and is attached to a control rod at B. Determine the maximum force P which can be safely applied at D if the maximum allowable value of the reaction at C is 500 N.

SOLUTION From f.b.d. of lever BCD ΣM C = 0: TAB ( 50 mm ) − P ( 75 mm ) = 0 ∴ TAB = 1.5P

(1)

ΣFx = 0: 0.6TAB + P − C x = 0 ∴ C x = P + 0.6TAB

(2)

Cx = P + 0.6 (1.5P ) = 1.9P

From Equation (1)

ΣFy = 0: 0.8TAB − C y = 0 ∴ C y = 0.8TAB

(3)

C y = 0.8 (1.5P ) = 1.2P

From Equation (1) From Equations (2) and (3) C =

C x2 + C y2 =

(1.9P )2 + (1.2P )2

= 2.2472 P

Since Cmax = 500 N, ∴ 500 N = 2.2472Pmax or

Pmax = 222.49 lb or P = 222 lb

PROBLEM 4.21 The required tension in cable AB is 800 N. Determine (a) the vertical force P which must be applied to the pedal, (b) the corresponding reaction at C.

SOLUTION (a) From f.b.d. of pedal ΣM C = 0: P ( 0.4 m ) − ( 800 N ) ( 0.18 m ) sin 60° = 0

∴ P = 311.77 N or P = 312 N (b) From f.b.d. of pedal ΣFx = 0: Cx − 800 N = 0 ∴ Cx = 800 N C x = 800 N

or

ΣFy = 0: C y − 311.77 N = 0 ∴ C y = 311.77 N

C y = 311.77 N

or Then

and

C =

C x2 + C y2 =

(800 )2 + ( 311.77 )2

= 858.60 N

 Cy  −1  311.77   = tan   = 21.291° C  800   x

θ = tan −1 

or C = 859 N

21.3°

PROBLEM 4.22 Determine the maximum tension which can be developed in cable AB if the maximum allowable value of the reaction at C is 1000 N.

SOLUTION Cmax = 1000 N

Have

C 2 = C x2 + C y2

Now ∴ Cy =

(1000 )2 − Cx2

(1)

From f.b.d. of pedal ΣFx = 0: C x − Tmax = 0 ∴ C x = Tmax

(2)

ΣM D = 0: C y ( 0.4 m ) − Tmax ( 0.18 m ) sin 60° = 0

∴ C y = 0.38971Tmax

(3)

Equating the expressions for C y in Equations (1) and (3), with Cx = Tmax from Equation (2) 2 (1000 )2 − Tmax

= 0.389711Tmax

2 ∴ Tmax = 868,150

and

Tmax = 931.75 N or Tmax = 932 N

PROBLEM 4.23 A steel rod is bent to form a mounting bracket. For each of the mounting brackets and loadings shown, determine the reactions at A and B.

SOLUTION (a) From f.b.d. of mounting bracket (a)

ΣM E = 0: A ( 8 in.) − 80 lb ⋅ in. − (10 lb )( 6 in.) − ( 20 lb )(12 in.) = 0 ∴ A = 47.5 lb

or A = 47.5 lb ΣFx = 0: Bx − 10 lb + 47.5 lb = 0 ∴ Bx = −37.5 lb

B x = 37.5 lb

or

ΣFy = 0: By − 20 lb = 0 ∴ By = 20 lb

B y = 20.0 lb

or Then and

B=

Bx2 + By2 =

( 37.5)2 + ( 20.0 )2

= 42.5 lb

 By  −1  20   = tan   = −28.072°  −37.5   Bx 

θ = tan −1 

or B = 42.5 lb

28.1°

(b) From f.b.d. of mounting bracket (b)

ΣM B = 0:

( A cos 45° )(8 in.) − 80 lb ⋅ in. − (10 lb )( 6 in.) − ( 20 lb )(12 in.) = 0 ∴ A = 67.175 lb

or A = 67.2 lb ΣFx = 0: Bx − 10 lb + 67.175cos 45° = 0 ∴ Bx = −37.500 lb

or

B x = 37.5 lb

45°

PROBLEM 4.23 CONTINUED ΣFy = 0: By − 20 lb + 67.175sin 45° = 0 ∴ By = −27.500 lb

B y = 27.5 lb

or Then and

B=

Bx2 + By2 =

( 37.5)2 + ( 27.5)2

= 46.503 lb

 By  −1  −27.5   = tan   = 36.254°  −37.5   Bx 

θ = tan −1 

or B = 46.5 lb

36.3°

PROBLEM 4.24 A steel rod is bent to form a mounting bracket. For each of the mounting brackets and loadings shown, determine the reactions at A and B.

SOLUTION (a)

(a) From f.b.d. of mounting bracket ΣM A = 0: − B ( 8 in.) − ( 20 lb )(12 in.) + (10 lb )( 2 in.) − 80 lb ⋅ in. = 0 ∴ B = −37.5 lb

or B = 37.5 lb ΣFx = 0: − 37.5 lb − 10 lb + Ax = 0 ∴ Ax = 47.5 lb

A x = 47.5 lb

or

ΣFy = 0: − 20 lb + Ay = 0 ∴ Ay = 20 lb

A y = 20.0 lb

or Ax2 + Ay2 =

Then

A=

and

θ = tan −1 

( 47.5)2 + ( 20 )2

= 51.539 lb

 Ay  −1  20   = tan   = 22.834°  47.5   Ax 

or A = 51.5 lb (b)

22.8°

(b) From f.b.d. of mounting bracket ΣM A = 0: − ( B cos 45° )( 8 in.) − ( 20 lb )( 2 in.) −80 lb ⋅ in. + (10 lb )( 2 in.) = 0 ∴ B = −53.033 lb

or B = 53.0 lb ΣFx = 0: Ax + ( −53.033 lb ) cos 45° − 10 = 0 ∴ Ax = 47.500 lb

or

A x = 47.5 lb

45°

PROBLEM 4.24 CONTINUED ΣFy = 0: Ay − ( 53.033 lb ) sin 45° − 20 = 0 ∴ Ay = −17.500 lb

A y = 17.50 lb

or Then and

A=

Ax2 + Ay2 =

( 47.5)2 + (17.5)2

= 50.621 lb

 Ay  −1  −17.5   = tan   = −20.225°  47.5   Ax 

θ = tan −1 

or A = 50.6 lb

20.2°

PROBLEM 4.25 A sign is hung by two chains from mast AB. The mast is hinged at A and is supported by cable BC. Knowing that the tensions in chains DE and FH are 50 lb and 30 lb, respectively, and that d = 1.3 ft, determine (a) the tension in cable BC, (b) the reaction at A.

SOLUTION

(8.4 )2 + (1.3)2

BC =

First note

= 8.5 ft

(a) From f.b.d. of mast AB  8.4   ΣM A = 0:   TBC  ( 2.5 ft ) − ( 30 lb )( 7.2 ft )  8.5   −50 lb ( 2.2 ft ) = 0 ∴ TBC = 131.952 lb or TBC = 132.0 lb (b) From f.b.d. of mast AB  8.4  ΣFx = 0: Ax −   (131.952 lb ) = 0  8.5  ∴ Ax = 130.400 lb

A x = 130.4 lb

or

 1.3  ΣFy = 0: Ay +   (131.952 lb ) − 30 lb − 50 lb = 0  8.5  ∴ Ay = 59.819 lb

A y = 59.819 lb

or Then

and

A=

Ax2 + Ay2 =

(130.4 )2 + ( 59.819 )2

= 143.466 lb

 Ay  −1  59.819   = tan   = 24.643° A  130.4   x

θ = tan −1 

or A = 143.5 lb

24.6°

PROBLEM 4.26 A sign is hung by two chains from mast AB. The mast is hinged at A and is supported by cable BC. Knowing that the tensions in chains DE and FH are 30 lb and 20 lb, respectively, and that d = 1.54 ft, determine (a) the tension in cable BC, (b) the reaction at A.

SOLUTION BC =

First note

(8.4 )2 + (1.54 )2

= 8.54 ft

(a) From f.b.d. of mast AB  8.4   ΣM A = 0:   TBC  ( 2.5 ft ) − 20 lb ( 7.2 ft )  8.54   − 30 lb ( 2.2 ft ) = 0 ∴ TBC = 85.401 lb or TBC = 85.4 lb (b) From f.b.d. of mast AB  8.4  ΣFx = 0: Ax −   ( 85.401 lb ) = 0  8.54  ∴ Ax = 84.001 lb

A x = 84.001 lb

or

 1.54  ΣFy = 0: Ay +   ( 85.401 lb ) − 20 lb − 30 lb = 0  8.54  ∴ Ay = 34.600 lb

A y = 34.600 lb

or Then and

A=

Ax2 + Ay2 =

(84.001)2 + ( 34.600 )2

= 90.848 lb

 Ay  −1  34.6   = tan   = 22.387°  84.001   Ax 

θ = tan −1 

or A = 90.8 lb

22.4°

PROBLEM 4.27 For the frame and loading shown, determine the reactions at A and E when (a) α = 30o , (b) α = 45o.

SOLUTION (a) Given α = 30° (a)

From f.b.d. of frame ΣM A = 0: − ( 90 N )( 0.2 m ) − ( 90 N )( 0.06 m ) + ( E cos 60° )( 0.160 m ) + ( E sin 60° )( 0.100 m ) = 0 ∴ E = 140.454 N or E = 140.5 N

60°

ΣFx = 0: Ax − 90 N + (140.454 N ) cos 60° = 0 ∴ Ax = 19.7730 N

A x = 19.7730 N

or

ΣFy = 0: Ay − 90 N + (140.454 N ) sin 60° = 0 ∴ Ay = −31.637 N

A y = 31.6 N

or Then

A=

Ax2 + Ay2 =

(19.7730 )2 + ( 31.637 )2

= 37.308 lb and

 Ay  −1  −31.637   = tan    19.7730   Ax 

θ = tan −1 

= −57.995° or A = 37.3 N

58.0°

PROBLEM 4.27 CONTINUED (b)

(b) Given α = 45° From f.b.d. of frame ΣM A = 0: − ( 90 N )( 0.2 m ) − ( 90 N )( 0.06 m ) + ( E cos 45° )( 0.160 m ) + ( E sin 45° )( 0.100 m ) = 0 ∴ E = 127.279 N or E = 127.3 N

45°

ΣFx = 0: Ax − 90 + (127.279 N ) cos 45° = 0 ∴ Ax = 0 ΣFy = 0: Ay − 90 + (127.279 N ) sin 45° = 0 ∴ Ay = 0 or A = 0

PROBLEM 4.28 A lever AB is hinged at C and is attached to a control cable at A. If the lever is subjected to a 300-N vertical force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION First x AC = ( 0.200 m ) cos 20° = 0.187 939 m y AC = ( 0.200 m ) sin 20° = 0.068 404 m Then yDA = 0.240 m − y AC = 0.240 m − 0.068404 m = 0.171596 m tan α =

and

yDA 0.171 596 = x AC 0.187 939

∴ α = 42.397°

β = 90° − 20° − 42.397° = 27.603°

and

(a) From f.b.d. of lever AB ΣM C = 0: T cos 27.603° ( 0.2 m ) − 300 N ( 0.3 m ) cos 20° = 0

∴ T = 477.17 N

or T = 477 N

(b) From f.b.d. of lever AB ΣFx = 0: C x + ( 477.17 N ) cos 42.397° = 0 ∴ C x = −352.39 N or

C x = 352.39 N ΣFy = 0: C y − 300 N − ( 477.17 N ) sin 42.397° = 0 ∴ C y = 621.74 N

or

C y = 621.74 N

PROBLEM 4.28 CONTINUED Then and

C =

C x2 + C y2 =

( 352.39 )2 + ( 621.74 )2

= 714.66 N

 Cy  −1  621.74   = tan   = −60.456° C  −352.39   x

θ = tan −1 

or C = 715 N

60.5°

PROBLEM 4.29 Neglecting friction and the radius of the pulley, determine the tension in cable BCD and the reaction at support A when d = 80 mm.

SOLUTION First  60   = 12.0948°  280 

α = tan −1 

 60 

β = tan −1   = 36.870°  80  From f.b.d. of object BAD ΣM A = 0:

( 40 N )( 0.18 m ) + (T cosα )( 0.08 m ) + (T sin α )( 0.18 m ) − (T cos β )( 0.08 m ) − (T sin β )( 0.18 m ) = 0

 7.2 N ⋅ m  ∴ T =  = 128.433 N  0.056061  or T = 128.4 N ΣFx = 0:

(128.433 N )( cos β

− cos α ) + Ax = 0

∴ Ax = 22.836 N

A x = 22.836 N

or

ΣFy = 0: Ay + (128.433 N )( sin β + sin α ) + 40 N = 0 ∴ Ay = −143.970 N

A y = 143.970 N

or Then and

A=

Ax2 + Ay2 =

( 22.836 )2 + (143.970 )2

= 145.770 N

 Ay  −1  −143.970   = tan   = −80.987°  22.836   Ax 

θ = tan −1 

or A = 145.8 N

81.0°

PROBLEM 4.30 Neglecting friction and the radius of the pulley, determine the tension in cable BCD and the reaction at support A when d = 144 mm.

SOLUTION First note  60 

α = tan −1   = 15.5241°  216   60 

β = tan −1   = 22.620°  144  From f.b.d. of member BAD ΣM A = 0:

( 40 N )( 0.18 m ) + (T cosα )( 0.08 m ) + (T sin α )( 0.18 m ) − (T cos β )( 0.08 m ) − (T sin β )( 0.18 m ) = 0

 7.2 N ⋅ m  ∴ T =  = 404.04 N  0.0178199 m  or T = 404 N ΣFx = 0: Ax + ( 404.04 N )( cos β − cos α ) = 0 ∴ Ax = 16.3402 N

A x = 16.3402 N

or

ΣFy = 0: Ay + ( 404.04 N )( sin β + sin α ) + 40 N = 0 ∴ Ay = −303.54 N

A y = 303.54 N

or Then and

A=

Ax2 + Ay2 =

(16.3402 )2 + ( 303.54 )2

= 303.98 N

 Ay  −1  −303.54   = tan   = −86.919° A  16.3402   x

θ = tan −1 

or A = 304 N

86.9°

PROBLEM 4.31 Neglecting friction, determine the tension in cable ABD and the reaction at support C.

SOLUTION From f.b.d. of inverted T-member ΣM C = 0: T ( 25 in.) − T (10 in.) − ( 30 lb )(10 in.) = 0

∴ T = 20 lb or T = 20.0 lb W ΣFx = 0: Cx − 20 lb = 0 ∴ C x = 20 lb C x = 20.0 lb

or

ΣFy = 0: C y + 20 lb − 30 lb = 0 ∴ C y = 10 lb C y = 10.00 lb

or Then

and or

C =

C x2 + C y2 =

( 20 )2 + (10 )2

= 22.361 lb

 Cy  −1  10   = tan   = 26.565°  20   Cx 

θ = tan −1 

C = 22.4 lb

26.6° W

PROBLEM 4.32 Rod ABC is bent in the shape of a circular arc of radius R. Knowing that θ = 35o , determine the reaction (a) at B, (b) at C.

SOLUTION For θ = 35° (a) From the f.b.d. of rod ABC ΣM D = 0: Cx( R ) − P( R ) = 0 ∴ Cx = P

Cx = P

or

ΣFx = 0: P − B sin 35° = 0 ∴ B=

P = 1.74345P sin 35° or B = 1.743P

55.0° W

(b) From the f.b.d. of rod ABC

ΣFy = 0: C y + (1.74345P ) cos 35° − P = 0 ∴ C y = −0.42815P

C y = 0.42815P

or Then

and

C =

C x2 + C y2 =

( P )2 + ( 0.42815P )2

= 1.08780 P

 Cy  −1  −0.42815P   = tan   = −23.178° P    Cx 

θ = tan −1 

or C = 1.088P

23.2° W

PROBLEM 4.33 Rod ABC is bent in the shape of a circular arc of radius R. Knowing that θ = 50o , determine the reaction (a) at B, (b) at C.

SOLUTION For θ = 50° (a) From the f.b.d. of rod ABC ΣM D = 0: Cx ( R ) − P ( R ) = 0 ∴ Cx = P

Cx = P

or

ΣFx = 0: P − B sin 50° = 0 ∴ B=

P = 1.30541P sin 50° or B = 1.305P

40.0° W

(b) From the f.b.d. of rod ABC ΣFy = 0: C y − P + (1.30541P ) cos 50° = 0 ∴ C y = 0.160900P

C y = 0.1609 P

or

C x2 + C y2 =

Then

C =

and

θ = tan −1 

( P )2 + ( 0.1609P )2

= 1.01286 P

 Cy  −1  0.1609 P   = tan   = 9.1405° C P    x

or C = 1.013P

9.14° W

PROBLEM 4.34 Neglecting friction and the radius of the pulley, determine (a) the tension in cable ABD, (b) the reaction at C.

SOLUTION First note  15 

α = tan −1   = 22.620°  36   15 

β = tan −1   = 36.870°  20  (a) From f.b.d. of member ABC ΣM C = 0:

( 30 lb )( 28 in.) − (T sin 22.620° )( 36 in.) − (T sin 36.870° )( 20 in.) = 0 ∴ T = 32.500 lb or T = 32.5 lb W

(b) From f.b.d. of member ABC ΣFx = 0: Cx + ( 32.500 lb )( cos 22.620° + cos 36.870° ) = 0 ∴ C x = −56.000 lb

C x = 56.000 lb

or

ΣFy = 0: C y − 30 lb + ( 32.500 lb )( sin 22.620° + sin 36.870° ) = 0 ∴ C y = −2.0001 lb

C y = 2.0001 lb

or Then and

C =

C x2 + C y2 =

( 56.0 )2 + ( 2.001)2

= 56.036 lb

 Cy  −1  −2.0   = tan   = 2.0454°  −56.0   Cx 

θ = tan −1 

or C = 56.0 lb

2.05° W

PROBLEM 4.35 Neglecting friction, determine the tension in cable ABD and the reaction at C when θ = 60o.

SOLUTION From f.b.d. of bent ACD ΣM C = 0:

(T cos30° )( 2a sin 60° ) + (T sin 30° )( a + 2a cos 60° ) −T (a) − P (a) = 0 ∴ T =

P 1.5 or T =

2P W 3

or C = 0.577P

W

 2P  ΣFx = 0: C x −   cos 30° = 0  3  ∴ Cx =

3 P = 0.57735P 3

C x = 0.577 P

or ΣFy = 0: C y +

2  2P  P−P+  cos 60° = 0 3  3 

∴ Cy = 0

PROBLEM 4.36 Neglecting friction, determine the tension in cable ABD and the reaction at C when θ = 30o.

SOLUTION From f.b.d. of bent ACD ΣM C = 0:

(T cos 60° )( 2a sin 30° ) + T sin 60° ( a + 2a cos 30° ) − P (a) − T (a) = 0 ∴ T =

P = 0.53590P 1.86603 or T = 0.536 P W

ΣFx = 0: C x − ( 0.53590P ) cos 60° = 0 ∴ Cx = 0.26795P or

C x = 0.268P ΣFy = 0: C y + 0.53590 P − P + ( 0.53590 P ) sin 60° = 0 ∴ Cy = 0 or C = 0.268P

W

PROBLEM 4.37 Determine the tension in each cable and the reaction at D.

SOLUTION

First note BE =

( 20 )2 + (8)2

in. = 21.541 in.

CF =

(10 )2 + (8)2

in. = 12.8062 in.

From f.b.d. of member ABCD ΣM C = 0:



 8   TBE  (10 in.) = 0  21.541  

(120 lb )( 20 in.) − 

∴ TBE = 646.24 lb or TBE = 646 lb W 8  8    ΣFy = 0: − 120 lb +   ( 646.24 lb ) −   TCF = 0 21.541 12.8062     ∴ TCF = 192.099 lb or TCF = 192.1 lb W  20   10  ΣFx = 0:   ( 646.24 lb ) +   (192.099 lb ) − D = 0  21.541   12.8062  ∴ D = 750.01 lb or D = 750 lb

W

PROBLEM 4.38 Rod ABCD is bent in the shape of a circular arc of radius 80 mm and rests against frictionless surfaces at A and D. Knowing that the collar at B can move freely on the rod and that θ = 45o. determine (a) the tension in cord OB, (b) the reactions at A and D.

SOLUTION (a) From f.b.d. of rod ABCD ΣM E = 0:

( 25 N ) cos 60° ( dOE ) − (T cos 45° ) ( dOE ) = 0 ∴ T = 17.6777 N or T = 17.68 N W

(b) From f.b.d. of rod ABCD ΣFx = 0: − (17.6777 N ) cos 45° + ( 25 N ) cos 60° + N D cos 45° − N A cos 45° = 0 ∴ N A − ND = 0 or

ND = N A

(1)

ΣFy = 0: N A sin 45° + N D sin 45° − (17.6777 N ) sin 45° − ( 25 N ) sin 60° = 0 ∴ N A + N D = 48.296 N

(2)

Substituting Equation (1) into Equation (2), 2 N A = 48.296 N N A = 24.148 N or N A = 24.1 N

45.0° W

and N D = 24.1 N

45.0° W

PROBLEM 4.39 Rod ABCD is bent in the shape of a circular arc of radius 80 mm and rests against frictionless surfaces at A and D. Knowing that the collar at B can move freely on the rod, determine (a) the value of θ for which the tension in cord OB is as small as possible, (b) the corresponding value of the tension, (c) the reactions at A and D.

SOLUTION (a) From f.b.d. of rod ABCD ΣM E = 0:

( 25 N ) cos 60° ( dOE ) − (T cosθ ) ( dOE ) = 0 T =

or

12.5 N cosθ

(1)

∴ T is minimum when cosθ is maximum,

or θ = 0° W (b) From Equation (1) T =

12.5 N = 12.5 N cos 0 or Tmin = 12.50 N W

ΣFx = 0: − N A cos 45° + N D cos 45° + 12.5 N

(c)

− ( 25 N ) cos 60° = 0 ∴ ND − N A = 0 or

ND = N A

(2)

ΣFy = 0: N A sin 45° + N D sin 45° − ( 25 N ) sin 60° = 0 ∴ N D + N A = 30.619 N

(3)

Substituting Equation (2) into Equation (3), 2 N A = 30.619 N A = 15.3095 N or N A = 15.31 N

45.0° W

and N D = 15.31 N

45.0° W

PROBLEM 4.40 Bar AC supports two 100-lb loads as shown. Rollers A and C rest against frictionless surfaces and a cable BD is attached at B. Determine (a) the tension in cable BD, (b) the reaction at A, (c) the reaction at C.

SOLUTION First note that from similar triangles yDB 10 = 6 20 and

BD =

∴ yDB = 3 in.

( 3)2 + (14 )2

in. = 14.3178 in.

Tx =

14 T = 0.97780T 14.3178

Ty =

3 T = 0.20953T 14.3178

(a) From f.b.d. of bar AC ΣM E = 0:

( 0.97780T )( 7 in.) − ( 0.20953T )( 6 in.) − (100 lb )(16 in.) − (100 lb )( 4 in.) = 0 ∴ T = 357.95 lb or T = 358 lb W

(b) From f.b.d. of bar AC ΣFy = 0: A − 100 − 0.20953 ( 357.95 ) − 100 = 0 ∴ A = 275.00 lb or A = 275 lb W (c) From f.b.d of bar AC ΣFx = 0: 0.97780 ( 357.95 ) − C = 0 ∴ C = 350.00 lb or C = 350 lb

W

PROBLEM 4.41 A parabolic slot has been cut in plate AD, and the plate has been placed so that the slot fits two fixed, frictionless pins B and C. The equation of the slot is y = x 2 /100, where x and y are expressed in mm. Knowing that the input force P = 4 N, determine (a) the force each pin exerts on the plate, (b) the output force Q. SOLUTION

y =

The equation of the slot is Now

x2 100

 dy    = slope of the slot at C  dx C  2x  = = 1.200  100  x = 60 mm

∴ α = tan −1 (1.200 ) = 50.194° and

θ = 90° − α = 90° − 50.194° = 39.806°

Coordinates of C are xC = 60 mm,

Also, the coordinates of D are

yC =

( 60 ) 2 100

= 36 mm

xD = 60 mm yD = 46 mm + ( 40 mm ) sin β

where

 120 − 66   = 12.6804°  240 

β = tan −1 

∴ yD = 46 mm + ( 40 mm ) tan12.6804° = 55.000 mm

PROBLEM 4.41 CONTINUED yED =

Also,

60 mm 60 mm = tan β tan12.6804°

= 266.67 mm From f.b.d. of plate AD ΣM E = 0:

( NC cosθ )  yED − ( yD − yC ) + ( NC sin θ )( xC ) − ( 4 N )( yED −

yD ) = 0

( NC cos 39.806° )  266.67 − ( 55.0 − 36.0 ) mm + NC sin ( 39.806° )( 60 mm ) − ( 4 N )( 266.67 − 55.0 ) mm = 0 ∴ NC = 3.7025 N or

NC = 3.70 N

39.8°

ΣFx = 0: − 4 N + NC cosθ + Q sin β = 0 −4 N + ( 3.7025 N ) cos 39.806° + Q sin12.6804° = 0 ∴ Q = 5.2649 N or

Q = 5.26 N

77.3°

ΣFy = 0: N B + NC sin θ − Q cos β = 0

N B + ( 3.7025 N ) sin 39.806° − ( 5.2649 N ) cos12.6804° = 0 ∴ N B = 2.7662 N or (a) (b)

N B = 2.77 N N B = 2.77 N , NC = 3.70 N Q = 5.26 N

39.8°

77.3° ( output )

PROBLEM 4.42 A parabolic slot has been cut in plate AD, and the plate has been placed so that the slot fits two fixed, frictionless pins B and C. The equation of the slot is y = x 2 /100, where x and y are expressed in mm. Knowing that the maximum allowable force exerted on the roller at D is 8.5 N, determine (a) the corresponding magnitude of the input force P, (b) the force each pin exerts on the plate.

SOLUTION y =

The equation of the slot is,

x2 100

 dy    = slope of slot at C  dx C

Now

 2x  = = 1.200  100  x = 60 mm

∴ α = tan −1 (1.200 ) = 50.194° and

θ = 90° − α = 90° − 50.194° = 39.806°

Coordinates of C are

xC = 60 mm, yC =

( 60 )2 100

= 36 mm

Also, the coordinates of D are

xD = 60 mm yD = 46 mm + ( 40 mm ) sin β where

 120 − 66   = 12.6804°  240 

β = tan −1 

∴ yD = 46 mm + ( 40 mm ) tan12.6804° = 55.000 mm

xE = 0

Note:

yE = yC + ( 60 mm ) tan θ = 36 mm + ( 60 mm ) tan 39.806° = 86.001 mm (a) From f.b.d. of plate AD ΣM E = 0: P ( yE ) − ( 8.5 N ) sin β  ( yE − yD ) − ( 8.5 N ) cos β  ( 60 mm ) = 0

PROBLEM 4.42 CONITNIUED P ( 86.001 mm ) − ( 8.5 N ) sin12.6804° ( 31.001 mm ) − ( 8.5 N ) cos12.6804° ( 60 mm ) = 0 ∴ P = 6.4581 N or P = 6.46 N (b)

ΣFx = 0: P − ( 8.5 N ) sin β − NC cosθ = 0 6.458 N − ( 8.5 N )( sin12.6804° ) − NC ( cos 39.806° ) = 0 ∴ NC = 5.9778 N or NC = 5.98 N

39.8°

ΣFy = 0: N B + NC sin θ − ( 8.5 N ) cos β = 0

N B + ( 5.9778 N ) sin 39.806° − ( 8.5 N ) cos12.6804° = 0 ∴ N B = 4.4657 N or N B = 4.47 N

PROBLEM 4.43 A movable bracket is held at rest by a cable attached at E and by frictionless rollers. Knowing that the width of post FG is slightly less than the distance between the rollers, determine the force exerted on the post by each roller when α = 20o.

SOLUTION From f.b.d. of bracket ΣFy = 0: T sin 20° − 60 lb = 0 ∴ T = 175.428 lb

Tx = (175.428 lb ) cos 20° = 164.849 lb Ty = (175.428 lb ) sin 20° = 60 lb Note: Ty and 60 lb force form a couple of 60 lb (10 in.) = 600 lb ⋅ in. ΣM B = 0: 164.849 lb ( 5 in.) − 600 lb ⋅ in. + FCD ( 8 in.) = 0 ∴ FCD = −28.030 lb or

FCD = 28.0 lb

ΣFx = 0: FCD + FAB − Tx = 0 −28.030 lb + FAB − 164.849 lb = 0 ∴ FAB = 192.879 lb or

FAB = 192.9 lb

Rollers A and C can only apply a horizontal force to the right onto the vertical post corresponding to the equal and opposite force to the left on the bracket. Since FAB is directed to the right onto the bracket, roller B will react FAB. Also, since FCD is acting to the left on the bracket, it will act to the right on the post at roller C.

PROBLEM 4.43 CONTINUED ∴ A=D=0

B = 192.9 lb C = 28.0 lb Forces exerted on the post are

A=D=0 B = 192.9 lb C = 28.0 lb

PROBLEM 4.44 Solve Problem 4.43 when α = 30o. P4.43 A movable bracket is held at rest by a cable attached at E and by frictionless rollers. Knowing that the width of post FG is slightly less than the distance between the rollers, determine the force exerted on the post by each roller when α = 20o.

SOLUTION From f.b.d. of bracket ΣFy = 0: T sin 30° − 60 lb = 0 ∴ T = 120 lb

Tx = (120 lb ) cos 30° = 103.923 lb Ty = (120 lb ) sin 30° = 60 lb Note: Ty and 60 lb force form a couple of

( 60 lb )(10 in.) = 600 lb ⋅ in. ΣM B = 0:

(103.923 lb )( 5 in.) − 600 lb ⋅ in. + FCD (8 in.) = 0 ∴ FCD = 10.0481 lb

or

FCD = 10.05 lb ΣFx = 0: FCD + FAB − Tx = 0 10.0481 lb + FAB − 103.923 lb = 0 ∴ FAB = 93.875 lb

or

FAB = 93.9 lb

Rollers A and C can only apply a horizontal force to the right on the vertical post corresponding to the equal and opposite force to the left on the bracket. The opposite direction apply to roller B and D. Since both FAB and FCD act to the right on the bracket, rollers B and D will react these forces. ∴ A=C=0

B = 93.9 lb D = 10.05 lb Forces exerted on the post are

A=C=0 B = 93.9 lb D = 10.05 lb

PROBLEM 4.45 A 20-lb weight can be supported in the three different ways shown. Knowing that the pulleys have a 4-in. radius, determine the reaction at A in each case.

SOLUTION (a) From f.b.d. of AB ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 20 lb = 0

Ay = 20.0 lb

or

and A = 20.0 lb ΣM A = 0: M A − ( 20 lb )(1.5 ft ) = 0 ∴ M A = 30.0 lb ⋅ ft or M A = 30.0 lb ⋅ ft  1 ft  4 in.   = 0.33333 ft  12 in. 

(b) Note: From f.b.d. of AB

ΣFx = 0: Ax − 20 lb = 0

Ax = 20.0 lb

or

ΣFy = 0: Ay − 20 lb = 0

Ay = 20.0 lb

or Then

A=

Ax2 + Ay2 =

( 20.0 )2 + ( 20.0 )2

= 28.284 lb

∴ A = 28.3 lb

45°

ΣM A = 0: M A + ( 20 lb )( 0.33333 ft ) − ( 20 lb )(1.5 ft + 0.33333 ft ) = 0 ∴ M A = 30.0 lb ⋅ ft or M A = 30.0 lb ⋅ ft

PROBLEM 4.45 CONTINUED (c) From f.b.d. of AB ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 20 lb − 20 lb = 0 or

Ay = 40.0 lb and A = 40.0 lb ΣM A = 0: M A − ( 20 lb )(1.5 ft − 0.33333 ft ) − ( 20 lb )(1.5 ft + 0.33333 ft ) = 0 ∴ M A = 60.0 lb ⋅ ft or M A = 60.0 lb ⋅ ft

PROBLEM 4.46 A belt passes over two 50-mm-diameter pulleys which are mounted on a bracket as shown. Knowing that M = 0 and Ti = TO = 24 N, determine the reaction at C.

SOLUTION From f.b.d. of bracket ΣFx = 0: Cx − 24 N = 0 ∴ Cx = 24 N ΣFy = 0: C y − 24 N = 0 ∴ C y = 24 N Then

C =

C x2 + C y2 =

( 24 )2 + ( 24 )2

= 33.941 N

∴ C = 33.9 N

45.0°

ΣM C = 0: M C − ( 24 N ) ( 45 − 25 ) mm  + ( 24 N ) ( 25 + 50 − 60 ) mm  = 0

∴ M C = 120 N ⋅ mm or M C = 0.120 N ⋅ m

PROBLEM 4.47 A belt passes over two 50-mm-diameter pulleys which are mounted on a bracket as shown. Knowing that M = 0.40 N ⋅ m m and that Ti and TO are equal to 32 N and 16 N, respectively, determine the reaction at C.

SOLUTION From f.b.d. of bracket ΣFx = 0: C x − 32 N = 0 ∴ C x = 32 N ΣFy = 0: C y − 16 N = 0 ∴ C y = 16 N Then and

C =

C x2 + C y2 =

( 32 )2 + (16 )2

= 35.777 N

 Cy  −1  16   = tan   = 26.565°  32   Cx 

θ = tan −1 

or C = 35.8 N

26.6°

ΣM C = 0: M C − ( 32 N )( 45 mm − 25 mm ) + (16 N )( 25 mm + 50 mm − 60 mm ) − 400 N ⋅ mm = 0 ∴ M C = 800 N ⋅ mm or M C = 0.800 N ⋅ m

PROBLEM 4.48 A 350-lb utility pole is used to support at C the end of an electric wire. The tension in the wire is 120 lb, and the wire forms an angle of 15° with the horizontal at C. Determine the largest and smallest allowable tensions in the guy cable BD if the magnitude of the couple at A may not exceed 200 lb ⋅ ft.

SOLUTION First note

LBD = Tmax :

( 4.5)2 + (10 )2

= 10.9659 ft

From f.b.d. of utility pole with M A = 200 lb ⋅ ft ΣM A = 0: − 200 lb ⋅ ft − (120 lb ) cos15° (14 ft )

 4.5   +   Tmax  (10 ft ) = 0  10.9659   ∴ Tmax = 444.19 lb or Tmax = 444 lb

Tmin :

From f.b.d. of utility pole with M A = 200 lb ⋅ ft ΣM A = 0: 200 lb ⋅ ft − (120 lb ) cos15°  (14 ft )

 4.5   +   Tmin  (10 ft ) = 0  10.9659   ∴ Tmin = 346.71 lb or Tmin = 347 lb

PROBLEM 4.49 In a laboratory experiment, students hang the masses shown from a beam of negligible mass. (a) Determine the reaction at the fixed support A knowing that end D of the beam does not touch support E. (b) Determine the reaction at the fixed support A knowing that the adjustable support E exerts an upward force of 6 N on the beam.

SOLUTION

(

)

WB = mB g = (1 kg ) 9.81 m/s 2 = 9.81 N

(

)

WC = mC g = ( 0.5 kg ) 9.81 m/s 2 = 4.905 N (a) From f.b.d. of beam ABCD ΣFx = 0: Ax = 0 ΣFy = 0: Ay − WB − WC = 0 Ay − 9.81 N − 4.905 N = 0 ∴ Ay = 14.715 N or A = 14.72 N ΣM A = 0: M A − WB ( 0.2 m ) − WC ( 0.3 m ) = 0 M A − ( 9.81 N )( 0.2 m ) − ( 4.905 N )( 0.3 m ) = 0 ∴ M A = 3.4335 N ⋅ m or M A = 3.43 N ⋅ m (b) From f.b.d. of beam ABCD ΣFx = 0: Ax = 0 ΣFy = 0: Ay − WB − WC + 6 N = 0 Ay − 9.81 N − 4.905 N + 6 N = 0 ∴ Ay = 8.715 N

or A = 8.72 N

ΣM A = 0: M A − WB ( 0.2 m ) − WC ( 0.3 m ) + ( 6 N )( 0.4 m ) = 0 M A − ( 9.81 N )( 0.2 m ) − ( 4.905 N )( 0.3 m ) + ( 6 N )( 0.4 m ) = 0 ∴ M A = 1.03350 N ⋅ m or M A = 1.034 N ⋅ m

PROBLEM 4.50 In a laboratory experiment, students hang the masses shown from a beam of negligible mass. Determine the range of values of the force exerted on the beam by the adjustable support E for which the magnitude of the couple at A does not exceed 2.5 N ⋅ m.

SOLUTION

(

)

WB = mB g = 1 kg 9.81 m/s 2 = 9.81 N

(

)

WC = mC g = 0.5 kg 9.81 m/s 2 = 4.905 N Maximum M A value is 2.5 N ⋅ m Fmin :

From f.b.d. of beam ABCD with M A = 2.5 N ⋅ m ΣM A = 0: 2.5 N ⋅ m − WB ( 0.2 m ) − WC ( 0.3 m ) + Fmin ( 0.4 m ) = 0

2.5 N ⋅ m − ( 9.81 N )( 0.2 m ) − ( 4.905 N )( 0.3 m ) + Fmin ( 0.4 m ) = 0 ∴ Fmin = 2.3338 N or Fmax :

Fmin = 2.33 N From f.b.d. of beam ABCD with M A = 2.5 N ⋅ m ΣM A = 0: − 2.5 N ⋅ m − WB ( 0.2 m ) − WC ( 0.3 m ) + Fmax ( 0.4 m ) = 0

−2.5 N ⋅ m − ( 9.81 N )( 0.2 m ) − ( 4.905 N )( 0.3 m ) + Fmax ( 0.4 m ) = 0 ∴ Fmax = 14.8338 N or

Fmax = 14.83 N or 2.33 N ≤ FE ≤ 14.83 N

PROBLEM 4.51 Knowing that the tension in wire BD is 300 lb, determine the reaction at fixed support C for the frame shown.

SOLUTION

From f.b.d. of frame with T = 300 lb

 5 ΣFx = 0: C x − 100 lb +   300 lb = 0  13 

∴ C x = −15.3846 lb

or

C x = 15.3846 lb

 12  ΣFy = 0: C y − 180 lb −   300 lb = 0  13  ∴ C y = 456.92 lb Then and

C =

C x2 + C y2 =

or

C y = 456.92 lb

(15.3846 )2 + ( 456.92 )2

= 457.18 lb

 Cy  −1  456.92   = tan   = −88.072° C  −15.3846   x

θ = tan −1 

or C = 457 lb

88.1°

 12   ΣM C = 0: M C + (180 lb )( 20 in.) + (100 lb )(16 in.) −   300 lb  (16 in.) = 0  13   ∴ M C = −769.23 lb ⋅ in. or M C = 769 lb ⋅ in.

PROBLEM 4.52 Determine the range of allowable values of the tension in wire BD if the magnitude of the couple at the fixed support C is not to exceed 75 lb ⋅ ft.

SOLUTION

Tmax

From f.b.d. of frame with M C = 75 lb ⋅ ft

= 900 lb ⋅ in.

 12   ΣM C = 0: 900 lb ⋅ in. + (180 lb )( 20 in.) + (100 lb )(16 in.) −   Tmax  (16 in.) = 0  13   ∴ Tmax = 413.02 lb Tmin

From f.b.d. of frame with

M C = 75 lb ⋅ ft

= 900 lb ⋅ in.

 12   ΣM C = 0: − 900 lb ⋅ in. + (180 lb )( 20 in.) + (100 lb )(16 in.) −   Tmin  (16 in.) = 0  13   ∴ Tmin = 291.15 lb ∴ 291 lb ≤ T ≤ 413 lb

PROBLEM 4.53 Uniform rod AB of length l and weight W lies in a vertical plane and is acted upon by a couple M. The ends of the rod are connected to small rollers which rest against frictionless surfaces. (a) Express the angle θ corresponding to equilibrium in terms of M, W, and l. (b) Determine the value of θ corresponding to equilibrium when M = 1.5 lb ⋅ ft, W = 4 lb, and l = 2 ft.

SOLUTION (a) From f.b.d. of uniform rod AB ΣFx = 0: − A cos 45° + B cos 45° = 0 ∴ −A + B = 0

or

B= A

(1)

ΣFy = 0: A sin 45° + B sin 45° − W = 0 ∴ A+B =

2W

(2)

From Equations (1) and (2) 2A =

2W

∴ A=

1 W 2

From f.b.d. of uniform rod AB  l   ΣM B = 0: W   cosθ  + M  2    1  − W  l cos ( 45° − θ )  = 0 2  

From trigonometric identity cos (α − β ) = cos α cos β + sin α sin β Equation (3) becomes  Wl   Wl    cosθ + M −   ( cosθ + sin θ ) = 0  2   2 

(3)

PROBLEM 4.53 CONTINUED or

 Wl   Wl   Wl    cosθ + M −   cosθ −   sin θ = 0  2   2   2  ∴ sin θ =

2M Wl  2M  or θ = sin −1    Wl 

(b)

 2 (1.5 lb ⋅ ft )   = 22.024°  ( 4 lb )( 2 ft ) 

θ = sin −1 

or θ = 22.0°

PROBLEM 4.54 A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by an elastic cord which passes over a pulley at C. (a) Express the tension in the cord in terms of W and θ . (b) Determine the value of θ for which the tension in the cord is equal to 3W.

SOLUTION (a) From f.b.d. of rod AB  l   ΣM C = 0: T ( l sin θ ) + W   cosθ  − T ( l cosθ ) = 0  2   ∴T =

W cosθ 2 ( cosθ − sin θ )

Dividing both numerator and denominator by cosθ , T =

W 1    2  1 − tan θ  W     2 or T = (1 − tan θ )

(b) For T = 3W , W     2 3W = (1 − tan θ ) ∴ 1 − tan θ = or

1 6

5

θ = tan −1   = 39.806° 6 or θ = 39.8°

PROBLEM 4.55 A thin, uniform ring of mass m and radius R is attached by a frictionless pin to a collar at A and rests against a small roller at B. The ring lies in a vertical plane, and the collar can move freely on a horizontal rod and is acted upon by a horizontal force P. (a) Express the angle θ corresponding to equilibrium in terms of m and P. (b) Determine the value of θ corresponding to equilibrium when m = 500 g and P = 5 N.

SOLUTION (a) From f.b.d. of ring ΣM C = 0: P ( R cosθ + R cosθ ) − W ( R sin θ ) = 0 2P = W tan θ where W = mg ∴ tan θ =

2P mg  2P  or θ = tan −1    mg 

(b) Have

m = 500 g = 0.500 kg and P = 5 N  2 (5 N ) ∴ θ = tan −1   ( 0.500 kg ) 9.81 m/s 2 

(

)

   

= 63.872° or θ = 63.9°

PROBLEM 4.56 Rod AB is acted upon by a couple M and two forces, each of magnitude P. (a) Derive an equation in θ , P, M, and l which must be satisfied when the rod is in equilibrium. (b) Determine the value of θ corresponding to equilibrium when M = 150 lb ⋅ in., P = 20 lb, and l = 6 in.

SOLUTION (a) From f.b.d. of rod AB ΣM C = 0: P ( l cosθ ) + P ( l sin θ ) − M = 0 or sin θ + cosθ = (b) For

M Pl

M = 150 lb ⋅ in., P = 20 lb, and l = 6 in. sin θ + cosθ =

150 lb ⋅ in. 5 = = 1.25 ( 20 lb )( 6 in.) 4 sin 2 θ + cos 2 θ = 1

Using identity

(

sin θ + 1 − sin 2 θ

(

1 − sin 2 θ

)

1 2

)

1 2

= 1.25

= 1.25 − sin θ

1 − sin 2 θ = 1.5625 − 2.5sin θ + sin 2 θ 2sin 2 θ − 2.5sin θ + 0.5625 = 0 Using quadratic formula sin θ = = or

− ( −2.5 ) ±

( 6.25) − 4 ( 2 )( 0.5625) 2 ( 2)

2.5 ± 1.75 4

sin θ = 0.95572 ∴ θ = 72.886°

and and

sin θ = 0.29428

θ = 17.1144° or θ = 17.11° and θ = 72.9°

PROBLEM 4.57 A vertical load P is applied at end B of rod BC. The constant of the spring is k, and the spring is unstretched when θ = 90o. (a) Neglecting the weight of the rod, express the angle θ corresponding to equilibrium in terms of P, k, and l. (b) Determine the value of θ corresponding to 1 equilibrium when P = kl. 4

SOLUTION First note T = tension in spring = ks s = elongation of spring

where

( )θ − ( AB )θ

= AB

= 90°

θ   90°  = 2l sin   − 2l sin   2  2 

  θ   1  = 2l sin   −     2   2    θ   1  ∴ T = 2kl sin   −     2   2  (a) From f.b.d. of rod BC   θ  ΣM C = 0: T l cos    − P ( l sin θ ) = 0  2   Substituting T From Equation (1)   θ   1    θ  2kl sin   −    l cos  2   − P ( l sin θ ) = 0 2       2     θ   1   θ  θ   θ  2kl 2 sin   −    cos  2  − Pl  2sin  2  cos  2   = 0 2            2  Factoring out

θ  2l cos   , leaves 2

(1)

PROBLEM 4.57 CONTINUED   θ   1  θ  kl sin   −   − P sin   = 0  2   2   2  or 1  kl  θ  sin   =   2  kl − P  2   kl ∴ θ = 2sin −1    2 ( kl − P )  (b) P =

kl 4 

kl  2 kl −

θ = 2sin −1 

(

kl 4

)

  kl  4   −1  4   = 2sin −1     = 2sin    3 2   2  3 kl  

= 2sin −1 ( 0.94281) = 141.058° or θ = 141.1°

PROBLEM 4.58 Solve Sample Problem 4.5 assuming that the spring is unstretched when θ = 90o.

SOLUTION First note T = tension in spring = ks s = deformation of spring

where

= rβ ∴ F = kr β From f.b.d. of assembly ΣM 0 = 0: W ( l cos β ) − F ( r ) = 0 Wl cos β − kr 2 β = 0

or

∴ cos β = For

kr 2 β Wl

k = 250 lb/in., r = 3 in., l = 8 in., W = 400 lb

cos β = or

( 250 lb/in.)( 3 in.)2 β ( 400 lb )(8 in.)

cos β = 0.703125β

Solving numerically,

β = 0.89245 rad or Then

β = 51.134° θ = 90° + 51.134° = 141.134° or θ = 141.1°

PROBLEM 4.59 A collar B of weight W can move freely along the vertical rod shown. The constant of the spring is k, and the spring is unstretched when θ = 0. (a) Derive an equation in θ , W, k, and l which must be satisfied when the collar is in equilibrium. (b) Knowing that W = 3 lb, l = 6 in., and k = 8 lb/ft, determine the value of θ corresponding to equilibrium.

SOLUTION T = ks

First note

k = spring constant

where

s = elongation of spring =

l l −l = (1 − cosθ ) cosθ cosθ

∴ T =

kl (1 − cosθ ) cosθ

(a) From f.b.d. of collar B ΣFy = 0: T sin θ − W = 0

kl (1 − cosθ ) sin θ − W = 0 cosθ

or

or tan θ − sin θ =

W kl

(b) For W = 3 lb, l = 6 in., k = 8 lb/ft

l =

6 in. = 0.5 ft 12 in./ft

tan θ − sin θ =

3 lb = 0.75 (8 lb/ft )( 0.5 ft )

Solving Numerically,

θ = 57.957° or θ = 58.0°

PROBLEM 4.60 A slender rod AB, of mass m, is attached to blocks A and B which move freely in the guides shown. The constant of the spring is k, and the spring is unstretched when θ = 0 . (a) Neglecting the mass of the blocks, derive an equation in m, g, k, l, and θ which must be satisfied when the rod is in equilibrium. (b) Determine the value of θ when m = 2 kg, l = 750 mm, and k = 30 N/m.

SOLUTION First note

Fs = spring force = ks k = spring constant

where

s = spring deformation = l − l cosθ = l (1 − cosθ ) ∴ Fs = kl (1 − cosθ ) (a) From f.b.d. of assembly l  ΣM D = 0: Fs ( l sin θ ) − W  cosθ  = 0 2  l  kl (1 − cosθ )( l sin θ ) − W  cosθ  = 0 2  W kl ( sin θ − cosθ sin θ ) −   2

  cosθ = 0 

Dividing by cosθ

kl ( tan θ − sin θ ) =

W 2

∴ tan θ − sin θ =

W 2kl or tan θ − sin θ =

(b) For m = 2 kg, l = 750 mm, k = 30 N/m

l = 750 mm = 0.750 m

mg 2kl

PROBLEM 4.60 CONTINUED Then

tan θ − sin θ =

( 2 kg ) ( 9.81 m/s2 ) 2 ( 30 N/m )( 0.750 m )

= 0.436

Solving Numerically,

θ = 50.328° or θ = 50.3°

PROBLEM 4.61 The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions assuming that the magnitude of the force P is 100 N.

SOLUTION 1. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From f.b.d. of bracket: ΣM A = 0: B (1 m ) − (100 N )( 0.6 m ) = 0

∴ B = 60.0 N ΣFx = 0: Ax − 60 N = 0 ∴ A x = 60.0 N ΣFy = 0: Ay − 100 N = 0 ∴ A y = 100 N Then

A=

( 60.0 )2 + (100 )2

= 116.619 N

 100 

θ = tan −1   = 59.036°  60.0 

and

∴ A = 116.6 N

59.0°

2. Four concurrent reactions through A (a)

Improperly constrained

(b)

Indeterminate

(c)

No equilibrium

3. Two reactions (a)

Partially constrained

(b)

Determinate

(c)

Equilibrium

PROBLEM 4.61 CONTINUED From f.b.d. of bracket ΣM A = 0: C (1.2 m ) − (100 N )( 0.6 m ) = 0 ∴ C = 50.0 N ΣFy = 0: A − 100 N + 50 N = 0 ∴ A = 50.0 N 4. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From f.b.d. of bracket  1.0 

θ = tan −1   = 39.8°  1.2  BC =

(1.2 )2 + (1.0 )2

= 1.56205 m

 1.2   ΣM A = 0:   B  (1 m ) − (100 N )( 0.6 m ) = 0  1.56205   ∴ B = 78.1 N

39.8°

ΣFx = 0: C − ( 78.102 N ) cos 39.806° = 0 ∴ C = 60.0 N

ΣFy = 0: A + ( 78.102 N ) sin 39.806° − 100 N = 0 ∴ A = 50.0 N 5. Four non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Indeterminate

(c)

Equilibrium

From f.b.d. of bracket ΣM C = 0:

(100 N )( 0.6 m ) − Ay (1.2 m ) = 0 ∴ Ay = 50 N

or A y = 50.0 N

6. Four non-concurrent non-parallel reactions (a)

Completely constrained

(b)

Indeterminate

(c)

Equilibrium

PROBLEM 4.61 CONTINUED From f.b.d. of bracket ΣM A = 0: − Bx (1 m ) − (100 N )( 0.6 m ) = 0 ∴ Bx = −60.0 N or B x = 60.0 N ΣFx = 0: − 60 + Ax = 0 ∴ Ax = 60.0 N or A x = 60.0 N 7. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From f.b.d. of bracket ΣFx = 0: Ax = 0 ΣM A = 0: C (1.2 m ) − (100 N )( 0.6 m ) = 0 ∴ C = 50.0 N or C = 50.0 N ΣFy = 0: Ay − 100 N + 50.0 N = 0 ∴ Ay = 50.0 N ∴ A = 50.0 N 8. Three concurrent, non-parallel reactions (a)

Improperly constrained

(b)

Indeterminate

(c)

No equilibrium

PROBLEM 4.62 Eight identical 20 × 30-in. rectangular plates, each weighing 50 lb, are held in a vertical plane as shown. All connections consist of frictionless pins, rollers, or short links. For each case, answer the questions listed in Problem 4.61, and, wherever possible, compute the reactions.

P6.1 The bracket ABC can be supported in the eight different ways shown. All connections consist of smooth pins, rollers, or short links. In each case, determine whether (a) the plate is completely, partially, or improperly constrained, (b) the reactions are statically determinate or indeterminate, (c) the equilibrium of the plate is maintained in the position shown. Also, wherever possible, compute the reactions assuming that the magnitude of the force P is 100 N.

SOLUTION 1. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From f.b.d. of plate ΣM A = 0: C ( 30 in.) − 50 lb (15 in.) = 0

C = 25.0 lb ΣFx = 0: Ax = 0 ΣFy = 0: Ay − 50 lb + 25 lb = 0

Ay = 25 lb

A = 25.0 lb

2. Three non-current, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From f.b.d. of plate ΣFx = 0: ΣM B = 0:

B=0

( 50 lb )(15 in.) − D ( 30 in.) = 0 D = 25.0 lb

ΣFy = 0: 25.0 lb − 50 lb + C = 0

C = 25.0 lb

PROBLEM 4.62 CONTINUED 3. Four non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Indeterminate

(c)

Equilibrium

From f.b.d. of plate ΣM D = 0: Ax ( 20 in.) − ( 50 lb )(15 in.) ∴ A x = 37.5 lb ΣFx = 0: Dx + 37.5 lb = 0 ∴ D x = 37.5 lb 4. Three concurrent reactions (a)

Improperly constrained

(b)

Indeterminate

(c)

No equilibrium

5. Two parallel reactions (a)

Partial constraint

(b)

Determinate

(c)

Equilibrium

From f.b.d. of plate ΣM D = 0: C ( 30 in.) − ( 50 lb )(15 in.) = 0

C = 25.0 lb ΣFy = 0: D − 50 lb + 25 lb = 0

D = 25.0 lb 6. Three non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Determinate

(c)

Equilibrium

From f.b.d. of plate ΣM D = 0: B ( 20 in.) − ( 50 lb )(15 in.) = 0

B = 37.5 lb ΣFx = 0: Dx + 37.5 lb = 0 ΣFy = 0: Dy − 50 lb = 0

D x = 37.5 lb D y = 50.0 lb or D = 62.5 lb

53.1°

PROBLEM 4.62 CONTINUED 7. Two parallel reactions (a)

Improperly constrained

(b)

Reactions determined by dynamics

(c)

No equilibrium

8. Four non-concurrent, non-parallel reactions (a)

Completely constrained

(b)

Indeterminate

(c)

Equilibrium

From f.b.d. of plate ΣM D = 0: B ( 30 in.) − ( 50 lb )(15 in.) = 0

B = 25.0 lb ΣFy = 0: Dy − 50 lb + 25.0 lb = 0

D y = 25.0 lb ΣFx = 0: Dx + C = 0

PROBLEM 4.63 Horizontal and vertical links are hinged to a wheel, and forces are applied to the links as shown. Knowing that a = 3.0 in., determine the value of P and the reaction at A.

SOLUTION As shown on the f.b.d., the wheel is a three-force body. Let point D be the intersection of the three forces. From force triangle A P 21 lb = = 5 4 3 ∴ P=

4 ( 21 lb ) = 28 lb 3 or P = 28.0 lb

and

A=

5 ( 21 lb ) = 35 lb 3 3  

θ = tan −1   = 36.870° 4 ∴ A = 35.0 lb

36.9°

PROBLEM 4.64 Horizontal and vertical links are hinged to a wheel, and forces are applied to the links as shown. Determine the range of values of the distance a for which the magnitude of the reaction at A does not exceed 42 lb.

SOLUTION Let D be the intersection of the three forces acting on the wheel. From the force triangle 21 lb = a

or

16 + a 2

A = 21

16 +1 a2

A = 42 lb

For

21 lb = a

a2 =

or

or

A

a=

42 lb 16 + a 2

16 + a 2 4

16 = 2.3094 in. 3 or a ≥ 2.31 in.

Since as a increases, A decreases

A = 21

16 +1 a2

PROBLEM 4.65 Using the method of Section 4.7, solve Problem 4.21. P4.21 The required tension in cable AB is 800 N. Determine (a) the vertical force P which must be applied to the pedal, (b) the corresponding reaction at C.

SOLUTION Let E be the intersection of the three forces acting on the pedal device. First note  (180 mm ) sin 60°   = 21.291° 400 mm  

α = tan −1  From force triangle (a)

P = ( 800 N ) tan 21.291°

= 311.76 N or P = 312 N (b)

C =

800 N cos 21.291°

= 858.60 N or C = 859 N

21.3°

PROBLEM 4.66 Using the method of Section 4.7, solve Problem 4.22. P4.22 Determine the maximum tension which can be developed in cable AB if the maximum allowable value of the reaction at C is 1000 N.

SOLUTION Let E be the intersection of the three forces acting on the pedal device. First note  (180 mm ) sin 60°   = 21.291° 400 mm  

α = tan −1  From force triangle

Tmax = (1000 N ) cos 21.291°

= 931.75 N or Tmax = 932 N

PROBLEM 4.67 To remove a nail, a small block of wood is placed under a crowbar, and a horizontal force P is applied as shown. Knowing that l = 3.5 in. and P = 30 lb, determine the vertical force exerted on the nail and the reaction at B.

SOLUTION Let D be the intersection of the three forces acting on the crowbar. First note  ( 36 in.) sin 50°   = 82.767° 3.5 in.  

θ = tan −1  From force triangle

FN = P tan θ = ( 30 lb ) tan 82.767°

= 236.381 lb ∴ on nail FN = 236 lb RB =

P 30 lb = = 238.28 lb cosθ cos82.767°

or R B = 238 lb

82.8°

PROBLEM 4.68 To remove a nail, a small block of wood is placed under a crowbar, and a horizontal force P is applied as shown. Knowing that the maximum vertical force needed to extract the nail is 600 lb and that the horizontal force P is not to exceed 65 lb, determine the largest acceptable value of distance l.

SOLUTION Let D be the intersection of the three forces acting on the crowbar. From force diagram tan θ =

FN 600 lb = = 9.2308 P 65 lb

∴ θ = 83.817° From f.b.d. tan θ =

∴ l =

( 36 in.) sin 50° l

( 36 in.) sin 50° tan 83.817°

= 2.9876 in. or l = 2.99 in.

PROBLEM 4.69 For the frame and loading shown, determine the reactions at C and D.

SOLUTION

Since member BD is acted upon by two forces, B and D, they must be colinear, have the same magnitude, and be opposite in direction for BD to be in equilibrium. The force B acting at B of member ABC will be equal in magnitude but opposite in direction to force B acting on member BD. Member ABC is a three-force body with member forces intersecting at E. The f.b.d.’s of members ABC and BD illustrate the above conditions. The force triangle for member ABC is also shown. The angles α and β are found from the member dimensions:  0.5 m   = 26.565°  1.0 m 

α = tan −1 

 1.5 m 

β = tan −1   = 56.310°  1.0 m  Applying the law of sines to the force triangle for member ABC, 150 N C B = = sin ( β − α ) sin ( 90° + α ) sin ( 90° − β ) or

150 N C B = = sin 29.745° sin116.565° sin 33.690° ∴ C =

and

(150 N ) sin116.565°

D= B=

sin 29.745°

= 270.42 N

(150 N ) sin 33.690° sin 29.745°

or C = 270 N

56.3°

or D = 167.7 N

26.6°

= 167.704 N

PROBLEM 4.70 For the frame and loading shown, determine the reactions at A and C.

SOLUTION

Since member AB is acted upon by two forces, A and B, they must be colinear, have the same magnitude, and be opposite in direction for AB to be in equilibrium. The force B acting at B of member BCD will be equal in magnitude but opposite in direction to force B acting on member AB. Member BCD is a three-force body with member forces intersecting at E. The f.b.d.’s of members AB and BCD illustrate the above conditions. The force triangle for member BCD is also shown. The angle β is found from the member dimensions:  60 m 

β = tan −1   = 30.964°  100 m  Applying of the law of sines to the force triangle for member BCD, 130 N B C = = sin ( 45° − β ) sin β sin135° or

130 N B C = = sin14.036° sin 30.964° sin135° ∴ A= B=

and

C =

(130 N ) sin 30.964° sin14.036°

(130 N ) sin135° sin14.036°

= 275.78 N or A = 276 N

45.0°

or C = 379 N

59.0°

= 379.02 N

PROBLEM 4.71 To remove the lid from a 5-gallon pail, the tool shown is used to apply an upward and radially outward force to the bottom inside rim of the lid. Assuming that the rim rests against the tool at A and that a 100-N force is applied as indicated to the handle, determine the force acting on the rim.

SOLUTION The three-force member ABC has forces that intersect at D, where 

α = tan −1 

 yDC

 90 mm  − yBC − 45 mm 

and yDC =

( 360 mm ) cos 35° xBC = tan 20° tan 20° = 810.22 mm

yBC = ( 360 mm ) sin 35°

= 206.49 mm  90  ∴ α = tan −1   = 9.1506°  558.73  Based on the force triangle, the law of sines gives

100 N A = sin α sin 20° ∴A= or

(100 N ) sin 20° sin 9.1506°

A = 215 N

= 215.07 N 80.8° on tool

and A = 215 N

80.8° on rim of can

PROBLEM 4.72 To remove the lid from a 5-gallon pail, the tool shown is used to apply an upward and radially outward force to the bottom inside rim of the lid. Assuming that the top and the rim of the lid rest against the tool at A and B, respectively, and that a 60-N force is applied as indicated to the handle, determine the force acting on the rim.

SOLUTION The three-force member ABC has forces that intersect at point D, where, from the law of sines ( ∆CDE ) 150 mm + (19 mm ) tan 35° L = sin 95° sin 30° ∴ L = 325.37 mm Then  45 mm    yBD 

α = tan −1  where

yBD = L − y AE − 22 mm = 325.37 mm −

19 mm − 22 mm cos 35°

= 280.18 mm  45 mm  ∴ α = tan −1   = 9.1246°  280.18 mm  Applying the law of sines to the force triangle, B 60 N = sin150° sin 9.1246° ∴ B = 189.177 N Or, on member and, on lid

B = 189.2 N

80.9° B = 189.2 N

80.9°

PROBLEM 4.73 A 200-lb crate is attached to the trolley-beam system shown. Knowing that a = 1.5 ft, determine (a) the tension in cable CD, (b) the reaction at B.

SOLUTION From geometry of forces  y



β = tan −1  BE   1.5 ft  where yBE = 2.0 − yDE

= 2.0 − 1.5 tan 35° = 0.94969 ft  0.94969  ∴ β = tan −1   = 32.339°  1.5  and

α = 90° − β = 90° − 32.339° = 57.661° θ = β + 35° = 32.339° + 35° = 67.339°

Applying the law of sines to the force triangle, 200 lb T B = = sin θ sin α sin 55° or (a)

( 200 lb ) sin 67.339° T =

=

T B = sin 57.661° sin 55°

( 200 lb )( sin 57.661° ) sin 67.339°

= 183.116 lb or T = 183.1 lb

(b)

B=

( 200 lb )( sin 55° ) sin 67.339°

= 177.536 lb or B = 177.5 lb

32.3°

PROBLEM 4.74 Solve Problem 4.73 assuming that a = 3 ft. P4.73 A 200-lb crate is attached to the trolley-beam system shown. Knowing that a = 1.5 ft, determine (a) the tension in cable CD, (b) the reaction at B.

SOLUTION From geometry of forces y



β = tan −1  BE   3 ft  where yBE = yDE − 2.0 ft

= 3tan 35° − 2.0 = 0.100623 ft  0.100623  ∴ β = tan −1   = 1.92103° 3   and

α = 90° + β = 90° + 1.92103° = 91.921° θ = 35° − β = 35° − 1.92103° = 33.079°

Applying the law of sines to the force triangle, 200 lb T B = = sin θ sin α sin 55° or

200 lb T B = = sin 33.079° sin 91.921° sin 55°

(a)

T =

( 200 lb )( sin 91.921° ) sin 33.079°

= 366.23 lb or T = 366 lb

(b)

B=

( 200 lb )( sin 55° ) sin 33.079°

= 300.17 lb or B = 300 lb

1.921°

PROBLEM 4.75 A 20-kg roller, of diameter 200 mm, which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 8 mm, determine the force P required to move the roller onto the tiles if the roller is pushed to the left.

SOLUTION Based on the roller having impending motion to the left, the only contact between the roller and floor will be at the edge of the tile. First note

(

)

W = mg = ( 20 kg ) 9.81 m/s 2 = 196.2 N

From the geometry of the three forces acting on the roller  92 mm   = 23.074°  100 mm 

α = cos −1  and

θ = 90° − 30° − α = 60° − 23.074 = 36.926°

Applying the law of sines to the force triangle, W P = sin θ sin α or

196.2 N P = sin 36.926° sin 23.074° ∴ P = 127.991 N or P = 128.0 N

30°

PROBLEM 4.76 A 20-kg roller, of diameter 200 mm, which is to be used on a tile floor, is resting directly on the subflooring as shown. Knowing that the thickness of each tile is 8 mm, determine the force P required to move the roller onto the tiles if the roller is pulled to the right.

SOLUTION Based on the roller having impending motion to the right, the only contact between the roller and floor will be at the edge of the tile. First note

(

W = mg = ( 20 kg ) 9.81 m/s 2

)

= 196.2 N From the geometry of the three forces acting on the roller  92 mm 

α = cos −1   = 23.074°  100 mm  and

θ = 90° + 30° − α = 120° − 23.074° = 96.926°

Applying the law of sines to the force triangle, W P = sin θ sin α or

196.2 N P = sin 96.926° sin 23.074 ∴ P = 77.460 N or P = 77.5 N

30°

PROBLEM 4.77 A small hoist is mounted on the back of a pickup truck and is used to lift a 120-kg crate. Determine (a) the force exerted on the hoist by the hydraulic cylinder BC, (b) the reaction at A.

SOLUTION First note

(

)

W = mg = (120 kg ) 9.81 m/s 2 = 1177.2 N

From the geometry of the three forces acting on the small hoist x AD = (1.2 m ) cos 30° = 1.03923 m y AD = (1.2 m ) sin 30° = 0.6 m and Then

yBE = x AD tan 75° = (1.03923 m ) tan 75° = 3.8785 m  yBE − 0.4 m  −1  3.4785   = tan   = 73.366° x AD  1.03923   

α = tan −1 

β = 75° − α = 75° − 73.366° = 1.63412° θ = 180° − 15° − β = 165° − 1.63412° = 163.366° Applying the law of sines to the force triangle, W B A = = sin β sin θ sin15° or (a)

1177.2 N B A = = sin1.63412° sin163.366° sin15° B = 11 816.9 N or B = 11.82 kN

(b)

75.0°

A = 10 684.2 N or A = 10.68 kN

73.4°

PROBLEM 4.78 The clamp shown is used to hold the rough workpiece C. Knowing that the maximum allowable compressive force on the workpiece is 200 N and neglecting the effect of friction at A, determine the corresponding (a) reaction at B, (b) reaction at A, (c) tension in the bolt.

SOLUTION From the geometry of the three forces acting on the clamp y AD = (105 mm ) tan 78° = 493.99 mm yBD = y AD − 70 mm = ( 493.99 − 70 ) mm = 423.99 mm Then



y



 423.99 

−1 BD θ = tan −1   = tan   = 65.301°  195 mm   195 

α = 90° − θ − 12° = 78° − 65.301° = 12.6987° (a) Based on the maximum allowable compressive force on the workpiece of 200 N,

( RB ) y

= 200 N

RB sin θ = 200 N

or ∴ RB =

200 N = 220.14 N sin 65.301° or R B = 220 N

65.3°

Applying the law of sines to the force triangle, RB NA T = = sin12° sin α sin ( 90° + θ ) or (b)

220.14 N NA T = = sin12° sin12.6987° sin155.301° N A = 232.75 N or N A = 233 N

(c)

T = 442.43 N or T = 442 N

PROBLEM 4.79 A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg. Knowing that θ = 45° and that the force exerted at C by the worker is perpendicular to the handle of the peavey, determine (a) the force exerted at C, (b) the reaction at A.

SOLUTION First note

(

)

W = mg = ( 36 kg ) 9.81 m/s 2 = 353.16 N

From the geometry of the three forces acting on the modified peavey 

1.1 m



β = tan −1   = 40.236°  1.1 m + 0.2 m  α = 45° − β = 45° − 40.236° = 4.7636° Applying the law of sines to the force triangle, W C A = = sin β sin α sin135° or (a)

(b)

353.16 N C A = = sin 40.236° sin 4.7636 sin135° C = 45.404 N or C = 45.4 N

45.0°

or A = 387 N

85.2°

A = 386.60 N

PROBLEM 4.80 A modified peavey is used to lift a 0.2-m-diameter log of mass 36 kg. Knowing that θ = 60° and that the force exerted at C by the worker is perpendicular to the handle of the peavey, determine (a) the force exerted at C, (b) the reaction at A.

SOLUTION First note

(

)

W = mg = ( 36 kg ) 9.81 m/s 2 = 353.16 N

From the geometry of the three forces acting on the modified peavey 



1.1 m

β = tan −1    DC + 0.2 m  where

DC = (1.1 m + a ) tan 30°  R  a= −R  tan 30°   0.1 m  =  − 0.1 m  tan 30°  = 0.073205 m ∴ DC = (1.173205 ) tan 30° = 0.67735 m

and



1.1



β = tan −1   = 51.424°  0.87735  α = 60° − β = 60° − 51.424° = 8.5756°

Applying the law of sines to the force triangle, W C A = = sin β sin α sin120° or (a)

353.16 N C A = = sin 51.424° sin 8.5756° sin120° C = 67.360 N or C = 67.4 N

(b)

30°

A = 391.22 N or A = 391 N

81.4°

PROBLEM 4.81 Member ABC is supported by a pin and bracket at B and by an inextensible cord at A and C and passing over a frictionless pulley at D. The tension may be assumed to be the same in portion AD and CD of the cord. For the loading shown and neglecting the size of the pulley, determine the tension in the cord and the reaction at B.

SOLUTION From the f.b.d. of member ABC, it is seen that the member can be treated as a three-force body. From the force triangle

T − 300 3 = T 4

3T = 4T − 1200 ∴ T = 1200 lb

B 5 = T 4

Also, ∴ B=

5 5 T = (1200 lb ) = 1500 lb 4 4 3

θ = tan −1   = 36.870° 4 and B = 1500 lb

36.9°

PROBLEM 4.82 Member ABCD is supported by a pin and bracket at C and by an inextensible cord attached at A and D and passing over frictionless pulleys at B and E. Neglecting the size of the pulleys, determine the tension in the cord and the reaction at C.

SOLUTION From the geometry of the forces acting on member ABCD  200 

β = tan −1   = 33.690°  300   375   = 61.928°  200 

α = tan −1 

α − β = 61.928° − 33.690° = 28.237° 180° − α = 180° − 61.928° = 118.072° Applying the law of sines to the force triangle, T − 80 N T C = = sin (α − β ) sin β sin (180° − α ) or

T − 80 N T C = = sin 28.237° sin 33.690° sin118.072°

Then

(T

− 80 N ) sin 33.690° = T sin 28.237° ∴ T = 543.96 N or T = 544 N

and

( 543.96 N ) sin118.072 = C sin 33.690° ∴ C = 865.27 N or C = 865 N

33.7°

PROBLEM 4.83 Using the method of Section 4.7, solve Problem 4.18. P4.18 Determine the reactions at A and B when (a) h = 0 , (b) h = 8 in.

SOLUTION (a) Based on symmetry

α = 30° From force triangle A = B = 40 lb or A = 40.0 lb

30°

and B = 40.0 lb

30°

(b) From geometry of forces  8 in. − (10 in.) tan 30°   = 12.5521° 10 in.  

α = tan −1  Also,

30° − α = 30° − 12.5521° = 17.4479° 90° + α = 90° + 12.5521° = 102.5521° Applying law of sines to the force triangle, 40 lb A B = = sin ( 30° − α ) sin 60° sin ( 90° + α ) or

40 lb A B = = sin17.4479° sin 60° sin102.5521 A = 115.533 lb or A = 115.5 lb

12.55°

B = 130.217 lb or B = 130.2 lb

30.0°

PROBLEM 4.84 Using the method of Section 4.7, solve Problem 4.28.

P4.28 A lever is hinged at C and is attached to a control cable at A. If the lever is subjected to a 300-N vertical force at B, determine (a) the tension in the cable, (b) the reaction at C.

SOLUTION From geometry of forces acting on lever  yDA    xDA 

α = tan −1  where

yDA = 0.24 m − y AC = 0.24 m − ( 0.2 m ) sin 20° = 0.171596 m xDA = ( 0.2 m ) cos 20° = 0.187939 m  0.171596  ∴ α = tan −1   = 42.397°  0.187939   y AC + yEA   xCE  

β = 90° − tan −1  where

xCE = ( 0.3 m ) cos 20° = 0.28191 m y AC = ( 0.2 m ) sin 20° = 0.068404 m yEA = ( xDA + xCE ) tan α = ( 0.187939 + 0.28191) tan 42.397° = 0.42898 m  0.49739  ∴ β = 90° − tan −1   = 29.544°  0.28191 

Also,

90° − (α + β ) = 90° − 71.941° = 18.0593° 90° + α = 90° + 42.397° = 132.397°

PROBLEM 4.84 CONTINUED Applying the law of sines to the force triangle, 300 N T C = = sin ( 90° + α ) sin 90° − (α + β )  sin β or

300 N T C = = sin18.0593° sin 29.544° sin132.397°

(a) T = 477.18 N (b) C = 714.67 N

or T = 477 N or C = 715 N

60.5°

PROBLEM 4.85 Knowing that θ = 35o , determine the reaction (a) at B, (b) at C.

SOLUTION From the geometry of the three forces applied to the member ABC y



α = tan −1  CD   R  where

yCD = R tan 55° − R = 0.42815R ∴ α = tan −1 ( 0.42815 ) = 23.178° Then

55° − α = 55° − 23.178° = 31.822° 90° + α = 90° + 23.178° = 113.178°

Applying the law of sines to the force triangle,

P B C = = sin ( 55° − α ) sin ( 90° + α ) sin 35° or (a)

P B C = = sin 31.822° sin113.178° sin 35° B = 1.74344P or B = 1.743P

(b)

55.0°

C = 1.08780P or C = 1.088P

23.2°

PROBLEM 4.86 Knowing that θ = 50o , determine the reaction (a) at B, (b) at C.

SOLUTION From the geometry of the three forces acting on member ABC y



α = tan −1  DC   R  where yDC = R − y AD = R 1 − tan ( 90° − 50° ) 

= 0.160900 R ∴ α = tan −1 ( 0.160900 ) = 9.1406° Then

90° − α = 90° − 9.1406° = 80.859° 40° + α = 40° + 9.1406° = 49.141°

Applying the law of sines to the force triangle,

P B C = = sin ( 40° + α ) sin ( 90° − α ) sin 50° or (a)

(b)

P B C = = sin 49.141° sin ( 80.859° ) sin 50° B = 1.30540P or B = 1.305P

40.0°

or C = 1.013P

9.14°

C = 1.01286P

PROBLEM 4.87 A slender rod of length L and weight W is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S. Derive an expression for the distance h in terms of L and S. Show that this position of equilibrium does not exist if S > 2L.

SOLUTION From the f.b.d of the three-force member AB, forces must intersect at D. Since the force T intersects point D, directly above G,

yBE = h For triangle ACE: S 2 = ( AE ) + ( 2h ) 2

2

(1)

For triangle ABE: L2 = ( AE ) + ( h ) 2

2

(2)

Subtracting Equation (2) from Equation (1)

S 2 − L2 = 3h 2

(3) or h =

S 2 − L2 3

As length S increases relative to length L, angle θ increases until rod AB is vertical. At this vertical position:

h+L=S

or

h≥S−L

Therefore, for all positions of AB

S 2 − L2 ≥S−L 3

or or or and For

h=S−L

(

)

S 2 − L2 ≥ 3 ( S − L ) = 3 S 2 − 2SL + L2 = 3S 2 − 6SL + 3L2 2

0 ≥ 2S 2 − 6SL + 4 L2 0 ≥ S 2 − 3SL + 2 L2 = ( S − L )( S − 2 L )

S−L=0

S = L

∴ Minimum value of S is L For

S − 2L = 0

S = 2L

∴ Maximum value of S is 2L Therefore, equilibrium does not exist if S > 2L

(4)

PROBLEM 4.88 A slender rod of length L = 200 mm is held in equilibrium as shown, with one end against a frictionless wall and the other end attached to a cord of length S = 300 mm. Knowing that the mass of the rod is 1.5 kg, determine (a) the distance h, (b) the tension in the cord, (c) the reaction at B.

SOLUTION From the f.b.d of the three-force member AB, forces must intersect at D. Since the force T intersects point D, directly above G,

yBE = h For triangle ACE: S 2 = ( AE ) + ( 2h ) 2

2

(1)

For triangle ABE: L2 = ( AE ) + ( h ) 2

2

(2)

Subtracting Equation (2) from Equation (1)

S 2 − L2 = 3h 2 or h = (a) For L = 200 mm and

h=

S 2 − L2 3

S = 300 mm

( 300 )2 − ( 200 )2 3

= 129.099 mm or h = 129.1 mm

(b) Have

(

)

W = mg = (1.5 kg ) 9.81 m/s 2 = 14.715 N  2h 

 2 (129.099 )   300 

θ = sin −1   = sin −1   s  

and

θ = 59.391° From the force triangle

T =

W 14.715 N = = 17.0973 N sin θ sin 59.391° or T = 17.10 N

(c)

B=

W 14.715 N = = 8.7055 N tan θ tan 59.391° or B = 8.71 N

PROBLEM 4.89 A slender rod of length L and weight W is attached to collars which can slide freely along the guides shown. Knowing that the rod is in equilibrium, derive an expression for the angle θ in terms of the angle β .

SOLUTION As shown in the f.b.d of the slender rod AB, the three forces intersect at C. From the force geometry tan β =

xGB y AB

where

y AB = L cosθ xGB =

and

∴ tan β =

1 2

1 L sin θ 2

L sin θ

L cosθ

=

1 tan θ 2 or tan θ = 2 tan β

PROBLEM 4.90 A 10-kg slender rod of length L is attached to collars which can slide freely along the guides shown. Knowing that the rod is in equilibrium and that β = 25°, determine (a) the angle θ that the rod forms with the vertical, (b) the reactions at A and B.

SOLUTION (a) As shown in the f.b.d. of the slender rod AB, the three forces intersect at C. From the geometry of the forces

xCB yBC

tan β = where

xCB =

1 L sin θ 2

yBC = L cosθ

and

∴ tan β =

1 tan θ 2

or

tan θ = 2 tan β

For

β = 25° tan θ = 2 tan 25° = 0.93262 ∴ θ = 43.003° or θ = 43.0°

(

)

W = mg = (10 kg ) 9.81 m/s 2 = 98.1 N

(b)

From force triangle

A = W tan β = ( 98.1 N ) tan 25° = 45.745 N or A = 45.7 N and

B=

W 98.1 N = = 108.241 N cos β cos 25° or B = 108.2 N

65.0°

PROBLEM 4.91 A uniform slender rod of mass 5 g and length 250 mm is balanced on a glass of inner diameter 70 mm. Neglecting friction, determine the angle θ corresponding to equilibrium.

SOLUTION From the geometry of the forces acting on the three-force member AB Triangle ACF

yCF = d tan θ xFE = yCF tan θ = d tan 2 θ

Triangle CEF Triangle AGE cosθ =

=

d + xFE d + d tan 2 θ = L L     2 2

(

2d 1 + tan 2 θ L

Now

(1 + tan θ ) = sec θ

Then

cosθ =

2

2

) 1 cosθ

2d 2d  1  sec 2 θ =   L L  cos 2 θ  ∴ cos3 θ =

For

secθ =

and

d = 70 mm and cos3 θ =

2d L

L = 250 mm

2 ( 70 ) = 0.56 250

∴ cosθ = 0.82426 and

θ = 34.487° or θ = 34.5°

PROBLEM 4.92 Rod AB is bent into the shape of a circular arc and is lodged between two pegs D and E. It supports a load P at end B. Neglecting friction and the weight of the rod, determine the distance c corresponding to equilibrium when a = 1 in. and R = 5 in.

SOLUTION yED = xED = a,

Since Slope of ED is ∴ slope of HC is

45° 45° DE =

Also and

2a

a 1 DH = HE =   DE = 2 2

For triangles DHC and EHC sin β =

a = 1 in. and sin β =

R = 5 in.

1 in. = 0.141421 2 ( 5 in.)

∴ β = 8.1301° and

a 2R

c = R sin ( 45° − β )

Now For

a/ 2 = R

or β = 8.13°

c = ( 5 in.) sin ( 45° − 8.1301° ) = 3.00 in. or c = 3.00 in.

PROBLEM 4.93 A uniform rod AB of weight W and length 2R rests inside a hemispherical bowl of radius R as shown. Neglecting friction determine the angle θ corresponding to equilibrium.

SOLUTION Based on the f.b.d., the uniform rod AB is a three-force body. Point E is the point of intersection of the three forces. Since force A passes through O, the center of the circle, and since force C is perpendicular to the rod, triangle ACE is a right triangle inscribed in the circle. Thus, E is a point on the circle. Note that the angle α of triangle DOA is the central angle corresponding to the inscribed angle θ of triangle DCA. ∴ α = 2θ The horizontal projections of AE , ( x AE ) , and AG, ( x AG ) , are equal. ∴ x AE = x AG = x A or

( AE ) cos 2θ

= ( AG ) cosθ

and

( 2R ) cos 2θ

Now

cos 2θ = 2cos 2 θ − 1

then

4cos 2 θ − 2 = cosθ

or

= R cosθ

4cos 2 θ − cosθ − 2 = 0

Applying the quadratic equation cosθ = 0.84307 ∴ θ = 32.534°

and

cosθ = −0.59307

and θ = 126.375°(Discard) or θ = 32.5°

PROBLEM 4.94 A uniform slender rod of mass m and length 4r rests on the surface shown and is held in the given equilibrium position by the force P. Neglecting the effect of friction at A and C, (a) determine the angle θ, (b) derive an expression for P in terms of m.

SOLUTION The forces acting on the three-force member intersect at D. (a) From triangle ACO  r  −1  1   = tan   = 18.4349°  3r   3

θ = tan −1 

tan θ =

(b) From triangle DCG ∴ DC = and

or θ = 18.43° r DC

r r = = 3r tan θ tan18.4349°

DO = DC + r = 3r + r = 4r  yDO    x AG 

α = tan −1  where

yDO = ( DO ) cosθ = ( 4r ) cos18.4349° = 3.4947r

and

x AG = ( 2r ) cosθ = ( 2r ) cos18.4349° = 1.89737r  3.4947r  ∴ α = tan −1   = 63.435°  1.89737r 

where

90° + (α − θ ) = 90° + 45° = 135.00°

Applying the law of sines to the force triangle, mg R = A   sin 90° + (α − θ )  sin θ ∴ RA = ( 0.44721) mg Finally,

P = RA cos α = ( 0.44721mg ) cos 63.435° = 0.20000mg

or P =

mg 5

PROBLEM 4.95 A uniform slender rod of length 2L and mass m rests against a roller at D and is held in the equilibrium position shown by a cord of length a. Knowing that L = 200 mm, determine (a) the angle θ, (b) the length a.

SOLUTION (a) The forces acting on the three-force member AB intersect at E. Since triangle DBC is isosceles, DB = a. From triangle BDE ED = DB tan 2θ = a tan 2θ From triangle GED ED = ∴ a tan 2θ =

L−a tan θ

tan θ a ( tan θ tan 2θ + 1) = L

or a=

From triangle BCD

( L − a)

1 (1.25L ) 2

cosθ

or

L = 1.6cosθ a

Substituting Equation (2) into Equation (1) yields 1.6cosθ = 1 + tan θ tan 2θ Now

tan θ tan 2θ =

sin θ sin 2θ cosθ cos 2θ

=

sin θ 2sin θ cosθ cosθ 2 cos 2 θ − 1

=

2 (1 − cos 2 θ ) 2cos 2 θ − 1 2 (1 − cos 2 θ ) 2cos 2 θ − 1

Then

1.6cosθ = 1 +

or

3.2cos3 θ − 1.6cosθ − 1 = 0

Solving numerically

θ = 23.515° or θ = 23.5°

(b) From Equation (2) for L = 200 mm and θ = 23.5° a=

5 ( 200 mm ) = 136.321 mm 8 cos 23.515° or a = 136.3 mm

(1) (2)

PROBLEM 4.96 Gears A and B are attached to a shaft supported by bearings at C and D. The diameters of gears A and B are 150 mm and 75 mm, respectively, and the tangential and radial forces acting on the gears are as shown. Knowing that the system rotates at a constant rate, determine the reactions at C and D. Assume that the bearing at C does not exert any axial force, and neglect the weights of the gears and the shaft.

SOLUTION Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft ΣFx = 0: ∴ Dx = 0 ΣM D( z -axis ) = 0: − C y (175 mm ) + ( 482 N ) ( 75 mm ) + ( 2650 N ) ( 50 mm ) = 0 ∴ C y = 963.71 N or

C y = ( 964 N ) j ΣM D( y -axis ) = 0: Cz (175 mm ) + (1325 N ) ( 75 mm ) + ( 964 N ) ( 50 mm ) = 0 ∴ C z = −843.29 N

or

C z = ( 843 N ) k and C = ( 964 N ) j − ( 843 N ) k ΣM C ( z -axis ) = 0: − ( 482 N ) (100 mm ) + Dy (175 mm ) + ( 2650 N ) ( 225 mm ) = 0 ∴ D y = −3131.7 N

or

D y = − ( 3130 N ) j ΣM C ( y -axis ) = 0: − (1325 N ) (100 mm ) − Dz (175 mm ) + ( 964 N ) ( 225 mm ) = 0 ∴ Dz = 482.29 N

or

D z = ( 482 N ) k and D = − ( 3130 N ) j + ( 482 N ) k

PROBLEM 4.97 Solve Problem 4.96 assuming that for gear A the tangential and radial forces are acting at E, so that FA = (1325 N)j + (482 N)k. P4.96 Gears A and B are attached to a shaft supported by bearings at C and D. The diameters of gears A and B are 150 mm and 75 mm, respectively, and the tangential and radial forces acting on the gears are as shown. Knowing that the system rotates at a constant rate, determine the reactions at C and D. Assume that the bearing at C does not exert any axial force, and neglect the weights of the gears and the shaft.

SOLUTION Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft ΣFx = 0: ∴ Dx = 0 ΣM D( z -axis ) = 0: − C y (175 mm ) − (1325 N ) ( 75 mm ) + ( 2650 N ) ( 50 mm ) = 0 ∴ C y = 189.286 N C y = (189.3 N ) j

or

ΣM D( y -axis ) = 0: C z (175 mm ) + ( 482 N ) ( 75 mm ) + ( 964 N ) ( 50 mm ) = 0 ∴ C z = −482.00 N C z = − ( 482 N ) k

or

and C = (189.3 N ) j − ( 482 N ) k ΣM C ( z -axis ) = 0:

(1325 N ) (100 mm ) + Dy (175 mm ) + ( 2650 N ) ( 225 mm ) = 0

∴ Dy = − 4164.3 N D y = − ( 4160 N ) j

or ΣM C ( y -axis ) = 0:

− ( 482 N )(100 mm ) − Dz (175 mm )

+ ( 964 N )( 225 mm ) = 0 ∴ Dz = 964.00 N or

D z = ( 964 N ) k and D = − ( 4160 N ) j + ( 964 N ) k

PROBLEM 4.98 Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The sheave at A has a radius of 50 mm, and the sheave at C has a radius of 40 mm. Knowing that the system rotates with a constant rate, determine (a) the tension T, (b) the reactions at B and D. Assume that the bearing at D does not exert any axial thrust and neglect the weights of the sheaves and the axle.

SOLUTION

Assume moment reactions at the bearing supports are zero. From f.b.d. of shaft (a)

ΣM x-axis = 0:

( 240 N − 180 N )( 50 mm ) + ( 300 N − T )( 40 mm ) = 0 ∴ T = 375 N

(b)

ΣFx = 0: Bx = 0 ΣM D( z -axis ) = 0:

( 300 N + 375 N )(120 mm ) − By ( 240 mm ) = 0 ∴ By = 337.5 N

ΣM D( y -axis ) = 0:

( 240 N + 180 N )( 400 mm ) + Bz ( 240 mm ) = 0 ∴ Bz = −700 N or B = ( 338 N ) j − ( 700 N ) k

ΣM B( z -axis ) = 0: − ( 300 N + 375 N )(120 mm ) + D y ( 240 mm ) = 0 ∴ D y = 337.5 N ΣM B( y -axis ) = 0:

( 240 N + 180 N )(160 mm ) + Dz ( 240 mm ) = 0 ∴ Dz = −280 N or D = ( 338 N ) j − ( 280 N ) k

PROBLEM 4.99 For the portion of a machine shown, the 4-in.-diameter pulley A and wheel B are fixed to a shaft supported by bearings at C and D. The spring of constant 2 lb/in. is unstretched when θ = 0, and the bearing at C does not exert any axial force. Knowing that θ = 180° and that the machine is at rest and in equilibrium, determine (a) the tension T, (b) the reactions at C and D. Neglect the weights of the shaft, pulley, and wheel.

SOLUTION

First, determine the spring force, FE , at θ = 180°.

FE = ks x ks = 2 lb/in.

where

x = ( yE )final − ( yE )initial = (12 in. + 3.5 in.) − (12 in. − 3.5 in.) = 7.0 in. ∴ FE = ( 2 lb/in.)( 7.0 in.) = 14.0 lb (a) From f.b.d. of machine part ΣM x = 0:

( 34 lb )( 2 in.) − T ( 2 in.) = 0

∴ T = 34 lb (b)

or T = 34.0 lb

ΣM D ( z -axis ) = 0: − C y (10 in.) − FE ( 2 in. + 1 in.) = 0 − C y (10 in.) − 14.0 lb ( 3 in.) = 0 ∴ C y = −4.2 lb

or

C y = − ( 4.20 lb ) j

ΣM D( y -axis ) = 0: C z (10 in.) + 34 lb ( 4 in.) + 34 lb ( 4 in.) = 0 ∴ C z = −27.2 lb

or

C z = − ( 27.2 lb ) k and C = − ( 4.20 lb ) j − ( 27.2 lb ) k

PROBLEM 4.99 CONTINUED ΣFx = 0: Dx = 0 ΣM C ( z -axis ) = 0: Dy (10 in.) − FE (12 in. + 1 in.) = 0 Dy (10 in.) − 14.0 (13 in.) = 0

or ∴ Dy = 18.2 lb

or

D y = (18.20 lb ) j

ΣM C ( y -axis ) = 0: − 2 ( 34 lb ) ( 6 in.) − Dz (10 in.) = 0 ∴ Dz = −40.8 lb

or

D z = − ( 40.8 lb ) k and D = (18.20 lb ) j − ( 40.8 lb ) k

PROBLEM 4.100 Solve Problem 4.99 for θ = 90°. P4.99 For the portion of a machine shown, the 4-in.-diameter pulley A and wheel B are fixed to a shaft supported by bearings at C and D. The spring of constant 2 lb/in. is unstretched when θ = 0, and the bearing at C does not exert any axial force. Knowing that θ = 180° and that the machine is at rest and in equilibrium, determine (a) the tension T, (b) the reactions at C and D. Neglect the weights of the shaft, pulley, and wheel.

SOLUTION

First, determine the spring force, FE , at θ = 90°. FE = ks x ks = 2 lb/in.

where and

 x = Lfinal − Linitial =  

( 3.5)2 + (12 )2  − (12 − 3.5) = 12.5 − 8.5 = 4.0 in. 

∴ FE = ( 2 lb/in.)( 4.0 in.) = 8.0 lb Then

FE =

−12.0 3.5 (8.0 lb ) j + (8.0 lb ) k = − ( 7.68 lb ) j + ( 2.24 lb ) k 12.5 12.5

(a) From f.b.d. of machine part ΣM x = 0:

( 34 lb ) ( 2 in.) − T ( 2 in.) − ( 7.68 lb )( 3.5 in.) = 0

∴ T = 20.56 lb (b)

or T = 20.6 lb

ΣM D ( z -axis ) = 0: − C y (10 in.) − ( 7.68 lb ) ( 3.0 in.) = 0 ∴ C y = −2.304 lb

or

C y = − ( 2.30 lb ) j

ΣM D ( y -axis ) = 0: Cz (10 in.) + ( 34 lb ) ( 4.0 in.) + ( 20.56 lb ) ( 4.0 in.) − ( 2.24 lb ) ( 3 in.) = 0 ∴ C z = −21.152 lb

or

C z = − ( 21.2 lb ) k

and C = − ( 2.30 lb ) j − ( 21.2 lb ) k

PROBLEM 4.100 CONTINUED ΣFx = 0: Dx = 0 ΣM C ( z -axis ) = 0: Dy (10 in.) − ( 7.68 lb )(13 in.) = 0 ∴ D y = 9.984 lb

or

D y = ( 9.98 lb ) j

ΣM C ( y -axis ) = 0: − ( 34 lb )( 6 in.) − ( 20.56 lb )( 6 in.) − Dz (10 in.) − ( 2.24 lb )(13 in.) = 0 ∴ Dz = −35.648 lb

or

D z = − ( 35.6 lb ) k

and D = ( 9.98 lb ) j − ( 35.6 lb ) k

PROBLEM 4.101 A 1.2 × 2.4-m sheet of plywood having a mass of 17 kg has been temporarily placed among three pipe supports. The lower edge of the sheet rests on small collars A and B and its upper edge leans against pipe C. Neglecting friction at all surfaces, determine the reactions at A, B, and C.

SOLUTION

(

)

W = mg = (17 kg ) 9.81 m/s 2 = 166.77 N

First note

h=

(1.2 )2 − (1.125)2

= 0.41758 m

From f.b.d. of plywood sheet  (1.125 m )  ΣM z = 0: C ( h ) − W   =0 2  

C ( 0.41758 m ) − (166.77 N ) ( 0.5625 m ) = 0

∴ C = 224.65 N

C = − ( 225 N ) i

or

ΣM B( y -axis ) = 0: − ( 224.65 N ) ( 0.6 m ) + Ax (1.2 m ) = 0 ∴ Ax = 112.324 N

or

A x = (112.3 N ) i

ΣM B( x-axis ) = 0: (166.77 N ) ( 0.3 m ) − Ay (1.2 m ) = 0 ∴ Ay = 41.693 N

or

A y = ( 41.7 N ) j

ΣM A( y -axis ) = 0: ( 224.65 N ) ( 0.6 m ) − Bx (1.2 m ) = 0 ∴ Bx = 112.325 N

or

B x = (112.3 N ) i

PROBLEM 4.101 CONTINUED ΣM A( x-axis ) = 0: By (1.2 m ) − (166.77 N ) ( 0.9 m ) = 0 ∴ By = 125.078 N

or

B y = (125.1 N ) j ∴ A = (112.3 N ) i + ( 41.7 N ) j B = (112.3 N ) i + (125.1 N ) j C = − ( 225 N ) i

PROBLEM 4.102 The 200 × 200-mm square plate shown has a mass of 25 kg and is supported by three vertical wires. Determine the tension in each wire.

SOLUTION

(

)

W = mg = ( 25 kg ) 9.81 m/s 2 = 245.25 N

First note From f.b.d. of plate

ΣM x = 0: ( 245.25 N ) (100 mm ) − TA (100 mm ) − TC ( 200 mm ) = 0 ∴ TA + 2TC = 245.25 N

(1)

ΣM z = 0: TB (160 mm ) + TC (160 mm ) − ( 245.25 N ) (100 mm ) = 0 ∴ TB + TC = 153.281 N

(2)

ΣFy = 0: TA + TB + TC − 245.25 N = 0 ∴ TB + TC = 245.25 − TA

(3)

TA = 245.25 N − 153.281 N = 91.969 N

(4)

Equating Equations (2) and (3) yields

TA = 92.0 N

or Substituting the value of TA into Equation (1)

TC =

( 245.25 N − 91.969 N ) 2

= 76.641 N

(5)

TC = 76.6 N

or Substituting the value of TC into Equation (2)

TB = 153.281 N − 76.641 N = 76.639 N

or

TB = 76.6 N TA = 92.0 N TB = 76.6 N TC = 76.6 N

PROBLEM 4.103 The 200 × 200-mm square plate shown has a mass of 25 kg and is supported by three vertical wires. Determine the mass and location of the lightest block which should be placed on the plate if the tensions in the three cables are to be equal.

SOLUTION

First note

(

)

WG = m p1g = ( 25 kg ) 9.81 m/s 2 = 245.25 N

(

)

W1 = mg = m 9.81 m/s 2 = ( 9.81m ) N From f.b.d. of plate ΣFy = 0: 3T − WG − W1 = 0 (1) ΣM x = 0: WG (100 mm ) + W1 ( z ) − T (100 mm ) − T ( 200 mm ) = 0 or − 300T + 100WG + W1z = 0

(2)

ΣM z = 0: 2T (160 mm ) − WG (100 mm ) − W1 ( x ) = 0 or 320T − 100WG − W1x = 0 Eliminate T by forming 100 ×  Eq. (1) + Eq. ( 2 )  −100W1 + W1z = 0 ∴ z = 100 mm

0 ≤ z ≤ 200 mm, ∴ okay

Now, 3 ×  Eq. ( 3)  − 320 ×  Eq. (1) yields 3 ( 320T ) − 3 (100 )WG − 3W1x − 320 ( 3T ) + 320WG + 320W1 = 0

(3)

PROBLEM 4.103 CONTINUED or

20WG + ( 320 − 3x )W1 = 0

or

W1 20 = WG 3 x − 320 ) (

The smallest value of

W1 will result in the smallest value of W1 since WG is given. WG ∴ Use x = xmax = 200 mm W1 20 1 = = WG 3 ( 200 ) − 320 14

and then

∴ W1 = and

WG 245.25 N = = 17.5179 N ( minimum ) 14 14

m=

W1 17.5179 N = = 1.78571 kg g 9.81 m/s 2 or m = 1.786 kg at x = 200 mm, z = 100 mm

PROBLEM 4.104 A camera of mass 240 g is mounted on a small tripod of mass 200 g. Assuming that the mass of the camera is uniformly distributed and that the line of action of the weight of the tripod passes through D, determine (a) the vertical components of the reactions at A, B, and C when θ = 0, (b) the maximum value of θ if the tripod is not to tip over.

SOLUTION

First note

Wtp = mtp For θ = 0

( ) g = ( 0.20 kg ) ( 9.81 m/s ) = 1.9620 N

WC = mC g = ( 0.24 kg ) 9.81 m/s 2 = 2.3544 N 2

xC = − ( 60 mm − 24 mm ) = −36 mm zC = 0

(a) From f.b.d. of camera and tripod as projected onto plane ABCD ΣFy = 0: Ay + By + C y − WC − Wtp = 0 ∴ Ay + By + C y = 2.3544 N + 1.9620 N = 4.3164 N ΣM x = 0: C y ( 38 mm ) − By ( 38 mm ) = 0

∴ C y = By

(1) (2)

ΣM z = 0: By ( 35 mm ) + C y ( 35 mm ) + ( 2.3544 N ) ( 36 mm ) − Ay ( 45 mm ) = 0 ∴ 9 Ay − 7By − 7C y = 16.9517

(3)

Substitute C y with By from Equation (2) into Equations (1) and (3), and solve by elimination

(

7 Ay + 2 By = 4.3164

)

9 Ay − 14By = 16.9517 16 Ay

= 47.166

PROBLEM 4.104 CONTINUED ∴ Ay = 2.9479 N or A y = 2.95 N Substituting Ay = 2.9479 N into Equation (1) 2.9479 N + 2 By = 4.3164 ∴ By = 0.68425 N

C y = 0.68425 N or B y = C y = 0.684 N (b) By = 0 for impending tipping

From f.b.d. of camera and tripod as projected onto plane ABCD ΣFy = 0: Ay + C y − WC − Wtp = 0 ∴ Ay + C y = 4.3164 N

(1)

ΣM x = 0: C y ( 38 mm ) − ( 2.3544 N ) ( 36 mm ) sin θ  = 0

∴ C y = 2.2305sin θ

(2)

ΣM z = 0: C y ( 35 mm ) − Ay ( 45 mm ) + ( 2.3544 N ) ( 36 mm ) cosθ  = 0

∴ 9 Ay − 7C y = (16.9517 N ) cosθ

(3)

Forming 7 ×  Eq. (1)  +  Eq. ( 3)  yields 16 Ay = 30.215 N + (16.9517 N ) cosθ

(4)

PROBLEM 4.104 CONTINUED Substituting Equation (2) into Equation (3) 9 Ay − (15.6134 N ) sin θ = (16.9517 N ) cosθ

(5)

Forming 9 ×  Eq. ( 4 )  − 16 ×  Eq. ( 5 )  yields

( 249.81 N ) sin θ or

= 271.93 N − (118.662 N ) cosθ

cos 2 θ =  2.2916 N − ( 2.1053 N ) sin θ 

2

cos 2 θ = 1 − sin 2 θ

Now

∴ 5.4323sin 2 θ − 9.6490sin θ + 4.2514 = 0 Using quadratic formula to solve, sin θ = 0.80981 and sin θ = 0.96641 ∴ θ = 54.078° and θ = 75.108° or θ max = 54.1° before tipping

PROBLEM 4.105 Two steel pipes AB and BC, each having a weight per unit length of 5 lb/ft, are welded together at B and are supported by three wires. Knowing that a = 1.25 ft, determine the tension in each wire.

SOLUTION

WAB = ( 5 lb/ft )( 2 ft ) = 10 lb

First note

WBC = ( 5 lb/ft )( 4 ft ) = 20 lb W = WAB + WBC = 30 lb To locate the equivalent force of the pipe assembly weight

rG/B × W = Σ ( ri × Wi ) = rG ( AB ) × WAB + rG ( BC ) × WBC

( xGi + zGk ) × ( −30 lb ) j = (1 ft ) k × ( −10 lb ) j + ( 2 ft ) i × ( −20 lb ) j

or

∴ From i-coefficient

k-coefficient

− ( 30 lb ) xGk + ( 30 lb ) zG i = (10 lb ⋅ ft ) i − ( 40 lb ⋅ ft ) k

zG =

10 lb ⋅ ft 1 = ft 30 lb 3

xG =

40 lb ⋅ ft 1 = 1 ft 30 lb 3

From f.b.d. of piping ΣM x = 0: W ( zG ) − TA ( 2 ft ) = 0 1  1  ∴ TA =  ft  30 lb  ft  = 5 lb 2  3 

or

TA = 5.00 lb

ΣFy = 0: 5 lb + TD + TC − 30 lb = 0 ∴ TD + TC = 25 lb

(1)

PROBLEM 4.105 CONTINUED 4  ΣM z = 0: TD (1.25 ft ) + TC ( 4 ft ) − 30 lb  ft  = 0 3  ∴ 1.25TD + 4TC = 40 lb ⋅ ft −4  Equation (1) 

−4TD − 4TC = −100

Equation (2) + Equation (3)

Results:

(3)

−2.75TD = −60 ∴ TD = 21.818 lb

From Equation (1)

(2)

or

TD = 21.8 lb

TC = 25 − 21.818 = 3.1818 lb

or

TC = 3.18 lb TA = 5.00 lb TC = 3.18 lb TD = 21.8 lb

PROBLEM 4.106 For the pile assembly of Problem 4.105, determine (a) the largest permissible value of a if the assembly is not to tip, (b) the corresponding tension in each wire.

P4.105 Two steel pipes AB and BC, each having a weight per unit length of 5 lb/ft, are welded together at B and are supported by three wires. Knowing that a = 1.25 ft, determine the tension in each wire.

SOLUTION

First note

WAB = ( 5 lb/ft )( 2 ft ) = 10 lb WBC = ( 5 lb/ft )( 4 ft ) = 20 lb

From f.b.d. of pipe assembly

ΣFy = 0: TA + TC + TD − 10 lb − 20 lb = 0 ∴ TA + TC + TD = 30 lb

(1)

ΣM x = 0: (10 lb )(1 ft ) − TA ( 2 ft ) = 0 or

TA = 5.00 lb

(2)

TC + TD = 25 lb

From Equations (1) and (2)

(3)

ΣM z = 0: TC ( 4 ft ) + TD ( amax ) − 20 lb ( 2 ft ) = 0 or

( 4 ft ) TC

+ TD amax = 40 lb ⋅ ft

(4)

PROBLEM 4.106 CONTINUED Using Equation (3) to eliminate TC 4 ( 25 − TD ) + TD amax = 40

amax = 4 −

or

60 TD

By observation, a is maximum when TD is maximum. From Equation (3), (TD )max occurs when TC = 0.

Therefore, (TD )max = 25 lb and

amax = 4 −

60 25

= 1.600 ft Results: (a)

amax = 1.600 ft

(b)

TA = 5.00 lb TC = 0 TD = 25.0 lb

PROBLEM 4.107 A uniform aluminum rod of weight W is bent into a circular ring of radius R and is supported by three wires as shown. Determine the tension in each wire.

SOLUTION From f.b.d. of ring

ΣFy = 0: TA + TB + TC − W = 0 ∴ TA + TB + TC = W

(1)

ΣM x = 0: TA ( R ) − TC ( R sin 30° ) = 0 ∴ TA = 0.5TC

(2)

ΣM z = 0: TC ( R cos 30° ) − TB ( R ) = 0 ∴ TB = 0.86603TC

(3)

Substituting TA and TB from Equations (2) and (3) into Equation (1) 0.5TC + 0.86603TC + TC = W

∴ TC = 0.42265W From Equation (2)

TA = 0.5 ( 0.42265W ) = 0.21132W From Equation (3)

TB = 0.86603 ( 0.42265W ) = 0.36603W or TA = 0.211W

TB = 0.366W TC = 0.423W

PROBLEM 4.108 A uniform aluminum rod of weight W is bent into a circular ring of radius R and is supported by three wires as shown. A small collar of weight W ′ is then placed on the ring and positioned so that the tensions in the three wires are equal. Determine (a) the position of the collar, (b) the value of W ′, (c) the tension in the wires.

SOLUTION

Let θ = angle from x-axis to small collar of weight W ′ From f.b.d. of ring

ΣFy = 0: 3T − W − W ′ = 0

(1)

ΣM x = 0: T ( R ) − T ( R sin 30° ) + W ′ ( R sin θ ) = 0 or

1 W ′ sin θ = − T 2

(2)

ΣM z = 0: T ( R cos 30° ) − W ′ ( R cosθ ) − T ( R ) = 0 or

 3 W ′ cosθ = − 1 − T  2  

(3)

Dividing Equation (2) by Equation (3)  1    3  tan θ =   1 −    2    2   ∴ θ = 75.000°

and

−1

= 3.7321

θ = 255.00°

Based on Equations (2) and (3), θ = 75.000° will give a negative value for W ′, which is not acceptable. (a)

∴ W ′ is located at θ = 255° from the x-axis or 15° from A towards B.

(b) From Equation (1) and Equation (2)

W ′ = 3 ( −2W ′ )( sin 255° ) − W ∴ W ′ = 0.20853W or W ′ = 0.209W (c) From Equation (1)

T = −2 ( 0.20853W ) sin 255° = 0.40285W or T = 0.403W

PROBLEM 4.109 An opening in a floor is covered by a 3 × 4-ft sheet of plywood weighing 12 lb. The sheet is hinged at A and B and is maintained in a position slightly above the floor by a small block C. Determine the vertical component of the reaction (a) at A, (b) at B, (c) at C.

SOLUTION

From f.b.d. of plywood sheet ΣM x = 0:

(12 lb )( 2 ft ) − C y ( 3.5 ft ) = 0

∴ C y = 6.8571 lb ΣM B( z -axis ) = 0:

or

C y = 6.86 lb

(12 lb )(1 ft ) + ( 6.8571 lb )( 0.5 ft ) − Ay ( 2 ft ) = 0

∴ Ay = 7.7143 lb

or

Ay = 7.71 lb

ΣM A( z -axis ) = 0: − (12 lb )(1 ft ) + By ( 2 ft ) + ( 6.8571 lb )( 2.5 ft ) = 0 ∴ By = 2.5714 lb

or

By = 2.57 lb

(a)

Ay = 7.71 lb

(b)

By = 2.57 lb

(c)

C y = 6.86 lb

PROBLEM 4.110 Solve Problem 4.109 assuming that the small block C is moved and placed under edge DE at a point 0.5 ft from corner E.

SOLUTION

rB/ A = ( 2 ft ) i

First,

rC/ A = ( 2 ft ) i + ( 4 ft ) k rG/ A = (1 ft ) i + ( 2 ft ) k

From f.b.d. of plywood sheet ΣM A = 0: rB/ A × ( B y j + Bzk ) + rC/ A × C y j + rG/ A × ( −Wj) = 0

( 2 ft ) i × By j + ( 2 ft ) i × Bzk + [( 2 ft ) i + ( 4 ft ) k ] × C y j + [(1 ft ) i + ( 2 ft ) k ] × ( −12 lb ) j = 0 2Byk − 2 Bz j + 2C yk − 4C y i − 12k + 24i = 0

i-coeff.

−4C y + 24 = 0

∴ C y = 6.00 lb

j-coeff.

−2 Bz = 0

∴ Bz = 0

k-coeff.

2By + 2C y − 12 = 0

or

2By + 2 ( 6 ) − 12 = 0

∴ By = 0

PROBLEM 4.110 CONTINUED ΣF = 0: Ay j + Azk + By j + Bzk + C y j − Wj = 0

Ay j + Az k + 0 j + 0k + 6 j − 12 j = 0 j-coeff.

Ay + 6 − 12 = 0

k-coeff.

Az = 0

∴ Ay = 6.00 lb

Az = 0 ∴ a) Ay = 6.00 lb

b) By = 0 c) C y = 6.00 lb

PROBLEM 4.111 The 10-kg square plate shown is supported by three vertical wires. Determine (a) the tension in each wire when a = 100 mm, (b) the value of a for which tensions in the three wires are equal.

SOLUTION First note (a)

(

)

W = mg = (10 kg ) 9.81 m/s 2 = 98.1 N

(a) From f.b.d. of plate ΣFy = 0: TA + TB + TC − W = 0

∴ TA + TB + TC = 98.1 N

(1)

ΣM x = 0: W (150 mm ) − TB( 300 mm ) − TC (100 mm ) = 0 ∴ 6TB + 2TC = 294.3

(2)

ΣM z = 0: TB(100 mm ) + TC ( 300 mm ) − ( 98.1 N )(150 mm ) = 0 ∴

− 6TB − 18TC = −882.9

(3)

Equation (2) + Equation (3)

−16TC = −588.6 ∴ TC = 36.788 N TC = 36.8 N W

or Substitution into Equation (2) 6TB + 2 ( 36.788 N ) = 294.3 N

∴ TB = 36.788 N

or

TB = 36.8 N W

From Equation (1) TA + 36.788 + 36.788 = 98.1 N

∴ TA = 24.525 N

or

TA = 24.5 N W

PROBLEM 4.111 CONTINUED (b)

(b) From f.b.d. of plate

ΣFy = 0: 3T − W = 0 ∴ T =

1 W 3

(1)

ΣM x = 0: W (150 mm ) − T ( a ) − T ( 300 mm ) = 0 ∴ T =

150W a + 300

(2)

Equating Equation (1) to Equation (2) 1 150W W = 3 a + 300 or

a + 300 = 3 (150 ) or a = 150.0 mm W

PROBLEM 4.112 The 3-m flagpole AC forms an angle of 30o with the z axis. It is held by a ball-and-socket joint at C and by two thin braces BD and BE. Knowing that the distance BC is 0.9 m, determine the tension in each brace and the reaction at C.

SOLUTION

TBE can be found from ΣM about line CE From f.b.d. of flagpole

(

)

ΣM CE = 0: λ CE ⋅ rB/C × TBD + λ CE ⋅ ( rA/C × FA ) = 0 where

λ CE =

( 0.9 m ) i + ( 0.9 m ) j ( 0.9 )2 + ( 0.9 )2 m

=

1 ( i + j) 2

rB/C = ( 0.9 m ) sin 30° j + ( 0.9 m ) cos 30°  k

= ( 0.45 m ) j + ( 0.77942 m ) k  − ( 0.9 m ) i + 0.9 m − ( 0.9 m ) sin 30°  j − ( 0.9 m ) cos 30° k      T TBD = λ BDTBD =   BD 2 2 2 ( 0.9 ) + ( 0.45 ) + ( 0.77942 ) m   T =  − ( 0.9 m ) i + ( 0.45 m ) j − ( 0.77942 m ) k  BD 1.62

= ( −0.70711i + 0.35355 j − 0.61237k ) TBD

rA/C = ( 3 m ) sin 30° j + ( 3 m ) cos30°k = (1.5 m ) j + ( 2.5981 m ) k FA = − ( 300 N ) j ∴

1 1 0 1 1 0  TBD   1  0 0.45 0.77942   + 0 1.5 2.5981  =0  2  2 −0.70711 0.35355 −0.61237 0 −300 0

PROBLEM 4.112 CONTINUED −1.10227TBD + 779.43 = 0

or

∴ TBD = 707.12 N

or TBD = 707 N W

TBE = TBD = 707.12 N

Based on symmetry with yz-plane,

or TBE = 707 N W

The reaction forces at C are found from ΣF = 0 ΣFx = 0: − (TBD ) x + (TBE ) x + C x = 0 ΣFy = 0:

or

Cx = 0

(TBD ) y + (TBE ) y + C y − 300 N = 0 C y = 300 N − 2 ( 0.35355)( 707.12 N ) ∴ C y = −200.00 N

ΣFz = 0: Cz − (TBD ) z − (TBE ) z = 0

Cz = 2 ( 0.61237 )( 707.12 N ) ∴ C z = 866.04 N or C = − ( 200 N ) j + ( 866 N ) k W

PROBLEM 4.113 A 3-m boom is acted upon by the 4-kN force shown. Determine the tension in each cable and the reaction at the ball-and-socket joint at A.

SOLUTION

From f.b.d. of boom

(

)

(

)

ΣM AE = 0: λ AE ⋅ rB/ A × TBD + λ AE ⋅ rC/ A × FC = 0 where

λ AE =

( 2.1 m ) j − (1.8 m ) k ( 2.1)2 + (1.8)2 m

= 0.27451j − 0.23529k

rB/ A = (1.8 m ) i TBD = λ BDTBD =

( −1.8 m ) i + ( 2.1 m ) j + (1.8 m ) k T BD (1.8)2 + ( 2.1)2 + (1.8)2 m

= ( −0.54545i + 0.63636 j + 0.54545k ) TBD

rC/ A = ( 3.0 m ) i FC = − ( 4 kN ) j

PROBLEM 4.113 CONTINUED ∴

0 0.27451 −0.23529 0 0.27451 −0.23529 1.8 0 0 0 0 =0 TBD + 3 −0.54545 0.63636 0.54545 0 −4 0

( −0.149731 − 0.149729 )1.8TBD + 2.82348 = 0 ∴ TBD = 5.2381 kN Based on symmetry,

or TBD = 5.24 kN W

TBE = TBD = 5.2381 kN ΣFz = 0: Az + (TBD ) z − (TBE ) z = 0

or TBE = 5.24 kN W

Az = 0

ΣFy = 0: Ay + (TBD ) y + (TBD ) y − 4 kN = 0

Ay + 2 ( 0.63636 )( 5.2381 kN ) − 4 kN = 0 ∴ Ay = −2.6666 kN ΣFx = 0: Ax − (TBD ) x − (TBE ) x = 0

Ax − 2 ( 0.54545 )( 5.2381 kN ) = 0 ∴ Ax = 5.7142 kN and A = ( 5.71 N ) i − ( 2.67 N ) j W

PROBLEM 4.114 An 8-ft-long boom is held by a ball-and-socket joint at C and by two cables AD and BE. Determine the tension in each cable and the reaction at C.

SOLUTION

From f.b.d. of boom

(

)

(

)

ΣM CE = 0: λ CE ⋅ rA/C × TAD + λ CE ⋅ rA/C × FA = 0 where

λ CE =

( 2 ft ) j − ( 3 ft ) k ( 2 )2 + ( 3)2 ft

=

1 ( 2 j − 3k ) 13

rA/C = ( 8 ft ) i TAD = λ ADTAD =

− ( 8 ft ) i + (1 ft ) j + ( 4 ft ) k

(8) + (1) + ( 4 ) ft 2

2

2

TAD

1 =   TAD ( −8i + j + 4k ) 9 FA = − (198 lb ) j ∴

0 2 −3 0 2 −3  TAD   198  8 0 0  + 8 0 0  =0  9 13   13  0 −1 0 −8 1 4

PROBLEM 4.114 CONTINUED

( −64 − 24 )

TAD 198 + ( 24 ) =0 9 13 13

∴ TAD = 486.00 lb or TAD = 486 lb W

(

)

(

ΣM CD = 0: λ CD ⋅ rB/C × TBE + λ CD ⋅ rA/C × FA where

λ CD =

(1 ft ) j + ( 4 ft ) k 17 ft

=

)

1 (1j + 4k ) 17

rB/C = ( 6 ft ) i TBE = λ BETBE =

− ( 6 ft ) i + ( 2 ft ) j − ( 3 ft ) k

(6) ∴

2

+ ( 2 ) + ( 3) ft 2

2

1 TBE =   TBE ( −6i + 2 j − 3k ) 7

0 1 4 0 1 4 198 TBE 6 0 0 + 8 0 0 =0 7 17 17 0 −1 0 −6 2 −3

(18 + 48)

TBE + ( −32 )198 = 0 7

∴ TBE = 672.00 lb or TBE = 672 lb W ΣFx = 0: C x − (TAD ) x − (TBE ) x = 0 8 6 Cx −   486 −   672 = 0 9   7 ∴ C x = 1008 lb ΣFy = 0: C y + (TAD ) y + (TBE ) y − 198 lb = 0 1 2 C y +   486 +   672 − 198 lb = 0 9 7 ∴ C y = −48.0 lb ΣFz = 0: Cz + (TAD ) z − (TBE ) z = 0 4 3 Cz +   486 −   ( 672 ) = 0 9   7 ∴ Cz = 72.0 lb or C = (1008 lb ) i − ( 48.0 lb ) j + ( 72.0 lb ) k W

PROBLEM 4.115 Solve Problem 4.114 assuming that the given 198-lb load is replaced with two 99-lb loads applied at A and B. P4.114 An 8-ft-long boom is held by a ball-and-socket joint at C and by two cables AD and BE. Determine the tension in each cable and the reaction at C.

SOLUTION

From f.b.d. of boom

(

)

(

)

(

)

ΣM CE = 0: λ CE ⋅ rA/C × TAD + λ CE ⋅ rA/C × FA + λ CE ⋅ rB/C × FB = 0 λ CE =

where

( 2 ft ) j − ( 3 ft ) k ( 2 )2 + ( 3)2 ft

=

1 ( 2 j − 3k ) 13

rA/C = ( 8 ft ) i rB/C = ( 6 ft ) i TAD = λ ADTAD =

− ( 8 ft ) i + (1 ft ) j + ( 4 ft ) k

(8) + (1) + ( 4 ) ft 2

2

2

TAD

1 =   TAD ( −8i + j + 4k ) 9 FA = − ( 99 lb ) j FB = − ( 99 lb ) j ∴

0 2 −3 0 2 −3 0 2 −3 99 99 TAD 8 0 0 + 8 0 0 + 6 0 0 =0 9 13 13 13 0 −1 0 0 −1 0 −8 1 4

PROBLEM 4.115 CONTINUED

( −64 − 24 )

TAD 99 + ( 24 + 18 ) =0 9 13 13

TAD = 425.25 lb

or

or TAD = 425 lb W

(

)

(

)

(

)

ΣM CD = 0: λ CD ⋅ rB/C × TBE + λ CD ⋅ rA/C × FA + λ CD ⋅ rB/C × FB = 0 where

λ CD =

(1 ft ) j + ( 4 ft ) k 17

=

1 ( j + 4k ) 17

rB/C = ( 6 ft ) i rA/C = ( 8 ft ) j TBE = λ BETBE =

∴

− ( 6 ft ) i + ( 2 ft ) j − ( 3 ft ) k

(6)

2

+ ( 2 ) + ( 3) ft 2

2

TBE =

TBE ( −6i + 2 j − 3k ) 7

0 1 4 0 1 4 0 1 4  TBE   99   99  6 0 0  + 8 0 0 + 6 0 0 =0  7 17   17   17  0 −1 0 0 −1 0 −6 2 −3

(18 + 48) 

TBE   99   + ( −32 − 24 )  =0  7 17   17 

or

TBE = 588.00 lb or TBE = 588 lb W ΣFx = 0: C x − (TAD ) x − (TBE ) x = 0 8 6 Cx −   425.25 −   588.00 = 0 9 7 ∴ C x = 882 lb ΣFy = 0: C y + (TAD ) y + (TBE ) y − 99 − 99 = 0 1 2 C y +   425.25 +   588.00 − 198 = 0 9   7 ∴ C y = −17.25 lb ΣFz = 0: Cz + (TAD ) z − (TBE ) z = 0 4 3 Cz +   425.25 −   588.00 = 0 9 7 ∴ Cz = 63.0 lb or C = ( 882 lb ) i − (17.25 lb ) j + ( 63.0 lb ) k W

PROBLEM 4.116 The 18-ft pole ABC is acted upon by a 210-lb force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a = 9 ft, determine the tension in each cable and the reaction at A.

SOLUTION

From f.b.d. of pole ABC

(

)

(

)

ΣM AE = 0: λ AE ⋅ rB/ A × TBD + λ AE ⋅ rC/ A × FC = 0

where

λ AE =

( 4.5 ft ) i + ( 9 ft ) k ( 4.5 )2 + ( 9 )2 ft

=

1 ( 4.5i + 9k ) 101.25

rB/ A = ( 9 ft ) j rC/ A = (18 ft ) j TBD = λ BDTBD =

( 4.5 ft ) i − ( 9 ft ) j − ( 9 ft ) k T BD ( 4.5)2 + ( 9 )2 + ( 9 )2 ft

T  =  BD  ( 4.5i − 9 j − 9k )  13.5  FC = λ CF ( 210 lb ) =

∴

−9i − 18 j + 6k

( 9 )2 + (18)2 + ( 6 )2

( 210 lb ) = 10 lb ( −9i − 18j + 6k )

4.5 0 9 4.5 0 9 TBD    10 lb  0 9 0   + 0 18 0  =0  13.5 101.25   101.25  −9 −18 6 4.5 −9 −9

PROBLEM 4.116 CONTINUED

( −364.5 − 364.5) T

BD

13.5 101.25

+

( 486 + 1458) (10 lb ) = 0 101.25

TBD = 360.00 lb

and

or TBD = 360 lb

(

)

(

)

ΣM AD = 0: λ AD ⋅ rB/ A × TBE + λ AD ⋅ rC/ A × FC = 0 where

λ AD =

( 4.5 ft ) i − ( 9 ft ) k ( 4.5 )2 + ( 9 )2 ft

1 ( 4.5i − 9k ) 101.25

=

rB/ A = ( 9 ft ) j rC/ A = (18 ft ) j TBE = λ BETBE =

∴

( 4.5 ft ) i − ( 9 ft ) j + ( 9 ft ) k T 2

2

( 4.5 ) + ( 9 ) + ( 9 ) ft

=

TBE ( 4.5i − 9 j + 9k ) 13.5

4.5 0 −9 4.5 0 −9 TBE    10 lb  0 9 0   + 0 18 0  =0  13.5 101.25   101.25  −9 −18 6 4.5 −9 9

( 364.5 + 364.5) T 13.5 101.25

or

BE

2

BE

+

( 486 − 1458)10 lb 101.25

=0

TBE = 180.0 lb or TBE = 180.0 lb ΣFx = 0: Ax + (TBD ) x + (TBE ) x − ( FC ) x = 0  4.5   4.5   9  Ax +   360 +  180 −   210 = 0  13.5   13.5   21  ∴ Ax = −90.0 lb ΣFy = 0: Ay − (TBD ) y − (TBE ) y − ( FC ) y = 0  9   9   18  Ay −   360 −  180 −   210 = 0  13.5   13.5   21  ∴ Ay = 540 lb ΣFz = 0: Az − (TBD ) z + (TBE ) z + ( FC ) z = 0  9   9   6  Az −   360 +  180 +   210 = 0 13.5 13.5      21  ∴ Az = 60.0 lb or A = − ( 90.0 lb ) i + ( 540 lb ) j + ( 60.0 lb ) k

PROBLEM 4.117 Solve Problem 4.116 for a = 4.5 ft. P4.116 The 18-ft pole ABC is acted upon by a 210-lb force as shown. The pole is held by a ball-and-socket joint at A and by two cables BD and BE. For a = 9 ft, determine the tension in each cable and the reaction at A.

SOLUTION

From f.b.d. of pole ABC

(

)

(

)

ΣM AE = 0 : λ AE ⋅ rB/ A × TBD + λ AE ⋅ rC/ A × FC = 0 λ AE =

where

( 4.5 ft ) i + ( 9 ft ) k ( 4.5 )2 + ( 9 )2 ft

=

1 ( 4.5i + 9k ) 101.25

rB/ A = ( 9 ft ) j rC/ A = (18 ft ) j TBD = λ BDTBD =

( 4.5 ft ) i − ( 9 ft ) j − ( 9 ft ) k T BD ( 4.5)2 + ( 9 )2 + ( 9 )2 ft

T  =  BD  ( 4.5i − 9 j − 9k )  13.5  FC = λ CF ( 210 lb ) =

−4.5i − 18 j + 6k

( 4.5)2 + (18)2 + ( 6 )2

( 210 lb )

 210 lb  =  ( −4.5i − 18 j + 6k )  19.5  ∴

4.5 0 9 4.5 0 9 TBD    210 lb  0 9 0  18 0  + 0 =0  13.5 101.25   19.5 101.25  −4.5 −18 6 4.5 −9 −9

PROBLEM 4.117 CONTINUED

( −364.5 − 364.5) T

+

BD

13.5 101.25

( 486 + 729 ) 19.5 101.25

( 210 lb ) = 0

TBD = 242.31 lb

or

(

or TBD = 242 lb

)

(

)

ΣM AD = 0: λ AD ⋅ rB/ A × TBE + λ AD ⋅ rC/ A × FC = 0 where

λ AD =

( 4.5 ft ) i − ( 9 ft ) k ( 4.5 )2 + ( 9 )2 ft

1 ( 4.5i − 9k ) , 101.25

=

rB/ A = ( 9 ft ) j rC/ A = (18 ft ) j TBE = λ BETBE =

∴

( 4.5 ft ) i − ( 9 ft ) j + ( 9 ft ) k T 2

2

( 4.5 ) + ( 9 ) + ( 9 ) ft

=

TBE ( 4.5i − 9 j + 9k ) 13.5

4.5 0 −9 4.5 0 −9 TBE    210 lb  0 9 0  18 0  + 0 =0  13.5 101.25   19.5 101.25  4.5 −9 9 −4.5 −18 6

( 364.5 + 364.5 ) T 13.5 101.25

or

BE

2

BE

+

( 486 − 729 )( 210 lb ) 19.5 101.25

=0

TBE = 48.462 lb

or TBE = 48.5 lb ΣFx = 0: Ax + (TBD ) x + (TBE ) x − ( FC ) x = 0  4.5   4.5   4.5  Ax +   242.31 +   48.462 −   210 = 0  13.5   13.5   19.5  ∴ Ax = −48.459 lb ΣFy = 0: Ay − (TBD ) y − (TBE ) y − ( FC ) y = 0  9   9   18  Ay −   242.31 −   48.462 −   210 =  13.5   13.5   19.5  ∴ Ay = 387.69 lb ΣFz = 0: Az − (TBD ) z + (TBE ) z + ( FC ) z = 0  9   9   6  Az −   242.31 +   48.462 +  2  13.5   13.5   19.5  ∴ Az = 64.591 lb or A = − ( 48.5 lb ) i + ( 388 lb ) j + ( 64.6 lb ) k

PROBLEM 4.118 Two steel pipes ABCD and EBF are welded together at B to form the boom shown. The boom is held by a ball-and-socket joint at D and by two cables EG and ICFH; cable ICFH passes around frictionless pulleys at C and F. For the loading shown, determine the tension in each cable and the reaction at D.

SOLUTION

From f.b.d. of boom

(

)

(

)

ΣM z = 0: k ⋅ rC/D × TCI + k ⋅ rA/D × FA = 0 where

rC/D = (1.8 m ) i TCI = λ CI TCI =

− (1.8 m ) i + (1.12 m ) j

(1.8) + (1.12 ) m 2

2

TCI

 T  =  CI  ( −1.8i + 1.12 j)  2.12  rA/D = ( 3.5 m ) i FA = − ( 560 N ) j

0 0 1 0 0 1  TCI  ∴ ΣM z = 1.8 0 0  + 3.5 0 0 ( 560 N ) = 0  2.12  −1.8 1.12 0 0 −1 0

( 2.016 ) or

TCI + ( −3.5 ) 560 = 0 2.12

TCI = TFH = 2061.1 N TICFH = 2.06 kN

PROBLEM 4.118 CONTINUED

(

)

(

)

ΣM y = 0: j ⋅ rG/D × TEG + j ⋅ rH /D × TFH = 0 where

rG/D = ( 3.4 m ) k rH /D = − ( 2.5 m ) k TEG =

− ( 3.0 m ) i + ( 3.15 m ) k

( 3)2 + ( 3.15)2

TFH = λ FH TFH =

∴

m

− ( 3.0 m ) i − ( 2.25 m ) k

( 3)

2

+ ( 2.25 ) m 2

( 2061.1 N ) =

2061.1 N ( −3i − 2.25k ) 3.75

0 1 0 0 1 0  TEG   2061.1 N  0 0 3.4   + 0 0 −2.5  =0  4.35   3.75  −3 0 3.15 −3 0 −2.25 − (10.2 )

or

T  TEG =  EG  ( −3i + 3.15k )  4.35 

TEG 2061.1 N + ( 7.5 ) =0 4.35 3.75 TEG = 1758.00 N TEG = 1.758 kN

ΣFx = 0: Dx − (TCI ) x − (TFH ) x − (TEG ) x = 0  1.8  (  3.0  (  3 ( Dx −   2061.1 N ) −   2061.1 N ) −   1758 N ) = 0  2.12   3.75   4.35  ∴ Dx = 4611.3 N ΣFy = 0: Dy + (TCI ) y − 560 N = 0  1.12  ( Dy +   2061.1 N ) − 560 N = 0  2.12  ∴ D y = −528.88 N ΣFz = 0: Dz + (TEG ) z − (TFH ) z = 0  3.15  (  2.25  ( Dz +   1758 N ) −   2061.1 N ) = 0  4.35   3.75  ∴ Dz = −36.374 N and D = ( 4610 N ) i − ( 529 N ) j − ( 36.4 N ) k

PROBLEM 4.119 Solve Problem 4.118 assuming that the 560-N load is applied at B. P4.118 Two steel pipes ABCD and EBF are welded together at B to form the boom shown. The boom is held by a ball-and-socket joint at D and by two cables EG and ICFH; cable ICFH passes around frictionless pulleys at C and F. For the loading shown, determine the tension in each cable and the reaction at D.

SOLUTION

From f.b.d. of boom

(

)

(

)

ΣM z = 0: k ⋅ rC/D × TCI + k ⋅ rB/D × FB = 0 rC/D = (1.8 m ) i

where

TCI = λ CI TCI =

− (1.8 m ) i + (1.12 m ) j

(1.8)2 + (1.12 )2

m

TCI

 T  =  CI  ( −1.8i + 1.12 j)  2.12  rB/D = ( 3.0 m ) i FB = − ( 560 N ) j

∴

0 0 1 0 0 1  TCI  1.8 0 0  + 3 0 0 ( 560 N ) = 0  2.12  −1.8 1.12 0 0 −1 0

( 2.016 ) or

TCI + ( −3) 560 = 0 2.12

TCI = TFH = 1766.67 N TICFH = 1.767 kN

PROBLEM 4.119 CONTINUED

(

)

(

)

ΣM y = 0: j ⋅ rG/D × TEG + j ⋅ rH /D × TFH = 0 where

rG/D = ( 3.4 m ) k rH /D = − ( 2.5 m ) k TEG = λ EGTEG =

TFH = λ FH TFH =

∴

− ( 3.0 m ) i + ( 3.15 m ) k

( 3)2 + ( 3.15)2

m

− ( 3.0 m ) i − ( 2.25 m ) k

( 3)

2

+ ( 2.25 ) m 2

TEG =

TEG ( −3i + 3.15k ) 4.35

TFH =

1766.67 N ( −3i − 2.25k ) 3.75

0 1 0 0 1 0  TEG   1766.67  0 0 3.4   + 0 0 −2.5  =0  4.35   3.75  −3 0 3.15 −3 0 −2.25 − (10.2 )

TEG 1766.67 + ( 7.5 ) =0 4.35 3.75 TEG = 1506.86 N

or

TEG = 1.507 kN ΣFx = 0: Dx − (TCI ) x − (TFH ) x − (TEG ) x = 0  1.8  (  3 (  3 ( Dx −   1766.67 N ) −   1766.67 N ) −   1506.86 N ) = 0  2.12   3.75   4.35  ∴ Dx = 3952.5 N ΣFy = 0: Dy + (TCI ) y − 560 N = 0  1.12  ( Dy +   1766.67 N ) − 560 N = 0  2.12  ∴ D y = −373.34 N ΣFz = 0: Dz + (TEG ) z − (TFH ) z = 0  3.15  (  2.25  ( Dz +   1506.86 N ) −   1766.67 N ) = 0  4.35   3.75  ∴ Dz = −31.172 N D = ( 3950 N ) i − ( 373 N ) j − ( 31.2 N ) k

PROBLEM 4.120 The lever AB is welded to the bent rod BCD which is supported by bearings at E and F and by cable DG. Knowing that the bearing at E does not exert any axial thrust, determine (a) the tension in cable DG, (b) the reactions at E and F.

SOLUTION

(a) From f.b.d. of assembly  − ( 0.12 m ) j − ( 0.225 m ) k  TDG TDG = λ DGTDG =   = 0.255  − ( 0.12 ) j − ( 0.225 ) k  2 2  ( 0.12 ) + ( 0.225 ) m    0.225   ΣM y = 0: − ( 220 N )( 0.24 m ) + TDG    ( 0.16 m ) = 0  0.255   

∴ TDG = 374.00 N or TDG = 374 N (b) From f.b.d. of assembly ΣM F ( z -axis ) = 0:



0.120     ( 0.16 m ) = 0  0.255  

( 220 N )( 0.19 m ) − Ex ( 0.13 m ) − 374 N  

∴ Ex = 104.923 N ΣFx = 0: Fx + 104.923 N − 220 N = 0 ∴ Fx = 115.077 N   0.225   ΣM F ( x-axis ) = 0: Ez ( 0.13 m ) + 374 N    ( 0.06 m ) = 0  0.255    ∴ Ez = −152.308 N

PROBLEM 4.120 CONTINUED  0.225  ΣFz = 0: Fz − 152.308 N − ( 374 N )  =0  0.255  ∴ Fz = 482.31 N  0.12  ΣFy = 0: Fy − ( 374 N )  =0  0.255  ∴ Fy = 176.0 N E = (104.9 N ) i − (152.3 N ) k F = (115.1 N ) i + (176.0 N ) j + ( 482 N ) k

PROBLEM 4.121 A 30-kg cover for a roof opening is hinged at corners A and B. The roof forms an angle of 30o with the horizontal, and the cover is maintained in a horizontal position by the brace CE. Determine (a) the magnitude of the force exerted by the brace, (b) the reactions at the hinges. Assume that the hinge at A does not exert any axial thrust.

SOLUTION

(

)

W = mg = ( 30 kg ) 9.81 m/s2 = 294.3 N

First note

FEC = λ EC FEC = ( sin15° ) i + ( cos15° ) j FEC

From f.b.d. of cover ΣM z = 0:

(a) or

( FEC cos15° ) (1.0 m ) − W ( 0.5 m ) = 0 FEC cos15° (1.0 m ) − ( 294.3 N )( 0.5 m ) = 0

∴ FEC = 152.341 N

or FEC = 152.3 N

ΣM x = 0: W ( 0.4 m ) − Ay ( 0.8 m ) − ( FEC cos15° ) ( 0.8 m ) = 0

(b)

or

( 294.3 N )( 0.4 m ) − Ay ( 0.8 m ) − (152.341 N ) cos15° ( 0.8 m ) = 0 ∴ Ay = 0 ΣM y = 0: Ax ( 0.8 m ) + ( FEC sin15° ) ( 0.8 m ) = 0

or

Ax ( 0.8 m ) + (152.341 N ) sin15° ( 0.8 m ) = 0

∴ Ax = −39.429 N ΣFx = 0: Ax + Bx + FEC sin15° = 0 −39.429 N + Bx + (152.341 N ) sin15° = 0 ∴ Bx = 0

PROBLEM 4.121 CONTINUED ΣFy = 0: FEC cos15° − W + By = 0 or

(152.341 N ) cos15° − 294.3 N + By

=0

∴ By = 147.180 N or A = − ( 39.4 N ) i B = (147.2 N ) j

PROBLEM 4.122 The rectangular plate shown has a mass of 15 kg and is held in the position shown by hinges A and B and cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION

(

)

W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N

First note

 ( 0.08 m ) i + ( 0.25 m ) j − ( 0.2 m ) k  TEF TEF = λ EFTEF =   TEF = 0.33 ( 0.08i + 0.25j − 0.2k ) 2 2 2 ( 0.08 ) + ( 0.25) + ( 0.2 ) m   From f.b.d. of rectangular plate ΣM x = 0: or or

(147.15 N )( 0.1 m ) − (TEF ) y ( 0.2 m ) = 0  0.25   14.715 N ⋅ m −   TEF  ( 0.2 m ) = 0  0.33  

TEF = 97.119 N or TEF = 97.1 N ΣFx = 0: Ax + (TEF ) x = 0

 0.08  Ax +   ( 97.119 N ) = 0  0.33  ∴ Ax = −23.544 N

PROBLEM 4.122 CONTINUED ΣM B( z -axis ) = 0: − Ay ( 0.3 m ) − (TEF ) y ( 0.04 m ) + W ( 0.15 m ) = 0 or

 0.25   − Ay ( 0.3 m ) −   97.119 N  ( 0.04 m ) + 147.15 N ( 0.15 m ) = 0  0.33   ∴ Ay = 63.765 N ΣM B( y -axis ) = 0: Az ( 0.3 m ) + (TEF ) x ( 0.2 m ) + (TEF ) z ( 0.04 m ) = 0  0.08    0.2   Az ( 0.3 m ) +   TEF  ( 0.2 m ) −   TEF  ( 0.04 m ) = 0  0.33    0.33   ∴ Az = −7.848 N and A = − ( 23.5 N ) i + ( 63.8 N ) j − ( 7.85 N ) k ΣFy = 0: Ay − W + (TEF ) y + By = 0  0.25  63.765 N − 147.15 N +   ( 97.119 N ) + By = 0  0.33  ∴ By = 9.81 N ΣFz = 0: Az − (TEF ) z + Bz = 0  0.2  −7.848 N −   ( 97.119 N ) + Bz = 0  0.33  ∴ Bz = 66.708 N and B = ( 9.81 N ) j + ( 66.7 N ) k

PROBLEM 4.123 Solve Problem 4.122 assuming that cable EF is replaced by a cable attached at points E and H. P4.122 The rectangular plate shown has a mass of 15 kg and is held in the position shown by hinges A and B and cable EF. Assuming that the hinge at B does not exert any axial thrust, determine (a) the tension in the cable, (b) the reactions at A and B.

SOLUTION

(

)

W = mg = (15 kg ) 9.81 m/s 2 = 147.15 N

First note

TEH = λ EH TEH

  − ( 0.3 m ) i + ( 0.12 m ) j − ( 0.2 m ) k  T  = T = EH  − ( 0.3) i + ( 0.12 ) j − ( 0.2 ) k    EH 2 2 2 0.38 ( 0.3) + ( 0.12 ) + ( 0.2 ) m  

From f.b.d. of rectangular plate ΣM x = 0: or or

(147.15 N )( 0.1 m ) − (TEH ) y ( 0.2 m ) = 0  0.12    TEH  ( 0.2 m ) = 0  0.38  

(147.15 N )( 0.1 m ) − 

TEH = 232.99 N or TEH = 233 N ΣFx = 0: Ax + (TEH ) x = 0  0.3  Ax −   ( 232.99 N ) = 0  0.38  ∴ Ax = 183.938 N

PROBLEM 4.123 CONTINUED ΣM B( z -axis ) = 0: − Ay ( 0.3 m ) − (TEH ) y ( 0.04 m ) + W ( 0.15 m ) = 0 or

 0.12 − Ay ( 0.3 m ) −  ( 232.99 N ) ( 0.04 m ) + (147.15 N )( 0.15 m ) = 0  0.38  ∴ Ay = 63.765 N ΣM B( y -axis ) = 0: Az ( 0.3 m ) + (TEH ) x ( 0.2 m ) + (TEH ) z ( 0.04 m ) = 0

or

 0.3    0.2   Az ( 0.3 m ) −   ( 232.99 N )  ( 0.2 m ) −   ( 232.99 )  ( 0.04 m ) = 0  0.38    0.38   ∴ Az = 138.976 N and A = (183.9 N ) i + ( 63.8 N ) j + (139.0 N ) k ΣFy = 0: Ay + By − W + (TEH ) y = 0  0.12  63.765 N + By − 147.15 N +   ( 232.99 N ) = 0  0.38  ∴ By = 9.8092 N ΣFz = 0: Az + Bz − (TEH ) z = 0  0.2  138.976 N + Bz −   ( 232.99 N ) = 0  0.38  ∴ Bz = −16.3497 N and B = ( 9.81 N ) j − (16.35 N ) k

PROBLEM 4.124 A small door weighing 16 lb is attached by hinges A and B to a wall and is held in the horizontal position shown by rope EFH. The rope passes around a small, frictionless pulley at F and is tied to a fixed cleat at H. Assuming that the hinge at A does not exert any axial thrust, determine (a) the tension in the rope, (b) the reactions at A and B.

SOLUTION

T = λ EFT =

First note

=

(12 in.) i + ( 54 in.) j − ( 28 in.) k T (12 )2 + ( 54 )2 + ( 28)2 in. T T (12i + 54 j − 28k ) = ( 6i + 27 j − 14k ) 62 31

W = − (16 lb ) j at G From f.b.d. of door ABCD (a)

ΣM x = 0: Ty ( 28 in.) − W (14 in.) = 0   27   T  31   ( 28 in.) − (16 lb )(14 in.) = 0   

∴ T = 9.1852 lb or T = 9.19 lb (b)

ΣM B( z -axis ) = 0: − Ay ( 30 in.) + W (15 in.) − Ty ( 4 in.) = 0   27   − Ay ( 30 in.) + (16 lb )(15 in.) − ( 9.1852 lb )    ( 4 in.) = 0  31    ∴ Ay = 6.9333 lb

PROBLEM 4.124 CONTINUED ΣM B( y -axis ) = 0: Az ( 30 in.) + Tx ( 28 in.) − Tz ( 4 in.) = 0    6   14   Az ( 30 in.) + ( 9.1852 lb )    ( 28 in.) − ( 9.1852 lb )    ( 4 in.) = 0  31    31     ∴ Az = −1.10617 lb or A = ( 6.93 lb ) j − (1.106 lb ) k  6 ΣFx = 0: Bx + Tx = Bx + ( 9.1852 lb )   = 0  31  ∴ Bx = −1.77778 lb ΣFy = 0: By + Ty − W + Ay = 0  27  By + ( 9.1852 lb )   − 16 lb + 6.9333 lb = 0  31  ∴ By = 1.06666 lb ΣFz = 0: Az − Tz + Bz = 0  14  −1.10617 lb − ( 9.1852 lb )   + Bz = 0  31  ∴ Bz = 5.2543 lb or B = − (1.778 lb ) i + (1.067 lb ) j + ( 5.25 lb ) k

PROBLEM 4.125 Solve Problem 4.124 assuming that the rope is attached to the door at I. P4.124 A small door weighing 16 lb is attached by hinges A and B to a wall and is held in the horizontal position shown by rope EFH. The rope passes around a small, frictionless pulley at F and is tied to a fixed cleat at H. Assuming that the hinge at A does not exert any axial thrust, determine (a) the tension in the rope, (b) the reactions at A and B.

SOLUTION

T = λ IF T =

First note

=

( 3 in.) i + ( 54 in.) j − (10 in.) k T ( 3)2 + ( 54 )2 + (10 )2 in. T ( 3i + 54 j − 10k ) 55

W = − (16 lb ) j

From f.b.d. of door ABCD (a)

ΣM x = 0: W (14 in.) − Ty (10 in.) = 0

(16 lb )(14 in.) − 

54   T (10 in.) = 0  55 

∴ T = 22.815 lb or T = 22.8 lb (b)

ΣM B( z -axis ) = 0: − Ay ( 30 in.) + W (15 in.) + Ty ( 5 in.) = 0  54  − Ay ( 30 in.) + (16 lb )(15 in.) + ( 22.815 lb )   ( 5 in.) = 0  55  ∴ Ay = 11.7334 lb

PROBLEM 4.125 CONTINUED ΣM B( y -axis ) = 0: Az ( 30 in.) + Tx (10 in.) + Tz ( 5 in.) = 0    3   10   Az ( 30 in.) + ( 22.815 lb )    (10 in.) + ( 22.815 lb )    ( 5 in.) = 0  55    55     ∴ Az = −1.10618 lb or A = (11.73 lb ) j − (1.106 lb ) k ΣFx = 0: Bx + Tx = 0  3  Bx +   ( 22.815 lb ) = 0  55  ∴ Bx = −1.24444 lb ΣFy = 0: Ay − W + Ty + By = 0  54  11.7334 lb − 16 lb + ( 22.815 lb )   + By = 0  55  ∴ By = −18.1336 lb ΣFz = 0: Az − Tz + Bz = 0  10  −1.10618 lb − ( 22.815 lb )   + Bz = 0  55  ∴ Bz = 5.2544 lb or B = − (1.244 lb ) i − (18.13 lb ) j + ( 5.25 lb ) k

PROBLEM 4.126 A 285-lb uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE, which passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.

SOLUTION

λ CD =

First note

= λ CE = =

− ( 23 in.) i + ( 22.5 in.) j − (15 in.) k 35.5 in.

1 ( −23i + 22.5 j − 15k ) 35.5

( 9 in.) i + ( 22.5 in.) j − (15 in.) k 28.5 in. 1 ( 9i + 22.5 j − 15k ) 28.5

W = − ( 285 lb ) j

From f.b.d. of plate (a)

ΣM x = 0:

 22.5    22.5    T  (15 in.) −   T  (15 in.) = 0  35.5    28.5  

( 285 lb )( 7.5 in.) − 

∴ T = 100.121 lb or T = 100.1 lb

PROBLEM 4.126 CONTINUED  23   9  ΣFx = 0: Ax − T   +T =0  35.5   28.5 

(b)

 23  (  9  Ax − (100.121 lb )   + 100.121 lb )  =0  35.5   28.5 

∴ Ax = 33.250 lb   22.5   ΣM B( z -axis ) = 0: − Ay ( 26 in.) + W (13 in.) − T    ( 6 in.) −   35.5   or

  22.5   T  28.5   ( 6 in.) = 0   

  22.5   − Ay ( 26 in.) + ( 285 lb )(13 in.) − (100.121 lb )    ( 6 in.)  35.5      22.5   − (100.121 lb )    ( 6 in.) = 0  28.5    ∴ Ay = 109.615 lb   15   ΣM B( y -axis ) = 0: Az ( 26 in.) − T    ( 6 in.) −   35.5     15   − T    ( 6 in.) +   28.5  

or

  23   T  35.5   (15 in.)   

  9  T  28.5   (15 in.) = 0   

1  −1 Az ( 26 in.) +  ( 90 + 345) − ( 90 − 135) (100.121 lb ) = 0 35.5 28.5   ∴ Az = 41.106 lb or A = ( 33.3 lb ) i + (109.6 lb ) j + ( 41.1 lb ) k  22.5   22.5  ΣFy = 0: By − W + T   +T  + Ay = 0  35.5   28.5   22.5 22.5  By − 285 lb + (100.121 lb )  +  + 109.615 lb = 0  35.5 28.5  ∴ By = 32.885 lb  15   15  ΣFz = 0: Bz + Az − T   −T =0  35.5   28.5  15   15 Bz + 41.106 lb − (100.121 lb )  + =0  35.5 28.5  ∴ Bz = 53.894 lb or B = ( 32.9 lb ) j + ( 53.9 lb ) k

PROBLEM 4.127 Solve Problem 4.126 assuming that cable DCE is replaced by a cable attached to point E and hook C. P4.126 A 285-lb uniform rectangular plate is supported in the position shown by hinges A and B and by cable DCE, which passes over a frictionless hook at C. Assuming that the tension is the same in both parts of the cable, determine (a) the tension in the cable, (b) the reactions at A and B. Assume that the hinge at B does not exert any axial thrust.

SOLUTION

First note

λ CE = =

( 9 in.) i + ( 22.5 in.) j − (15 in.) k 28.5 in. 1 ( 9i + 22.5 j − 15k ) 28.5

W = − ( 285 lb ) j

From f.b.d. of plate (a)

ΣM x = 0:

 22.5    T  (15 in.) = 0  28.5  

( 285 lb )( 7.5 in.) −  ∴ T = 180.500 lb

or T = 180.5 lb (b)

 9  ΣFx = 0: Ax + T  =0  28.5   9  Ax + 180.5 lb  =0  28.5  ∴ Ax = −57.000 lb

PROBLEM 4.127 CONTINUED   22.5   ΣM B( z -axis ) = 0: − Ay ( 26 in.) + W (13 in.) − T    ( 6 in.) = 0   28.5     22.5   − Ay ( 26 in.) + ( 285 lb )(13 in.) − (180.5 lb )    ( 6 in.) = 0  28.5    ∴ Ay = 109.615 lb   15   ΣM B( y -axis ) = 0: Az ( 26 in.) − T    ( 6 in.) +   28.5  

  9  T  28.5   (15 in.) = 0   

 45  Az ( 26 in.) + (180.5 lb )  =0  28.5  ∴ Az = −10.9615 lb or A = − ( 57.0 lb ) i + (109.6 lb ) j − (10.96 lb ) k  22.5  ΣFy = 0: By − W + T   + Ay = 0  28.5   22.5  By − 285 lb + (180.5 lb )   − 109.615 lb = 0  28.5  ∴ By = 32.885 lb  15  ΣFz = 0: Bz + Az − T  =0  28.5   15  Bz − 10.9615 lb − 180.5 lb  =0  28.5  ∴ Bz = 105.962 lb or B = ( 32.9 lb ) j + (106.0 lb ) k

PROBLEM 4.128 The tensioning mechanism of a belt drive consists of frictionless pulley A, mounting plate B, and spring C. Attached below the mounting plate is slider block D which is free to move in the frictionless slot of bracket E. Knowing that the pulley and the belt lie in a horizontal plane, with portion F of the belt parallel to the x axis and portion G forming an angle of 30° with the x axis, determine (a) the force in the spring, (b) the reaction at D.

SOLUTION

From f.b.d. of plate B (a)

ΣFx = 0: 12 N + (12 N ) cos 30° − T = 0 ∴ T = 22.392 N

(b)

or T = 22.4 N

ΣFy = 0: Dy = 0 ΣFz = 0: Dz − (12 N ) sin 30° = 0 ∴ Dz = 6 N

or D = ( 6.00 N ) k

ΣM x = 0: M Dx − (12 N ) sin 30°  ( 22 mm ) = 0

∴ M Dx = 132.0 N ⋅ mm ΣM D( y -axis ) = 0: M Dy + ( 22.392 N )( 30 mm ) − (12 N )( 75 mm ) − (12 N ) cos 30° ( 75 mm ) = 0

∴ M Dy = 1007.66 N ⋅ mm ΣM D( z -axis ) = 0: M Dz + ( 22.392 N )(18 mm ) − (12 N )( 22 mm ) − (12 N ) cos 30° ( 22 mm ) = 0 ∴ M Dz = 89.575 N ⋅ mm or M D = ( 0.1320 N ⋅ m ) i + (1.008 N ⋅ m ) j + ( 0.0896 N ⋅ m ) k

PROBLEM 4.129 The assembly shown is welded to collar A which fits on the vertical pin shown. The pin can exert couples about the x and z axes but does not prevent motion about or along the y axis. For the loading shown, determine the tension in each cable and the reaction at A.

SOLUTION

TCF = λ CFTCF =

First note

− ( 0.16 m ) i + ( 0.12 m ) j

( 0.16 ) + ( 0.12 ) m 2

2

TCF

= TCF ( −0.8i + 0.6 j) TDE = λ DETDE =

( 0.24 m ) j − ( 0.18 m ) k T DE ( 0.24 )2 + ( 0.18)2 m

= TDE ( 0.8j − 0.6k ) (a) From f.b.d. of assembly ΣFy = 0: 0.6TCF + 0.8TDE − 800 N = 0 or

0.6TCF + 0.8TDE = 800 N

(1)

ΣM y = 0: − ( 0.8TCF ) ( 0.27 m ) + ( 0.6TDE )( 0.16 m ) = 0 or

TDE = 2.25TCF

(2)

PROBLEM 4.129 CONTINUED Substituting Equation (2) into Equation (1) 0.6TCF + 0.8 ( 2.25 ) TCF  = 800 N

∴ TCF = 333.33 N and from Equation (2)

or TCF = 333 N

TDE = 2.25 ( 333.33 N ) = 750.00 N

or TDE = 750 N

(b) From f.b.d. of assembly ΣFz = 0: Az − ( 0.6 )( 750.00 N ) = 0

∴ Az = 450.00 N

ΣFx = 0: Ax − ( 0.8 )( 333.33 N ) = 0

∴ Ax = 266.67 N or A = ( 267 N ) i + ( 450 N ) k

ΣM x = 0: M Ax + ( 800 N )( 0.27 m ) − ( 333.33 N )( 0.6 )  ( 0.27 m ) − ( 750 N )( 0.8 )  ( 0.18 m ) = 0

∴ M Ax = −54.001 N ⋅ m ΣM z = 0: M Az − ( 800 N )( 0.16 m ) + ( 333.33 N )( 0.6 )  ( 0.16 m ) + ( 750 N )( 0.8 )  ( 0.16 m ) = 0

∴ M Az = 0 or M A = − ( 54.0 N ⋅ m ) i

PROBLEM 4.130 The lever AB is welded to the bent rod BCD which is supported by bearing E and by cable DG. Assuming that the bearing can exert an axial thrust and couples about axes parallel to the x and z axes, determine (a) the tension in cable DG, (b) the reaction at E.

SOLUTION

TDG = λ DGTDG =

First note

=

− ( 0.12 m ) j − ( 0.225 m ) k

( 0.12 ) + ( 0.225) m 2

2

TDG

TDG ( −0.12 j − 0.225k ) 0.255

(a) From f.b.d. of weldment  0.225   ΣM y = 0:   TDG  ( 0.16 m ) − ( 220 N )( 0.24 m ) = 0  0.255   ∴ TDG = 374.00 N (b) From f.b.d. of weldment ΣFx = 0: Ex − 220 N = 0 ∴ Ex = 220.00 N  0.12  ΣFy = 0: E y − ( 374.00 N )  =0  0.255  ∴ E y = 176.000 N

or TDG = 374 N

PROBLEM 4.130 CONTINUED  0.225  ΣFz = 0: Ez − ( 374.00 N )  =0  0.255  ∴ Ez = 330.00 N or E = ( 220 N ) i + (176.0 N ) j + ( 330 N ) k ΣM x = 0: M Ex + ( 330.00 N )( 0.19 m ) = 0 ∴ M Ex = −62.700 N ⋅ m ΣM z = 0:

( 220 N )( 0.06 m ) + M Ez

  0.12   − ( 374.00 N )    ( 0.16 m ) = 0  0.255   

∴ M Ez = −14.9600 N ⋅ m or M E = − ( 62.7 N ⋅ m ) i − (14.96 N ⋅ m ) k

PROBLEM 4.131 Solve Problem 4.124 assuming that the hinge at A is removed and that the hinge at B can exert couples about the y and z axes. P4.124 A small door weighing 16 lb is attached by hinges A and B to a wall and is held in the horizontal position shown by rope EFH. The rope passes around a small, frictionless pulley at F and is tied to a fixed cleat at H. Assuming that the hinge at A does not exert any axial thrust, determine (a) the tension in the rope, (b) the reactions at A and B.

SOLUTION From f.b.d. of door ΣM B = 0: rG/B × W + rE/B × TEF + M B = 0

(a) where

W = − (16 lb ) j

M B = M By j + M Bz k TEF = λ EFTEF =

=

(12 in.) i + ( 54 in.) j − ( 28 in.) k T EF (12 )2 + ( 54 )2 + ( 28)2 in. TEF ( 6i + 27 j − 14k ) 31

rG/B = − (15 in.) i + (14 in.) k rE/B = − ( 4 in.) i + ( 28 in.) k ∴

i j k i j k T  −15 0 14 (16 lb ) + −4 0 28  EF  + M By j + M Bz k = 0  31  0 −1 0 6 27 −14

(

or

( 224 − 24.387TEF ) i + ( 3.6129TEF

(

)

)

+ M By j

)

+ 240 − 3.4839TEF + M Bz k = 0 From i-coefficient

224 − 24.387TEF = 0 ∴ TEF = 9.1852 lb or TEF = 9.19 lb W

(b) From j-coefficient

3.6129 ( 9.1852 ) + M By = 0

∴ M By = −33.185 lb ⋅ in.

PROBLEM 4.131 CONTINUED From k-coefficient

240 − 3.4839 ( 9.1852 ) + M Bz = 0 ∴ M Bz = −208.00 lb ⋅ in. or M B = − ( 33.2 lb ⋅ in.) j − ( 208 lb ⋅ in.) k W

ΣFx = 0: Bx +

6 ( 9.1852 lb ) = 0 31

∴ Bx = −1.77778 lb ΣFy = 0: By − 16 lb +

27 ( 9.1852 lb ) = 0 31

∴ By = 8.0000 lb ΣFz = 0: Bz −

14 ( 9.1852 lb ) = 0 31

∴ Bz = 4.1482 lb or B = − (1.778 lb ) i + ( 8.00 lb ) j + ( 4.15 lb ) k W

PROBLEM 4.132 The frame shown is supported by three cables and a ball-and-socket joint at A. For P = 0, determine the tension in each cable and the reaction at A.

SOLUTION First note TDI = λ DI TDI =

= TEH = λ EH TEH =

= TFG = λ FGTFG =

=

− ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k

( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m

TDI

TDI ( −0.65i + 0.2 j − 0.44k ) 0.81 − ( 0.45 m ) i + ( 0.24 m ) j

( 0.45)2 + ( 0.24 )2 m

TEH

TEH ( −0.45i ) + ( 0.24 j) 0.51 − ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k

( 0.45

)

2

+ ( 0.2 ) + ( 0.36 ) m 2

2

TFG

TFG ( −0.45i + 0.2 j + 0.36k ) 0.61

From f.b.d. of frame ΣM A = 0: rD/ A × TDI + rC/ A × ( −280 N ) j + rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j = 0 or

i j k i j k i j k i j k  TDI   TEH   TFG  0.65 0.2 0  0.32 0   + 0.65 0 0 ( 280 N ) + 0  + 0.45 0 0.06    0.81   0.51   0.61  −0.65 0.2 −0.44 −0.45 0.24 0 −0.45 0.2 0.36 0 −1 0 i j k + 0.45 0 0.06 ( 360 N ) = 0 0 −1 0

or

( −0.088i + 0.286 j + 0.26k )

TDI T + ( −0.65k ) 280 N + ( 0.144k ) EH 0.81 0.51

+ ( −0.012i − 0.189 j + 0.09k )

TFG + ( 0.06i − 0.45k )( 360 N ) = 0 0.61

PROBLEM 4.132 CONTINUED From i-coefficient

T  T  −0.088  DI  − 0.012  FG  + 0.06 ( 360 N ) = 0  0.81   0.61 

∴ 0.108642TDI + 0.0196721TFG = 21.6 From j-coefficient

(1)

T  T  0.286  DI  − 0.189  FG  = 0 0.81    0.61  ∴ TFG = 1.13959TDI

(2)

From k-coefficient T  T  T  0.26  DI  − 0.65 ( 280 N ) + 0.144  EH  + 0.09  FG   0.81   0.51   0.61  − 0.45 ( 360 N ) = 0

∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344 N

(3)

Substitution of Equation (2) into Equation (1) 0.108642TDI + 0.0196721(1.13959TDI ) = 21.6 ∴ TDI = 164.810 N TDI = 164.8 N W

or Then from Equation (2)

TFG = 1.13959 (164.810 N ) = 187.816 N TFG = 187.8 N W

or And from Equation (3)

0.32099 (164.810 N ) + 0.28235TEH + 0.147541(187.816 N ) = 344 N ∴ TEH = 932.84 N TEH = 933 N W

or The vector forms of the cable forces are: TDI =

164.810 N ( −0.65i + 0.2 j − 0.44k ) 0.81

= − (132.25 N ) i + ( 40.694 N ) j − ( 89.526 N ) k

TEH =

932.84 N ( −0.45i + 0.24 j) = − (823.09 N ) i + ( 438.98 N ) j 0.51

TFG =

187.816 N ( −0.45i + 0.2 j + 0.36k ) 0.61

= − (138.553 N ) i + ( 61.579 N ) j + (110.842 N ) k

PROBLEM 4.132 CONTINUED Then, from f.b.d. of frame ΣFx = 0: Ax − 132.25 − 823.09 − 138.553 = 0 ∴ Ax = 1093.89 N ΣFy = 0: Ay + 40.694 + 438.98 + 61.579 − 360 − 280 = 0 ∴ Ay = 98.747 N ΣFz = 0: Az − 89.526 + 110.842 = 0 ∴ Az = −21.316 N or

A = (1094 N ) i + ( 98.7 N ) j − ( 21.3 N ) k W

PROBLEM 4.133 The frame shown is supported by three cables and a ball-and-socket joint at A. For P = 50 N, determine the tension in each cable and the reaction at A.

SOLUTION First note TDI = λ DI TDI =

= TEH = λ EH TEH =

= TFG = λ FGTFG =

=

− ( 0.65 m ) i + ( 0.2 m ) j − ( 0.44 m ) k

( 0.65)2 + ( 0.2 )2 + ( 0.44 )2 m

TDI

TDI ( −65i + 20 j − 44k ) 81 − ( 0.45 m ) i + ( 0.24 m ) j

( 0.45)2 + ( 0.24 )2 m

TEH

TEH ( −15i + 8j) 17 − ( 0.45 m ) i + ( 0.2 m ) j + ( 0.36 m ) k

( 0.45) + ( 0.2 ) + ( 0.36 ) m 2

2

2

TFG

TFG ( −45i + 20 j + 36k ) 61

From f.b.d. of frame ΣM A = 0: rD/ A × TDI + rC/ A ×  − ( 280 N ) j + ( 50 N ) k 

+ rH / A × TEH + rF / A × TFG + rF / A × ( −360 N ) j or

i j k i j k i j k  TDI  T  0.65 0.2 0  0 + 0 0.32 0  EH   + 0.65 0  81   17  0 −280 50 −65 20 −44 −15 8 0 i j k i j k  TFG  + 0.45 0 0.06   + 0.45 0 0.06 ( 360 N ) = 0  61  0 −1 0 −45 20 36

and

( −8.8i + 28.6 j + 26k ) 

TDI   TEH   + ( −32.5 j − 182k ) + ( 4.8k )    81   17 

T  + ( −1.2i − 18.9 j + 9.0k )  FG  + ( 0.06i − 0.45k ) ( 360 ) = 0  61 

PROBLEM 4.133 CONTINUED T  T −8.8  DI  − 1.2  FG  81   61

From i-coefficient

  + 0.06 ( 360 ) = 0 

∴ 0.108642TDI + 0.0196721TFG = 21.6

(1)

T  T  From j-coefficient 28.6  DI  − 32.5 − 18.9  FG  = 0  81   61  ∴ 0.35309TDI − 0.30984TFG = 32.5

(2)

From k-coefficient T  T  T  26  DI  − 182 + 4.8  EH  + 9.0  FG  − 0.45 ( 360 ) = 0  81   17   61  ∴ 0.32099TDI + 0.28235TEH + 0.147541TFG = 344 −3.25 × Equation (1) Add Equation (2)

(3)

−0.35309TDI − 0.063935TFG = −70.201 0.35309TDI − 0.30984TFG = −0.37378TFG

32.5

= −37.701

∴ TFG = 100.864 N TFG = 100.9 N W

or Then from Equation (1)

0.108642TDI + 0.0196721(100.864 ) = 21.6 ∴ TDI = 180.554 N TDI = 180.6 N W

or and from Equation (3)

0.32099 (180.554 ) + 0.28235TEH + 0.147541(100.864 ) = 344 ∴ TEH = 960.38 N TEH = 960 N W

or The vector forms of the cable forces are: TDI =

180.554 N ( −65i + 20 j − 44k ) 81

= − (144.889 N ) i + ( 44.581 N ) j − ( 98.079 N ) k

TEH =

960.38 N ( −15i + 8j) = − (847.39 N ) i + ( 451.94 N ) j 17

TFG =

100.864 N ( −45i + 20 j + 36k ) 61

= − ( 74.409 N ) i + ( 33.070 N ) j + ( 59.527 N ) k

PROBLEM 4.133 CONTINUED Then from f.b.d. of frame ΣFx = 0: Ax − 144.889 − 847.39 − 74.409 = 0 ∴ Ax = 1066.69 N ΣFy = 0: Ay + 44.581 + 451.94 + 33.070 − 360 − 280 = 0 ∴ Ay = 110.409 N ΣFz = 0: Az − 98.079 + 59.527 + 50 = 0 ∴ Az = −11.448 N Therefore,

A = (1067 N ) i + (110.4 N ) j − (11.45 N ) k W

PROBLEM 4.134 The rigid L-shaped member ABF is supported by a ball-and-socket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A.

SOLUTION First note TBG = λ BGTBG =

− (18 in.) i + (13.5 in.) k

(18)

2

+ (13.5 ) in. 2

TBG

= TBG ( −0.8i + 0.6k ) TDH = λ DH TDH =

− (18 in.) i + ( 24 in.) j

(18)2 + ( 24 )2 in.

TDH

= TDH ( −0.6i + 0.8 j) Since λ FJ = λ DH , TFJ = TFJ ( −0.6i + 0.8 j)

From f.b.d. of member ABF ΣM A( x-axis ) = 0:

( 0.8TFJ ) ( 48 in.) + ( 0.8TDH )( 24 in.) − (120 lb )( 36 in.) − (120 lb )(12 in.) = 0 ∴ 3.2TFJ + 1.6TDH = 480

ΣM A( z -axis ) = 0:

( 0.8TFJ ) (18 in.) + ( 0.8TDH )(18 in.) − (120 lb )(18 in.) − (120 lb )(18 in.) = 0 ∴

− 3.2TFJ − 3.2TDH = −960

Equation (1) + Equation (2) Substituting in Equation (1)

ΣM A( y -axis ) = 0:

(1)

(2) TDH = 300 lb W TFJ = 0 W

( 0.6TFJ ) ( 48 in.) + 0.6 ( 300 lb ) ( 24 in.) − ( 0.6TBG ) (18 in.) = 0 ∴ TBG = 400 lb W

PROBLEM 4.134 CONTINUED ΣFx = 0: − 0.6TFJ − 0.6TDH − 0.8TBG + Ax = 0 −0.6 ( 300 lb ) − 0.8 ( 400 lb ) + Ax = 0 ∴ Ax = 500 lb ΣFy = 0: 0.8TFJ + 0.8TDH − 240 lb + Ay = 0 0.8 ( 300 lb ) − 240 + Ay = 0 ∴ Ay = 0 ΣFz = 0: 0.6TBG + Az = 0 0.6 ( 400 lb ) + Az = 0 ∴ Az = −240 lb Therefore,

A = ( 500 lb ) i − ( 240 lb ) k W

PROBLEM 4.135 Solve Problem 4.134 assuming that the load at C has been removed. P4.134 The rigid L-shaped member ABF is supported by a ball-andsocket joint at A and by three cables. For the loading shown, determine the tension in each cable and the reaction at A.

SOLUTION First TBG = λ BGTBG =

− (18 in.) i + (13.5 in.) k

(18)

2

+ (13.5 ) in. 2

TBG

= TBG ( −0.8i + 0.6k ) TDH = λ DH TDH =

− (18 in.) i + ( 24 in.) j

(18)2 + ( 24 )2 in.

TDH

= TDH ( −0.6i + 0.8 j) λ FJ = λ DH

Since

TFJ = TFJ ( −0.6i + 0.8 j)

From f.b.d. of member ABF ΣM A( x-axis ) = 0:

( 0.8TFJ ) ( 48 in.) + ( 0.8TDH )( 24 in.) − (120 lb )( 36 in.) = 0 ∴ 3.2TFJ + 1.6TDH = 360

ΣM A( z -axis ) = 0:

(1)

( 0.8TFJ ) (18 in.) + ( 0.8TDH )(18 in.) − (120 lb )(18 in.) = 0 ∴

− 3.2TFJ − 3.2TDH = −480

(2)

Equation (1) + Equation (2)

TDH = 75.0 lb W

Substituting into Equation (2)

TFJ = 75.0 lb W

ΣM A( y -axis ) = 0:

or

( 0.6TFJ ) ( 48 in.) + ( 0.6TDH )( 24 in.) − ( 0.6TBG ) (18 in.) = 0 ( 75.0 lb )( 48 in.) + ( 75.0 lb )( 24 in.) = TBG (18 in.) TBG = 300 lb W

PROBLEM 4.135 CONTINUED ΣFx = 0: − 0.6TFJ − 0.6TDH − 0.8TBG + Ax = 0 −0.6 ( 75.0 + 75.0 ) − 0.8 ( 300 ) + Ax = 0 ∴ Ax = 330 lb ΣFy = 0: 0.8TFJ + 0.8TDH − 120 lb + Ay = 0 0.8 (150 lb ) − 120 lb + Ay = 0 ∴ Ay = 0 ΣFz = 0: 0.6TBG + Az = 0 0.6 ( 300 lb ) + Az = 0 ∴ Az = −180 lb Therefore

A = ( 330 lb ) i − (180 lb ) k W

PROBLEM 4.136 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor which rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = − ( 60 N ) k , M = − (108 N ⋅ m ) k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe.

SOLUTION From f.b.d. of pipe assembly ABCD

ΣFx = 0: Bx = 0 ΣM D( x-axis ) = 0:

( 60 N )( 2.5 m ) − Bz ( 2 m ) = 0 ∴ Bz = 75.0 N and B = ( 75.0 N ) k

ΣM D( z -axis ) = 0: C y ( 3 m ) − 108 N ⋅ m = 0 ∴ C y = 36.0 N ΣM D( y -axis ) = 0: − Cz ( 3 m ) − ( 75 N )( 4 m ) + ( 60 N )( 4 m ) = 0 ∴ C z = −20.0 N and C = ( 36.0 N ) j − ( 20.0 N ) k ΣFy = 0: Dy + 36.0 = 0 ∴ D y = −36.0 N ΣFz = 0: Dz − 20.0 N + 75.0 N − 60 N = 0 ∴ Dz = 5.00 N and D = − ( 36.0 N ) j + ( 5.00 N ) k

PROBLEM 4.137 Solve Problem 4.136 assuming that the plumber exerts a force F = − ( 60 N ) k and that the motor is turned off ( M = 0 ) . P4.136 In order to clean the clogged drainpipe AE, a plumber has disconnected both ends of the pipe and inserted a power snake through the opening at A. The cutting head of the snake is connected by a heavy cable to an electric motor which rotates at a constant speed as the plumber forces the cable into the pipe. The forces exerted by the plumber and the motor on the end of the cable can be represented by the wrench F = − ( 60 N ) k , M = − (108 N ⋅ m ) k. Determine the additional reactions at B, C, and D caused by the cleaning operation. Assume that the reaction at each support consists of two force components perpendicular to the pipe.

SOLUTION From f.b.d. of pipe assembly ABCD ΣFx = 0: Bx = 0 ΣM D( x-axis ) = 0:

( 60 N )( 2.5 m ) − Bz ( 2 m ) = 0 ∴ Bz = 75.0 N and B = ( 75.0 N ) k

ΣM D( z -axis ) = 0: C y ( 3 m ) − Bx ( 2 m ) = 0 ∴ Cy = 0 ΣM D( y -axis ) = 0: Cz ( 3 m ) − ( 75.0 N )( 4 m ) + ( 60 N )( 4 m ) = 0 ∴ C z = −20 N and C = − ( 20.0 N ) k

ΣFy = 0: Dy + C y = 0 ∴ Dy = 0 ΣFz = 0: Dz + Bz + C z − F = 0 Dz + 75 N − 20 N − 60 N = 0 ∴ Dz = 5.00 N and D = ( 5.00 N ) k

PROBLEM 4.138 Three rods are welded together to form a “corner” which is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 N, a = 120 mm, b = 80 mm, and c = 100 mm.

SOLUTION From f.b.d. of weldment ΣM O = 0: rA/O × A + rB/O × B + rC/O × C = 0 i j k i j k i j k 120 0 0 + 0 80 0 + 0 0 100 = 0 0 Ay Az Bx 0 Bz Cx Cy 0

( −120 Az j + 120 Ayk ) + (80Bzi − 80Bxk ) + ( −100C yi + 100Cx j) = 0 From i-coefficient

80Bz − 100C y = 0 Bz = 1.25C y

or j-coefficient

−120 Az + 100Cx = 0 Cx = 1.2 Az

or k-coefficient

(1)

(2)

120 Ay − 80Bx = 0 Bx = 1.5 Ay

or

(3)

ΣF = 0: A + B + C − P = 0

( Bx + Cx ) i + ( Ay + C y − 240 N ) j + ( Az + Bz ) k = 0

or

Bx + Cx = 0

From i-coefficient

Cx = − Bx

or j-coefficient

Ay + C y − 240 N = 0 Ay + C y = 240 N

or

(5)

Az + Bz = 0

k-coefficient

or

(4)

Az = − Bz

(6)

PROBLEM 4.138 CONTINUED Substituting Cx from Equation (4) into Equation (2) − Bz = 1.2 Az

(7)

Using Equations (1), (6), and (7) Cy =

Bz − Az 1  Bx  Bx = =  = 1.25 1.25 1.25  1.2  1.5

(8)

From Equations (3) and (8) Cy =

1.5 Ay 1.5

C y = Ay

or

and substituting into Equation (5) 2 Ay = 240 N ∴ Ay = C y = 120 N

(9)

Using Equation (1) and Equation (9) Bz = 1.25 (120 N ) = 150.0 N Using Equation (3) and Equation (9) Bx = 1.5 (120 N ) = 180.0 N From Equation (4)

Cx = −180.0 N

From Equation (6)

Az = −150.0 N

Therefore

A = (120.0 N ) j − (150.0 N ) k B = (180.0 N ) i + (150.0 N ) k C = − (180.0 N ) i + (120.0 N ) j

PROBLEM 4.139 Solve Problem 4.138 assuming that the force P is removed and is replaced by a couple M = + ( 6 N ⋅ m ) j acting at B. P4.138 Three rods are welded together to form a “corner” which is supported by three eyebolts. Neglecting friction, determine the reactions at A, B, and C when P = 240 N, a = 120 mm, b = 80 mm, and c = 100 mm.

SOLUTION From f.b.d. of weldment ΣM O = 0: rA/O × A + rB/O × B + rC/O × C + M = 0 i j k i j k i j k 0.12 0 0 + 0 0.08 0 + 0 0 0.1 + ( 6 N ⋅ m ) j = 0 0 Ay Az Bx 0 Bz Cx Cy 0

( −0.12 Az j + 0.12 Ayk ) + ( 0.08Bz j − 0.08Bxk ) (

)

+ −0.1C y i + 0.1Cx j + ( 6 N ⋅ m ) j = 0 From i-coefficient

0.08Bz − 0.1C y = 0 C y = 0.8Bz

or j-coefficient

(1)

−0.12 Az + 0.1Cx + 6 = 0 Cx = 1.2 Az − 60

or k-coefficient

(2)

0.12 Ay − 0.08Bx = 0 Bx = 1.5 Ay

or

(3)

ΣF = 0: A + B + C = 0

( Bx + Cx ) i + ( Ay + C y ) j + ( Az

+ Bz ) k = 0

From i-coefficient

Cx = − Bx

(4)

j-coefficient

C y = − Ay

(5)

k-coefficient

Az = − Bz

(6)

Substituting Cx from Equation (4) into Equation (2) B  Az = 50 −  x   1.2 

(7)

PROBLEM 4.139 CONTINUED Using Equations (1), (6), and (7) 2 C y = 0.8Bz = −0.8 Az =   Bx − 40 3

(8)

From Equations (3) and (8) C y = Ay − 40 2 Ay = 40

Substituting into Equation (5) ∴ Ay = 20.0 N From Equation (5)

C y = −20.0 N

Equation (1)

Bz = −25.0 N

Equation (3)

Bx = 30.0 N

Equation (4)

Cx = −30.0 N

Equation (6)

Az = 25.0 N

Therefore

A = ( 20.0 N ) j + ( 25.0 N ) k B = ( 30.0 N ) i − ( 25.0 N ) k C = − ( 30.0 N ) i − ( 20.0 N ) j

PROBLEM 4.140 The uniform 10-lb rod AB is supported by a ball-and-socket joint at A and leans against both the rod CD and the vertical wall. Neglecting the effects of friction, determine (a) the force which rod CD exerts on AB, (b) the reactions at A and B. (Hint: The force exerted by CD on AB must be perpendicular to both rods.)

SOLUTION (a) The force acting at E on the f.b.d. of rod AB is perpendicular to AB and CD. Letting λ E = direction cosines for force E, λE =

rB/ A × k rB/ A × k

 − ( 32 in.) i + ( 24 in.) j − ( 40 in.) k  × k = 2 2 ( 32 ) + ( 24 ) in. = 0.6i + 0.8 j Also,

W = − (10 lb ) j

B = Bk E = E ( 0.6i + 0.8 j)

From f.b.d. of rod AB ΣM A = 0: rG/ A × W + rE/ A × E + rB/ A × B = 0 i j k i j k i j k ∴ −16 12 −20 (10 lb ) + −24 18 −30 E + −32 24 −40 B = 0 0 −1 0 0.6 0.8 0 0 0 1

( −20i + 16k )(10 lb ) + ( 24i − 18 j − 30k ) E + ( 24i + 32 j) B = 0 From k-coefficient

160 − 30 E = 0 ∴ E = 5.3333 lb

and

E = 5.3333 lb ( 0.6i + 0.8 j) E = ( 3.20 lb ) i + ( 4.27 lb ) j

or (b) From j-coefficient

−18 ( 5.3333 lb ) + 32 B = 0

∴ B = 3.00 lb or

B = ( 3.00 lb ) k

PROBLEM 4.140 CONTINUED From f.b.d. of rod AB ΣF = 0: A + W + E + B = 0

Ax i + Ay j + Az k − (10 lb ) j + ( 3.20 lb ) i + ( 4.27 lb ) j + ( 3.00 lb ) k = 0 From i-coefficient

Ax + 3.20 lb = 0 ∴ Ax = −3.20 lb

j-coefficient

Ay − 10 lb + 4.27 lb = 0 ∴ Ay = 5.73 lb

k-coefficient

Az + 3.00 lb = 0 ∴ Az = −3.00 lb

Therefore

A = − ( 3.20 lb ) i + ( 5.73 lb ) j − ( 3.00 lb ) k

PROBLEM 4.141 A 21-in.-long uniform rod AB weighs 6.4 lb and is attached to a ball-andsocket joint at A. The rod rests against an inclined frictionless surface and is held in the position shown by cord BC. Knowing that the cord is 21 in. long, determine (a) the tension in the cord, (b) the reactions at A and B.

SOLUTION

First note W = − ( 6.4 lb ) j N B = N B ( 0.8j + 0.6k )

LAB = 21 in. =

( xB )2 + (13 + 3)2 + ( 4 )2

=

xB2 + (16 ) + ( 4 ) 2

2

∴ xB = 13 in. TBC = λ BCTBC =

=

(13 in.) i + (16 in.) j − ( 4 in.) k T 21 in.

BC

TBC (13i + 16 j − 4k ) 21

From f.b.d. of rod AB ΣM A = 0: rG/ A × W + rB/ A × N B + rC/ A × TBC = 0 i j k i j k i j k 26TBC 6.5 −8 2 + 13 −16 4 N B + 1 0 0 =0 21 0 −6.4 0 0 0.8 0.6 13 16 −4

(12.8i − 41.6k ) + ( −12.8i − 7.8j + 10.4k ) N B + ( 4 j + 16k )

26TBC =0 21

PROBLEM 4.141 CONTINUED 12.8 − 12.8N B = 0

From i-coeff.

∴ N B = 1.00 lb

N B = ( 0.800 lb ) j + ( 0.600 lb ) k

or

 26  −7.8 N B + 4   TBC = 0  21 

From j-coeff.

∴ TBC = 1.575 lb

From f.b.d. of rod AB ΣF = 0: A + W + N B + TBC = 0  ( Axi + Ay j + Azk ) − ( 6.4 lb ) j + ( 0.800 lb ) j + ( 0.600 lb ) k +  1.575  (13i + 16 j − 4k ) = 0 21  From i-coefficient

Ax = −0.975 lb

j-coefficient

Ay = 4.40 lb

k-coefficient

Az = −0.3 lb

∴ (a)

TBC = 1.575 lb

(b)

A = − ( 0.975 lb ) i + ( 4.40 lb ) j − ( 0.300 lb ) k N B = ( 0.800 lb ) j + ( 0.600 lb ) k

PROBLEM 4.142 While being installed, the 56-lb chute ABCD is attached to a wall with brackets E and F and is braced with props GH and IJ. Assuming that the weight of the chute is uniformly distributed, determine the magnitude of the force exerted on the chute by prop GH if prop IJ is removed.

SOLUTION

First note

 42 in. 

θ = tan −1   = 16.2602°  144 in.  xG = ( 50 in.) cos16.2602° = 48 in. yG = 78 in. − ( 50 in.) sin16.2602° = 64 in.

λ BA =

− (144 in.) i + ( 42 in.) j

(144 )

2

+ ( 42 ) in. 2

=

1 ( −24i + 7 j) 25

rK / A = ( 72 in.) i − ( 21 in.) j + ( 9 in.) k rG/ A = ( 48 in.) i − ( 78 in. − 64 in.) j + (18 in.) k = ( 48 in.) i − (14 in.) j + (18 in.) k W = − ( 56 lb ) j PHG = λ HG PHG

=

=

− ( 2 in.) i + ( 64 in.) j − (16 in.) k

( 2 ) + ( 64 ) + (16 ) in. 2

2

PHG ( −i + 32 j − 8k ) 33

2

PHG

PROBLEM 4.142 CONTINUED From the f.b.d. of the chute

ΣM BA = 0: λ BA ⋅ ( rK / A × W ) + λ BA ⋅ ( rG/ A × PHG ) = 0 −24 −24 7 0 7 0  56   P  72 −21 9   + 48 −14 18  HG  = 0 ( )  25   33 25  0 −1 0 −1 32 −8 −216 ( 56 ) P + [ −24 ( −14 ) ( −8 ) − ( −24 ) (18 )( 32 ) + 7 (18 ) ( −1) − ( 7 )( 48 )( −8 )] HG = 0 25 33 ( 25 ) ∴ PHG = 29.141 lb or PHG = 29.1 lb

PROBLEM 4.143 While being installed, the 56-lb chute ABCD is attached to a wall with brackets E and F and is braced with props GH and IJ. Assuming that the weight of the chute is uniformly distributed, determine the magnitude of the force exerted on the chute by prop IJ if prop GH is removed.

SOLUTION

First note  42 in. 

θ = tan −1   = 16.2602°  144 in.  xI = (100 in.) cos16.2602° = 96 in. yI = 78 in. − (100 in.) sin16.2602° = 50 in. λ BA =

− (144 in.) i + ( 42 in.) j

(144 )

2

+ ( 42 ) in. 2

=

1 ( −24i + 7 j) 25

rK / A = ( 72 in.) i − ( 21 in.) j + ( 9 in.) k rI / A = ( 96 in.) i − ( 78 in. − 50 in.) j + (18 in.) k = ( 96 in.) i − ( 28 in.) j + (18 in.) k W = − ( 56 lb ) j PJI = λ JI PJI =

=

− (1 in.) i + ( 50 in.) j − (10 in.) k

(1)

2

+ ( 50 ) + (10 ) in. 2

PJI ( −i + 50 j − 10k ) 51

2

PJI

PROBLEM 4.143 CONTINUED From the f.b.d. of the chute ΣM BA = 0: λ BA ⋅ ( rK / A × W ) + λ BA ⋅ ( rI / A × PJI ) = 0 −24 −24 7 7 0 0  56   P  72 −21 9   + 96 −28 18  JI  = 0  25   51( 25 )  0 −1 0 −1 50 −10 −216 ( 56 ) P + [ −24 ( −28 )( −10 ) − ( −24 ) (18 )( 50 ) + 7 (18 ) ( −1) − ( 7 )( 96 )( −10 )] JI = 0 25 51( 25 ) ∴ PJI = 28.728 lb or PJI = 28.7 lb

PROBLEM 4.144 To water seedlings, a gardener joins three lengths of pipe, AB, BC, and CD, fitted with spray nozzles and suspends the assembly using hinged supports at A and D and cable EF. Knowing that the pipe weighs 0.85 lb/ft, determine the tension in the cable.

SOLUTION First note rG/ A = (1.5 ft ) i WAB = − ( 0.85 lb/ft )( 3 ft ) j = − ( 2.55 lb ) j rF / A = ( 2 ft ) i T = λ FET =

− ( 2 ft ) i + ( 3 ft ) j − ( 4.5 ft ) k

( 2 )2 + ( 3)2 + ( 4.5)2

T

ft

 T  =  ( −2i + 3j − 4.5k )  33.25 

rB/ A = ( 3 ft ) i WBC = − ( 0.85 lb/ft )(1 ft ) j = − ( 0.85 lb ) j rH / A = ( 3 ft ) i − ( 2.25 ft ) k WCD = − ( 0.85 lb/ft )( 4.5 ft ) j = − ( 3.825 lb ) j λ AD =

( 3 ft ) i − (1 ft ) j − ( 4.5 ft ) k ( 3)2 + (1)2 + ( 4.5)2 ft

=

1 ( 3i − j − 4.5k ) 5.5

PROBLEM 4.144 CONTINUED From f.b.d. of the pipe assembly ΣM AD = 0:

λ AD ⋅ ( rG/ A × WAB ) + λ AD ⋅ ( rF / A × T ) + λ AD ⋅ ( rB/ A × WBC ) + λ AD ⋅ ( rH / A × WCD ) = 0

3 3 −1 −4.5 −1 −4.5 T  1    0 0  0  ∴ 1.5 + 2 0   5.5   5.5 33.25  0 −2.55 0 −2 3 −4.5

3 −1 −4.5 3 −1 −4.5  1   1  + 3 0 + − 0  3 0 2.25   =0  5.5   5.5  0 −0.85 0 0 −3.825 0 

T   + (11.475 ) + ( 25.819 ) = 0 33.25  

(17.2125) + ( −36 ) 

∴ T = 8.7306 lb or T = 8.73 lb

PROBLEM 4.145 Solve Problem 4.144 assuming that cable EF is replaced by a cable connecting E and C. P4.144 To water seedlings, a gardener joins three lengths of pipe, AB, BC, and CD, fitted with spray nozzles and suspends the assembly using hinged supports at A and D and cable EF. Knowing that the pipe weighs 0.85 lb/ft, determine the tension in the cable.

SOLUTION First note

rG/ A = (1.5 ft ) i WAB = − ( 0.85 lb/ft )( 3 ft ) j = − ( 2.55 lb ) j rC/ A = ( 3 ft ) i − (1 ft ) j T = λ CET =

− ( 3 ft ) i + ( 4 ft ) j − ( 4.5 ft ) k

( 3)2 + ( 4 )2 + ( 4.5)2

T

ft

 T  =  ( −3i + 4 j − 4.5k )  45.25 

rB/ A = ( 3 ft ) i WBC = − ( 0.85 lb/ft )(1 ft ) j = − ( 0.85 lb ) j rH / A = ( 3 ft ) i − ( 2.25 ft ) k WCD = − ( 0.85 lb/ft )(1 ft ) j = − ( 3.825 lb ) j λ AD =

( 3 ft ) i − (1 ft ) j − ( 4.5 ft ) k ( 3)2 + (1)2 + ( 4.5)2 ft

=

1 ( 3i − j − 4.5k ) 5.5

PROBLEM 4.145 CONTINUED From f.b.d. of the pipe assembly ΣM AD = 0:

λ AD ⋅ ( rG/ A × WAB ) + λ AD ⋅ ( rC/ A × T ) + λ AD ⋅ ( rB/ A × WBC ) + λ AD ⋅ ( rH / A × WCD ) = 0

3 −1 −4.5 3 −1 −4.5 T  1    ∴ 1.5 0 0   + 3 −1 0    5.5   5.5 45.25  −3 4 −4.5 0 −2.55 0 3 −1 −4.5 3 −1 −4.5  1   1  +3 −2.25  0 0  0 + 3 =0  5.5   5.5  0 −0.85 0 0 −3.825 0 

T   + (11.475 ) + ( 25.819 ) = 0  45.25 

(17.2125) + ( −40.5) 

∴ T = 9.0536 lb or T = 9.05 lb

PROBLEM 4.146 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 600-N load is applied at C as shown, determine the tension in the cable.

SOLUTION First note λ AE =

− ( 70 mm ) i + ( 240 mm ) k

( 70 )

2

=

+ ( 240 ) mm 2

1 ( −7i + 24k ) 25

rC/ A = ( 90 mm ) i + (100 mm ) k FC = − ( 600 N ) j rD/ A = ( 90 mm ) i + ( 240 mm ) k T = λ DFT =

=

− (160 mm ) i + (110 mm ) j − ( 80 mm ) k

(160 )2 + (110 )2 + (80 )2

T

mm

T ( −16 i + 11j − 8k ) 21

From the f.b.d. of the bend rod

(

)

(

)

ΣM AE = 0: λ AE ⋅ rC/ A × FC + λ AE ⋅ rD/ A × T = 0 ∴

−7 0 24 −7 0 24  600   T  90 0 100  =0  + 90 0 240  25 ( 21)   25   −16 11 −8 0 −1 0

( −700 − 2160 ) 

600   T   + (18 480 + 23 760 )   =0 25    25 ( 21) 

∴ T = 853.13 N or T = 853 N

PROBLEM 4.147 Solve Problem 4.146 assuming that cable DF is replaced by a cable connecting B and F. P4.146 The bent rod ABDE is supported by ball-and-socket joints at A and E and by the cable DF. If a 600-N load is applied at C as shown, determine the tension in the cable.

SOLUTION First note λ AE =

− ( 70 mm ) i + ( 240 mm ) k

( 70 )

2

=

+ ( 240 ) mm 2

1 ( −7i + 24k ) 25

rC/ A = ( 90 mm ) i + (100 mm ) k FC = − ( 600 N ) j rB/ A = ( 90 mm ) i T = λ BFT =

=

− (160 mm ) i + (110 mm ) j + (160 mm ) k

(160 )2 + (110 )2 + (160 )2

T

mm

1 ( −160 i + 110 j + 160k ) 251.59

From the f.b.d. of the bend rod

(

)

(

)

ΣM AE = 0: λ AE ⋅ rC/ A × FC + λ AE ⋅ rB/ A × T = 0 −7 0 24 −7 0 24 T  600    ∴ 90 0 100  0 0  =0  + 90 25 ( 251.59 )   25   0 −1 0 −160 110 160

( −700 − 2160 ) 

600  T   =0  + ( 237 600 )  25 251.59  25  )   ( ∴ T = 1817.04 N or T = 1817 N

PROBLEM 4.148 Two rectangular plates are welded together to form the assembly shown. The assembly is supported by ball-and-socket joints at B and D and by a ball on a horizontal surface at C. For the loading shown, determine the reaction at C.

SOLUTION

λ BD =

First note

=

− ( 80 mm ) i − ( 90 mm ) j + (120 mm ) k

(80 )2 + ( 90 )2 + (120 )2

mm

1 ( −8i − 9 j + 12k ) 17

rA/B = − ( 60 mm ) i P = ( 200 N ) k rC/D = ( 80 mm ) i C = (C ) j From the f.b.d. of the plates

(

)

(

)

ΣM BD = 0: λ BD ⋅ rA/B × P + λ BD ⋅ rC/D × C = 0 ∴

−8 −9 12 −8 −9 12  60 ( 200 )   C ( 80 )  1 0 0  −1 0 0  +   =0  17   17  0 0 1 0 1 0

( −9 )( 60 )( 200 ) + (12 )(80 ) C = 0 ∴ C = 112.5 N

or C = (112.5 N ) j

PROBLEM 4.149 Two 1 × 2-m plywood panels, each of mass 15 kg, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.

SOLUTION

(

)

W1 = W2 = − ( mg ) j = − (15 kg ) 9.81 m/s 2 j

Let

= − (147.15 N ) j From the f.b.d. of the panels

(

)

(

)

(

=

1 ( 2i − j − 2k ) 3

)

ΣM AF = 0: λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rT / A × W2 = 0 where

λ AF =

( 2 m ) i − (1 m ) j − ( 2 m ) k ( 2 )2 + (1)2 + ( 2 )2 m

rG/ A = (1 m ) i rB/ A = ( 2 m ) i rI / A = ( 2 m ) i − (1 m ) k

PROBLEM 4.149 CONTINUED λ BH =

( x − 2) i + ( y ) j − ( 2) k ( x − 2 ) 2 + y 2 + ( 2 )2

T = λ BH T =

∴

( x − 2) i + ( y ) j − ( 2) k ( x − 2 ) 2 + y 2 + ( 2 )2

2 −1 −2 2 −1 −2 T  147.15   1 0 0  0 0  + 2 2 2 2  3   0 −1 0 x − 2 y −2  3 ( x − 2 ) + y + ( 2 )

2 −1 −2   147.15   + 2 0 −1  3  = 0   0 −1 0

2 (147.15 ) T 147.15 + ( −4 − 4 y ) + ( −2 + 4 ) =0 2 2 3 3 3 ( x − 2) + y 2 + ( 2) T =

or For x − 2 m, T = Tmin

147.15 ( x − 2 ) 2 + y 2 + ( 2 )2 1+ y ∴ Tmin =

1 147.15 2 ( y + 4) 2 (1 + y )

−1

The y-value for Tmin is found from

 dT   dy  = 0:  

Setting the numerator equal to zero,

(1 + y ) y

1

1 (1 + y ) ( y 2 + 4 ) 2 ( 2 y ) − ( y 2 + 4 ) 2 (1) 2

(1 + y )2

=0

= y2 + 4

y = 4m Then

T min =

147.15 ( 2 2 2 2 − 2 ) + ( 4 ) + ( 2 ) = 131.615 N (1 + 4 )

∴ (a)

x = 2.00 m, y = 4.00 m

(b)

Tmin = 131.6 N

PROBLEM 4.150 Solve Problem 4.149 subject to the restriction that H must lie on the y axis.

P4.149 Two 1 × 2-m plywood panels, each of mass 15 kg, are nailed together as shown. The panels are supported by ball-and-socket joints at A and F and by the wire BH. Determine (a) the location of H in the xy plane if the tension in the wire is to be minimum, (b) the corresponding minimum tension.

SOLUTION

(

)

W1 = W2 = − ( mg ) j = − (15 kg ) 9.81 m/s 2 j = − (147.15 N ) j

Let From the f.b.d. of the panels

(

)

(

)

(

=

1 ( 2i − j − 2k ) 3

)

ΣM AF = 0: λ AF ⋅ rG/ A × W1 + λ AF ⋅ rB/ A × T + λ AF ⋅ rI / A × W2 = 0 where

λ AF =

( 2 m ) i − (1 m ) j − ( 2 m ) k ( 2 )2 + (1)2 + ( 2 )2 m

rG/ A = (1 m ) i rB/ A = ( 2 m ) i rI / A = ( 2 m ) i − (1 m ) k T = λ BH T =

=

− (2 m) i + ( y) j − (2 m) k

( 2 )2 + ( y )2 + ( 2 )2 T 8 + y2

( −2i +

m

yj − 2k )

T

PROBLEM 4.150 CONTINUED ∴

2 −1 −2 2 −1 −2 T  147.15   1 0 0 + 2 0 0  2   3  0 −1 0 −2 y −2  3 8 + y

2 −1 −2   147.15   + 2 0 −1  3  = 0  0 −1 0

)

(

2 (147.15 ) + ( −4 − 4 y ) T 8 + y 2 + ( 2 )147.15 = 0 ∴ T =

For Tmin ,

 dT   =0  dy 

∴

(147.15) 8 + (1 + y ) (1 + y ) 12 (8 +

y2

) ( 2 y ) − (8 + (1 + y )2

y2

− 12

y2

)

1 2

(1)

=0

Setting the numerator equal to zero,

(1 + y ) y

= 8 + y2

∴ y = 8.00 m and

Tmin =

(147.15) 8 + (8)2 (1 + 8)

= 138.734 N

∴ (a)

x = 0, y = 8.00 m

(b)

Tmin = 138.7 N

PROBLEM 4.151 A uniform 20 × 30-in. steel plate ABCD weighs 85 lb and is attached to ball-and-socket joints at A and B. Knowing that the plate leans against a frictionless vertical wall at D, determine (a) the location of D, (b) the reaction at D.

SOLUTION (a) Since rD/ A is perpendicular to rB/ A , rD/ A ⋅ rB/ A = 0 where coordinates of D are ( 0, y, z ) , and

rD/ A = − ( 4 in.) i + ( y ) j + ( z − 28 in.) k rB/ A = (12 in.) i − (16 in.) k ∴ rD/ A ⋅ rB/ A = −48 − 16 z + 448 = 0 or

z = 25 in.

Since

LAD = 30 in. 30 =

( 4 )2 + ( y )2 + ( 25 − 28)2

900 = 16 + y 2 + 9 y =

or

875 in. = 29.580 in.

∴ Coordinates of D :

x = 0, y = 29.6 in., z = 25.0 in.

(b) From f.b.d. of steel plate ABCD

ΣM AB = 0: where

λ AB =

λ AB ⋅ ( rD/ A × N D ) + λ AB ⋅ ( rG/B × W ) = 0

(12 in.) i − (16 in.) k (12 )2 + (16 )2 in.

=

1 ( 3i − 4k ) 5

rD/ A = − ( 4 in.) i + ( 29.580 in.) j − ( 3 in.) k

N D = N Di

PROBLEM 4.151 CONTINUED rG/B =

1 1 rD/B =  − (16 in.) i + ( 29.580 in.) j + ( 25 in. − 12 in.) k  2 2

W = − ( 85 lb ) j 3 0 3 0 −4 −4  ND   85  ∴ −4 29.580 −3  =0  + −16 29.580 13  2 ( 5 )   5   1 0 0 0 0 −1 118.32 N D + ( 39 − 64 ) 42.5 = 0 ∴ N D = 8.9799 lb or N D = ( 8.98 lb ) i

PROBLEM 4.152 Beam AD carries the two 40-lb loads shown. The beam is held by a fixed support at D and by the cable BE which is attached to the counter-weight W. Determine the reaction at D when (a) W = 100 lb, (b) W = 90 lb.

SOLUTION

(a) W = 100 lb From f.b.d. of beam AD ΣFx = 0: Dx = 0 ΣFy = 0: Dy − 40 lb − 40 lb + 100 lb = 0 ∴ Dy = −20.0 lb or D = 20.0 lb M D − (100 lb )( 5 ft ) + ( 40 lb )( 8 ft )

ΣM D = 0:

+ ( 40 lb )( 4 ft ) = 0 ∴ M D = 20.0 lb ⋅ ft or M D = 20.0 lb ⋅ ft (b) W = 90 lb From f.b.d. of beam AD ΣFx = 0: Dx = 0 ΣFy = 0: Dy + 90 lb − 40 lb − 40 lb = 0 ∴ Dy = −10.00 lb or D = 10.00 lb ΣM D = 0:

M D − ( 90 lb )( 5 ft ) + ( 40 lb )( 8 ft ) + ( 40 lb )( 4 ft ) = 0 ∴ M D = −30.0 lb ⋅ ft or M D = 30.0 lb ⋅ ft

PROBLEM 4.153 For the beam and loading shown, determine the range of values of W for which the magnitude of the couple at D does not exceed 40 lb ⋅ ft.

SOLUTION For Wmin ,

M D = −40 lb ⋅ ft

From f.b.d. of beam AD ΣM D = 0:

( 40 lb )(8 ft ) − Wmin ( 5 ft ) + ( 40 lb )( 4 ft ) − 40 lb ⋅ ft

=0

∴ Wmin = 88.0 lb For Wmax ,

M D = 40 lb ⋅ ft

From f.b.d. of beam AD ΣM D = 0:

( 40 lb )(8 ft ) − Wmax ( 5 ft ) + ( 40 lb )( 4 ft ) + 40 lb ⋅ ft

=0

∴ Wmax = 104.0 lb or 88.0 lb ≤ W ≤ 104.0 lb

PROBLEM 4.154 Determine the reactions at A and D when β = 30°.

SOLUTION From f.b.d. of frame ABCD ΣM D = 0: − A ( 0.18 m ) + (150 N ) sin 30°  ( 0.10 m ) + (150 N ) cos 30° ( 0.28 m ) = 0

∴ A = 243.74 N or A = 244 N ΣFx = 0:

( 243.74 N ) + (150 N ) sin 30° + Dx

=0

∴ Dx = −318.74 N ΣFy = 0: Dy − (150 N ) cos 30° = 0 ∴ Dy = 129.904 N Then and

D=

( Dx )2 + Dx2

=

( 318.74 )2 + (129.904 )2

= 344.19 N

 Dy  −1  129.904   = tan   = −22.174° D  −318.74   x

θ = tan −1 

or D = 344 N

22.2°

PROBLEM 4.155 Determine the reactions at A and D when β = 60°.

SOLUTION From f.b.d. of frame ABCD ΣM D = 0: − A ( 0.18 m ) + (150 N ) sin 60° ( 0.10 m ) + (150 N ) cos 60° ( 0.28 m ) = 0

∴ A = 188.835 N or A = 188.8 N

(188.835 N ) + (150 N ) sin 60° + Dx

ΣFx = 0:

=0

∴ Dx = −318.74 N ΣFy = 0: Dy − (150 N ) cos 60° = 0 ∴ Dy = 75.0 N Then and

D=

( Dx )2 + ( Dy )

2

=

( 318.74 )2 + ( 75.0 )2

= 327.44 N

 Dy  −1  75.0   = tan   = −13.2409°  −318.74   Dx 

θ = tan −1 

or D = 327 N

13.24°

PROBLEM 4.156 A 2100-lb tractor is used to lift 900 lb of gravel. Determine the reaction at each of the two (a) rear wheels A, (b) front wheels B.

SOLUTION

(a) From f.b.d. of tractor ΣM B = 0:

( 2100 lb )( 40 in.) − ( 2 A)( 60 in.) − ( 900 lb )( 50 in.) = 0 ∴ A = 325 lb

or A = 325 lb

(b) From f.b.d. of tractor ΣM A = 0:

( 2B )( 60 in.) − ( 2100 lb )( 20 in.) − ( 900 lb )(110 in.) = 0 ∴ B = 1175 lb

or B = 1175 lb

PROBLEM 4.157 A tension of 5 lb is maintained in a tape as it passes the support system shown. Knowing that the radius of each pulley is 0.4 in., determine the reaction at C.

SOLUTION

From f.b.d. of system ΣFx = 0: C x + ( 5 lb ) = 0 ∴ Cx = −5 lb ΣFy = 0: C y − ( 5 lb ) = 0 ∴ C y = 5 lb Then and

C =

( Cx )2 + ( C y )

2

=

( 5 )2 + ( 5 )2

= 7.0711 lb

 +5 

θ = tan −1   = −45°  −5  or C = 7.07 lb

45.0°

ΣM C = 0: M C + ( 5 lb )( 6.4 in.) + ( 5 lb )( 2.2 in.) = 0 ∴ M C = −43.0 lb ⋅ in

or M C = 43.0 lb ⋅ in.

PROBLEM 4.158 Solve Problem 4.157 assuming that 0.6-in.-radius pulleys are used. P4.157 A tension of 5 lb is maintained in a tape as it passes the support system shown. Knowing that the radius of each pulley is 0.4 in., determine the reaction at C.

SOLUTION From f.b.d of system ΣFx = 0: C x + ( 5 lb ) = 0

∴ Cx = −5 lb ΣFy = 0: C y − ( 5 lb ) = 0 ∴ C y = 5 lb Then and

C =

( C x )2 + ( C y )

2

=

( 5 )2 + ( 5 )2

= 7.0711 lb

 5 

θ = tan −1   = −45.0°  −5  or C = 7.07 lb

45.0°

ΣM C = 0: M C + ( 5 lb )( 6.6 in.) + ( 5 lb )( 2.4 in.) = 0 ∴ M C = −45.0 lb ⋅ in. or M C = 45.0 lb ⋅ in.

PROBLEM 4.159 The bent rod ABEF is supported by bearings at C and D and by wire AH. Knowing that portion AB of the rod is 250 mm long, determine (a) the tension in wire AH, (b) the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

SOLUTION (a) From f.b.d. of bent rod

(

)

(

)

ΣM CD = 0: λ CD ⋅ rH /B × T + λ CD ⋅ rF /E × F = 0 where

λ CD = i rH /B = ( 0.25 m ) j T = λ AH T

=

( y AH ) j − ( z AH ) k T ( y AH )2 + ( z AH )2

y AH = ( 0.25 m ) − ( 0.25 m ) sin 30°

= 0.125 m z AH = ( 0.25 m ) cos 30°

= 0.21651 m ∴T=

T ( 0.125j − 0.21651k ) 0.25

rF /E = ( 0.25 m ) k F = −400 N j 1 0 0 1 0 0  T  1 0 0.25 ∴ 0 + ( )  0 0 1 ( 0.25 )( 400 N ) = 0  0.25  0 0.125 −0.21651 0 −1 0 −0.21651T + 0.25 ( 400 N ) = 0 ∴ T = 461.88 N or T = 462 N

PROBLEM 4.159 CONTINUED (b) From f.b.d. of bent rod ΣFx = 0: C x = 0

ΣM D( z -axis ) = 0: − ( 461.88 N ) sin 30°  ( 0.35 m ) − C y ( 0.3 m ) − ( 400 N )( 0.05 m ) = 0 ∴ C y = −336.10 N ΣM D( y -axis ) = 0: Cz ( 0.3 m ) − ( 461.88 N ) cos 30°  ( 0.35 m ) = 0 ∴ Cz = 466.67 N or C = − ( 336 N ) j + ( 467 N ) k ΣFy = 0: Dy − 336.10 N + ( 461.88 N ) sin 30° − 400 N = 0 ∴ Dy = 505.16 N ΣFz = 0: Dz + 466.67 N − ( 461.88 N ) cos30° = 0 ∴ Dz = −66.670 N or D = ( 505 N ) j − ( 66.7 N ) k

PROBLEM 4.160 For the beam and loading shown, determine (a) the reaction at A, (b) the tension in cable BC.

SOLUTION (a) From f.b.d of beam ΣFx = 0: Ax = 0 ΣM B = 0:

(15 lb )( 28 in.) + ( 20 lb )( 22 in.) + ( 35 lb )(14 in.) + ( 20 lb )( 6 in.) − Ay ( 6 in.) = 0 ∴ Ay = 245 lb or A = 245 lb

(b) From f.b.d of beam ΣM A = 0:

(15 lb )( 22 in.) + ( 20 lb )(16 in.) + ( 35 lb )(8 in.) − (15 lb )( 6 in.) − TB ( 6 in.) = 0 ∴ TB = 140.0 lb or TB = 140.0 lb

Check: ΣFy = 0: −15 lb − 20 lb − 35 lb − 20 lb − 15 lb − 140 lb + 245 lb = 0? 245 lb − 245 lb = 0 ok

PROBLEM 4.161 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. For a = 150 mm, determine the tension in each cable and the reaction at A.

SOLUTION First note

TDG = λ DGTDG =

− ( 0.48 m ) i + ( 0.14 m ) j

( 0.48)2 + ( 0.14 )2

=

−0.48i + 0.14 j TDG 0.50

=

TDG ( 24i + 7 j) 25

TBE = λ BETBE =

m

− ( 0.48 m ) i + ( 0.2 m ) k

( 0.48)2 + ( 0.2 )2 m

=

−0.48i + 0.2k TBE 0.52

=

TBE ( −12 j + 5k ) 13

TDG

TBE

From f.b.d. of frame ABCD  7  ΣM x = 0:  TDG  ( 0.3 m ) − ( 350 N )( 0.15 m ) = 0  25  or TDG = 625 N  24   5  ΣM y = 0:  × 625 N  ( 0.3 m ) −  TBE  ( 0.48 m ) = 0 13  25    or TBE = 975 N  7  ΣM z = 0: TCF ( 0.14 m ) +  × 625 N  ( 0.48 m )  25  − ( 350 N )( 0.48 m ) = 0 or TCF = 600 N

PROBLEM 4.161 CONTINUED ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0  12   24  Ax − 600 N −  × 975 N  −  × 625 N  = 0  13   25  ∴ Ax = 2100 N ΣFy = 0: Ay + (TDG ) y − 350 N = 0  7  Ay +  × 625 N  − 350 N = 0 25   ∴ Ay = 175.0 N ΣFz = 0: Az + (TBE ) z = 0 5  Az +  × 975 N  = 0  13  ∴ Az = −375 N Therefore

A = ( 2100 N ) i + (175.0 N ) j − ( 375 N ) k

PROBLEM 4.162 Frame ABCD is supported by a ball-and-socket joint at A and by three cables. Knowing that the 350-N load is applied at D (a = 300 mm), determine the tension in each cable and the reaction at A.

SOLUTION First note

TDG = λ DGTDG =

− ( 0.48 m ) i + ( 0.14 m ) j

( 0.48)2 + ( 0.14 )2 m

=

−0.48i + 0.14 j TDG 0.50

=

TDG ( 24i + 7 j) 25

TBE = λ BETBE =

− ( 0.48 m ) i + ( 0.2 m ) k

( 0.48)2 + ( 0.2 )2 m

=

−0.48i + 0.2k TBE 0.52

=

TBE ( −12i + 5k ) 13

TDG

TBE

From f.b.d of frame ABCD  7  ΣM x = 0:  TDG  ( 0.3 m ) − ( 350 N )( 0.3 m ) = 0  25  or TDG = 1250 N  24   5  ΣM y = 0:  × 1250 N  ( 0.3 m ) −  TBE  ( 0.48 m ) = 0 13  25    or TBE = 1950 N  7  ΣM z = 0: TCF ( 0.14 m ) +  × 1250 N  ( 0.48 m )  25  − ( 350 N )( 0.48 m ) = 0 or TCF = 0

PROBLEM 4.162 CONTINUED ΣFx = 0: Ax + TCF + (TBE ) x + (TDG ) x = 0  12   24  Ax + 0 −  × 1950 N  −  × 1250 N  = 0  13   25  ∴ Ax = 3000 N ΣFy = 0: Ay + (TDG ) y − 350 N = 0  7  Ay +  × 1250 N  − 350 N = 0 25   ∴ Ay = 0 ΣFz = 0: Az + (TBE ) z = 0 5  Az +  × 1950 N  = 0  13  ∴ Az = −750 N Therefore

A = ( 3000 N ) i − ( 750 N ) k

PROBLEM 4.163 In the problems listed below, the rigid bodies considered were completely constrained and the reactions were statically determinate. For each of these rigid bodies it is possible to create an improper set of constraints by changing a dimension of the body. In each of the following problems determine the value of a which results in improper constraints. (a) Problem 4.81, (b) Problem 4.82.

SOLUTION (a)

ΣM B = 0:

(a)

( 300 lb )(16 in.) − T (16 in.) + T ( a ) = 0 T =

or

( 300 lb )(16 in.) (16 − a ) in.

∴ T becomes infinite when 16 − a = 0 or a = 16.00 in.

ΣM C = 0:

(b) (b)

(T

 8  − 80 N )( 0.2 m ) −  T  ( 0.175 m )  17 

 15  −  T  ( 0.4 m − a ) = 0  17  0.2T − 16.0 − 0.82353T − 0.35294T + 0.88235Ta = 0 or

T =

16.0 0.88235a − 0.23529

∴ T becomes infinite when 0.88235a − 0.23529 = 0

a = 0.26666 m or a = 267 mm