Friction_flat_belts.pdf

7.4 Flat Belts

7.4 Flat Belts Example 1, page 1 of 2 1. Determine the minimum number of turns of rope that will allow the 5-lb force to support the 600-lb block, if the coefficient of static friction is 0.15.

1

Tensions in the rope

5 lb 5 lb

}

n turns

600 lb 600 lb

2

Impending motion (The motion can't be up, since the 5-lb force is too small to lift the 600-lb block.)

7.4 Flat Belts Example 1, page 2 of 2 3

Apply the equation for belt friction: T2 = T1e

4 (1)

In this equation, T2 is the tension in the direction of impending motion T2 = 600 lb The other tension, T1, is in the direction opposite the impending motion, so T1 = 5 lb

Since

= 0.15, Eq. 1 becomes 600 lb = (5 lb) e

Solving gives = 31.917 radians. If n is the number of turns of rope, then the number of radians is 2 n. Equating this to gives an equation for n: 2 n= = 31.917 rad Solving for n gives n = 5.08 turns Rounding off to the next higher integer then gives n=6

Ans.

7.4 Flat Belts Example 2, page 1 of 2 2. If the coefficient of static friction between the rope and the fixed circular drums A and B is 0.2, determine the largest value of the force P that can be applied without moving the 150-lb weight upwards.

1

Tensions in the rope on either side of drum A TAB Impending motion (The rope must move to the right if the 150-lb weight moves up)

=

P 60°

A 150 lb

B

2

Because slip impends between the rope and the drum, we can apply the equation for belt friction: T2 = T1e

(1)

where T2 is the tension in the direction of impending motion. In our particular problem, A

Rope is horizontal. T2 = TAB and the tension opposing the impending motion is

150 lb

T1 = 150 lb Using these results and

= 0.2 and

TAB = (150)e 0.2( /2) = 205.4 lb

=

2 in Eq. 1 gives

7.4 Flat Belts Example 2, page 2 of 2 3 Tensions in the rope on either side of drum B

4

Flat-belt friction equation:

Impending motion

T2 = T1e so,

P

P = (205.4 lb)e

(2)

60° B

5

Geometry

TAB = 205.4 lb

= 60° =

60 60° 30° 60° 6

Using

= 60° =

rad in Eq. 2 gives

P = (205.4)e = 253 lb

/ Ans.

rad

7.4 Flat Belts Example 3, page 1 of 3 3. Determine the smallest force P applied to the handle of the band brake that will prevent the drum from rotating when the 15 lb·ft moment is applied. The coefficient of static friction is 0.25, and the weight of lever arm ABC can be neglected.

1

Free-body diagram of drum 15 lb·ft = 15 lb·ft = 180 lb·in.

6 in. D

15 lb-ft

Dx

6 in.

TA

D 2 P 80

o

B

10 in.

Impending slip of band relative to drum (An observer on the drum would see the belt move down.)

TB

C

15 in.

3

+

A

o

80

Dy

Equilibrium equation for drum MD = 0: TA(6 in.)

TB(6 in.)

180 lb·in. = 0

(1)

12 in./ft

7.4 Flat Belts Example 3, page 2 of 3 4 The belt-friction equation is in general T2 = T1e

5

Geometry = 180° + 10°

(2)

where T2 is the tension in the direction of impending slip. In our particular problem,

= 190° 10° = (190/180) rad

10°

= 3.316 rad

80°

T2 = TA

80°

and the tension opposing the impending motion is T1 = TB

Eq. 3 becomes, with

3.316 and

0.25

TA = TBe0.25(3.316) Solving Eqs. 1 and 4 simultaneously gives

Eq. (2) becomes TA = TBe

(3)

TA = 53.237 lb TB = 23.237 lb

7.4 Flat Belts Example 3, page 3 of 3 6

Free-body diagram of lever arm ABC TA = 53.237 lb

TB sin 80° = (23.237 lb) sin 80°

A

B

P

TB cos 80°

Ax Ay 10 in.

+

7

15 in.

Equilibrium equation for lever arm MA = 0: [(23.237 lb) sin 80°](10 in.)

P(10 in. + 15 in.) = 0

Solving gives P = 9.15 lb

Ans.

(5)

7.4 Flat Belts Example 4, page 1 of 3 4. The uniform beam ABC weighs 40 lb. The coefficient of static friction between the cord and the fixed drum D is 0.3. Determine the smallest value of the weight W for which the beam will remain horizontal.

1

Free-body diagram of block B

D

TB (Tension in the cord)

2

C

B W

4 ft

W

N=0

4 ft 3

+

A

If end A of the beam is about to move down, then block B is about to lose contact with the beam. Thus the normal force N is zero.

Equilibrium equation for block B Fy = 0: TB

W=0

(1)

7.4 Flat Belts Example 4, page 2 of 3 4

Free-body diagram of beam y

TA is the tension in the cord.

5

N=0 A

Cy C

B

Cx

x

40 lb 4 ft

8

4 ft

Tensions in cord on either side of drum

6 The 40-lb weight of the beam acts through the center of the span. TA = 20 lb

Equilibrium equation for the beam

+

7

MC = 0: (40 lb)(4 ft)

TA(4 ft + 4 ft) = 0

(2)

TB Impending motion

Solving gives TA = 20 lb

9 Because the cord is on the verge of slipping, we can apply the equation for belt friction: T2 = T1e

(3)

where T2 = 20 lb is the tension in the direction of impending slip and T1 = TB is the tension opposite the direction of impending motion

7.4 Flat Belts Example 4, page 3 of 3 10 Geometry

=

TB

TA 11

With

and

0.3, Eq. 3 becomes

T2 = T1e 20 lb

(Eq. 3 repeated) 0.3

Solving gives TB = 7.79 lb Eq. 1 then gives W = TB = 7.79 lb

Ans.

7.4 Flat Belts Example 5, page 1 of 4 5. Determine the largest value W of the weight of block B for which neither block will move. The coefficients of static friction are 0.2 between the blocks and the planes, and 0.25 between the cord and the drum.

1

Free-body diagram of block A x

2

Impending motion of block A (Since we are to determine the "largest value of the weight" of block B, block B must be about to move down the plane. Thus block A must be about to move up the plane.)

TA A

y

fA B A 80 lb

80 lb

NA 70°

W 3

Equilibrium equations for block A:

50° Fx = 0: TA

+

+

70°

fA

F = 0: NA

y

(80 lb) sin

(80 lb) cos

=0

(1) (2)

Slip impends, so f A = f A-max = NA = (0.2)NA

(3)

7.4 Flat Belts Example 5, page 2 of 4 4

6 Tensions in cord on either side of drum

Geometry = 70°

20°

TB

A TA = 80.65 lb 20° Impending motion (Block B moves down plane)

70° 5

Solving Eqs. 1, 2, and 3, with

= 70° gives 7

TA = 80.65 lb f A = 5.47 lb NA = 27.36 lb

Because the cord is about to slip over the drum, we can apply the equation for belt friction: T2 = T1 e drum

(4)

where T2 = TB is the tension in the direction of impending slip, and T1 = 80.65 lb is the tension opposite to the direction of impending motion.

7.4 Flat Belts Example 5, page 3 of 4 8

Geometry

50° 70° 40° 20° 50° 70°

= 70° + 50° 70°

= 120° =

9

2 rad 3

Substituting = 2 /3, T2 = TB , T1 = 80.65 lb, and drum 0.25 in Eq. 4 gives TB = (80.65 lb) e Evaluating the right hand side yields TB = 136.1 lb

50°

7.4 Flat Belts Example 5, page 4 of 4 10 Free-body diagram of block B y

12 Equilibrium equations for block B: +

+

TB = 136.1 lb x

Fx' =0: NB

W cos

Fy' = 0: 136.1 lb + f B

=0 W sin

(5) =0

(6)

f B = f B-max

B fB

= NB W 50°

NB 11 Geometry

Impending motion

= (0.2)NB 13 Solving Eqs. 5, 6, and 7, with

(7) = 50° gives

f B = 27.4 lb NB = 137.2 lb

B

W = 213 lb

= 50° 40° 50°

Ans.

7.4 Flat Belts Example 6, page 1 of 4 6. A motor attached to pulley A drives the pulley clockwise with a 200 lb-in. torque. The flat belt then overcomes the resisting torque T at pulley B and rotates the pulley B clockwise. Determine the minimum tension that can exist in the belt without causing the belt to slip at pulley A. Also determine the corresponding resisting torque T. The coefficient of static friction between the belt and the pulleys is 0.3.

T 200 lb·in. A 4 in.

B 7 in.

Driving pulley Driven pulley 22 in.

7.4 Flat Belts Example 6, page 2 of 4 Free-body diagram of pulley A TC

C

3

+

1

Equilibrium equation for pulley A MA = 0: TD(4 in.)

TC(4 in.)

200 lb·in. = 0

(1)

200 lb·in. 4 in. A Ax

Because the belt is about to slip on the pulley, the belt friction equation applies:

Ay D

TD

T2 = T1e where T2 = TD, the tension in the direction of impending motion, and T1 = TC, so TD = TC e

2

Impending motion of belt relative to pulley (Since pulley A is driven by a clockwise torque, the pulley would slip in a clockwise sense relative to the belt. Thus an observer at point D on the pulley would see the belt move in the direction shown.)

(2)

7.4 Flat Belts Example 6, page 3 of 4 4

Geometry Radius = 7 in. 6

4 in.

Free-body diagram of pulley B

C 4 in.

7 in. 22 in.

A

4 in. = 3 in.

TC = 36.65 lb

B

From triangle ABC,

T

3 in. cos ( 2 ) = 22 in. Solving gives

= 164.33° = 2.868 rad.

So Eq. 2, with

= 0.3, becomes 0.3(2.868)

TD = TC e

Solving Eqs. 1 and 3 simultaneously gives

By

TC = 36.65 lb TD = 86.65 lb

Ans.

7 in.

TD = 86.65 lb (3) 7

Equilibrium equation for pulley B

+

5

Bx

MB = 0: T + (36.65 lb)7 in.

(86.65 lb)7 in. = 0

Solving gives T = 350 lb·in.

Ans.

7.4 Flat Belts Example 6, page 4 of 4 8

Finally, we must check that pulley B does not slip. But this follows from the observation that pulley B has a larger angle of wrap than pulley A. Thus it must be able to carry a maximum possible tension larger than the 86.5 lb maximum tension that pulley A carries. 9

More precisely, because '> we know that

36.65 lb tension

e B

A 86.65 lb tension

'

'

>e .

Multiplying through by 36.65 lb gives (36.65 lb)e

'

Tmax-B , the maximum possible tension that pulley B could support without slipping.

>

(36.65 lb)e

86.65 lb, by Eq. 2

Thus Tmax-B > 86.65 lb and pulley B won't slip.

7.4 Flat Belts Example 7, page 1 of 2 7. If the coefficient of static friction between the fixed drums D and E and the ropes is 0.35, determine the largest weight W that can be supported.

1 D

Free-body diagram of block C TD

TE

C

E

W

A 100 lb 2

Equilibrium equation for block C

50 lb

+

B C W

Fy = 0: TE + TD

W=0

(1)

7.4 Flat Belts Example 7, page 2 of 2 3

Rope tensions acting on drum D

6

Rope tensions acting on drum E (not drum D) =

= E D

TE

50 lb TD

100 lb

Impending motion 7

4

5

Impending motion (Since we are to determine the largest value of weight of block C that can be supported, block C must be about to move down, so the rope connected to C must also be about to move down.)

Flat-belt friction equation: T2 = T1e or,

Because slip is about to occur, the belt-friction equation applies:

TE = (50 lb)e0.35 = 150.1 lb

T2 = T1e Here, T2 = TD, T1 = 100 lb,

8 0.35, and

TD = (100 lb)e = 300.3 lb

so

Using the results of Eqs. 2 and 3 in Eq. 1 gives W = 450 lb

(2)

(3)

Ans.

7.4 Flat Belts Example 8, page 1 of 3 8. Pulley A is rotating under the action of a 6 N-m torque. This motion is transmitted through a flat belt to drive pulley B, which is in turn acted upon by a resisting torque T (the "load" on pulley B). The coefficient of static friction between the belt and the pulleys is 0.45. Determine a) the maximum possible value of T, b) the maximum force in the belt, and c) the corresponding force required in the spring C.

T

6 N·m

80 mm A B Driven pulley

80 mm Driving pulley

C

7.4 Flat Belts Example 8, page 2 of 3 1

Free-body diagram of pulley A

3

Equilibrium equations for pulley A

T2

+

Fx = 0: Ax

+

Fy = 0: Ay = 0

D 6 N-m

=

A

Ax

Ay T1

+

80 mm 4

T1 T2 = 0

MA = 0: T2(0.08 m)

(1) (2)

T1(0.08 m)

6 N·m = 0

Flat-belt friction equation: T2 = T1e = T1e

(4)

Solving Eqs. 1-4 simultaneously gives 2

Impending motion (Since the driving torque acting on pulley A is clockwise, if slip is about to occur, then pulley A will slip clockwise relative to the belt. An observer located on point D on pulley A would see the belt move in the direction shown.)

(3)

Ax = 123.21 N Ay = 0 T1 = 24.11 N T2 = 99.11 N

Ans.

7.4 Flat Belts Example 8, page 3 of 3 5

Free-body diagram of member AC

7

Free-body diagram of pulley B T2 = 99.11 N

Fspring

Ax = 123.21 N

+

Equilibrium equations for member AC Fx = 0: Fspring

123.21 N = 0

80 mm (4)

Bx

T B

Solving gives Fspring = 123.21 N

By Ans.

T1 = 24.11 N 8

Equilibrium equations for pulley B

+

6

MB = 0: T + (24.11 N)(0.08 m)

99

(0.08 m) = 0

Solving gives T = 6 N·m

Ans.

That is, the maximum resisting torque equals the driving torque.