Section 5.5 Free Body Diagrams

Section 5.5 Free body diagrams

• Bodies connected by string • Motion on an inclined plane • Motion with a change in direction • Forces in equilibrium © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 84)

Free body diagrams F1

F4

F2 F3

Free body diagram — indicate all the forces acting on a body

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5.5 Free body diagram (SB p. 84)

Bodies connected by string Free body diagram:

• R – reaction force • T – tension in string • mg – weight • F – external force © Manhattan Press (H.K.) Ltd.

Consider net force in vertical and horizontal directions 3

5.5 Free body diagram (SB p. 85)

(1) Vertical direction no motion

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Consider m1

R1 = m1g

Consider m2

R2 = m2g 4

5.5 Free body diagram (SB p. 85)

Horizontal direction (2) Horizontal direction: Not equilibrium

By F = ma Consider m1

T = m1a

Consider m2

F - T = m2a The string is stretched, a of m1 and m2 are the same

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5.5 Free body diagram (SB p. 85)

Bodies connected by string The boxes as a whole system:

F = (m1 + m2 + m3) a

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5.5 Free body diagram (SB p. 86)

Example 2: Two blocks A and B connected by a string are moving on a smooth horizontal surface as shown in Fig. (a). Block A is pulled by a force of 10 N.

Fig. (a) (a) Find the tension in the string and the acceleration of the blocks. Solut ion © Manhattan Press (H.K.) Ltd. 7

5.5 Free body diagram (SB p. 86)

Example 2: (Cont) Let a and T be the acceleration of blocks and tension in the string respectively. (a) By Newton's second law of motion, F = ma, Consider block A, 10 − T = 1 × a .................... (1) Consider block B, T = 4 × a .................... (2) Solve (1) and (2), we have T = 8 N and a = 2 m s−2.

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5.5 Free body diagram (SB p. 86)

Example 2: (Cont) (b) If the positions of block A and block B are exchanged, and block B is pulled by a force of 10 N towards right. Find their acceleration and the tension in the string. Solut ion

Since the total forces acting on the blocks (10 N) and their masses remain the same, a does not change and is equal to 2 m s−2. Consider block A, T = ma =1× 2 =2N © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 87)

Motion on an inclined plane Free body diagram: y

CAL Workshop 2 Motion on a frictionless inclined plane

x

Consider net force direction (parallel and perpendicular to the plane) © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 87)

The block is stationary no net force y

x

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x:

Wx − f = 0

y:

Wy − R = 0 11

5.5 Free body diagram (SB p. 87)

The block slides down y

x

(i) In x direction (parallel to plane), Force acting on the body (F ) = ma mg sin θ − f = ma mg sin θ − f a= m

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5.5 Free body diagram (SB p. 88)

The block slides down y

(ii) in y direction (perpendicular to plane), F = ma

x

R − mg cosθ = 0 cosθ

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∴

R = mg

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5.5 Free body diagram (SB p. 88)

Class Practice 4: 4 Find the unknown forces (W and Wy) in the figure.

Ans wer Wx = W sin25o

15 = 35.5 N o W = ___________________________ sin25 o W cos25 = 32.2 N Wy = ___________________________ © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 88)

Example 3: A trolley of mass 0.6 kg runs down a friction-compensated runway at a constant speed as shown in Fig. (a).

Fig. (a)

(a) Find the friction (f ) acting on the trolley. Solut ion © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 88)

Example 3: (Cont) (a) The forces acting on the trolley along the direction of motion are shown in Fig. (b). Fig. (b)

When the trolley runs down the friction-compensated runway, its speed is constant, hence a = 0 m s−2. The net force acting on the trolley is zero. By F = mg sinθ − f 0 = mg sinθ − f f = mg sin 5° = 0.6 × 10 × sin 5° = 0.52 N © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 89)

Example 3: (Cont) (b) If the trolley is then pulled by a force (T ) of 3 N along the runway, find the acceleration of the trolley. Assume that Solut the friction remains the same. ion when it is pulled by a Fig. (c) shows the forces acting on the trolley force (T ) of 3 N. Fig. (c)

As it is a friction-compensated runway, the component mg sinθ is balanced by f . Therefore, only the pulling force 3 N is exerted on the trolley. By F = ma a=F/m = 3 / 0.6 ∴ a = 5 m s−2 © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 90)

Class Practice 5: 5 A trolley of mass 1 kg runs down a runway with an acceleration of 0.2 m s−2 as shown in the figure. (a) Find the friction acting on the trolley. Ans

mg sin θ − f = ma

wer

1 x 10 x sin 25o

1x ___________ − f = __________ _ 0.2 4.0 f = __________ _N 3N © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 90)

Class Practice 5 (Cont): (Cont) (b) Find the perpendicular force exerted on the trolley by the runway. Ans wer mg cos θ R = _________________________ 1 x 10 x cos25o = _________________________ 9.06 N = _________________________ © Manhattan Press (H.K.) Ltd.

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5.5 Free body diagram (SB p. 90)

Motion with a change in direction

F acted to left, -F = ma F a=- m By v = u + at = u - Ft m © Manhattan Press (H.K.) Ltd.

mu = F

The trolley stops at t After that, it moves towards left.

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5.5 Free body diagram (SB p. 90)

Motion with a change in direction

when opposite force acting on object ⇒ may change direction

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5.5 Free body diagram (SB p. 91)

Example 4: A 10-kg monkey hangs from the middle of a massless rope as shown. If the rope is at rest, find the tension in the rope.

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Solut ion

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5.5 Free body diagram (SB p. 91)

Example 4: (Cont) If the rope is at rest, the net forces acting on it in the directions of x-axis and y-axis are zero. Resolve the forces in these two directions. x component: T cosθ + (−T cosθ ) = 0 y component: T sinθ + T sinθ + (−mg) = 0 Consider y component, we have 2T sin15° − (10)(10) = 0 T = 193.2 N

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5.5 Free body diagram (SB p. 92)

Class Practice 6 : A 18 kg bag is hung by two light ropes as shown. If the bag is stationary, find the tension in the rope. Consider y component, we have 2T sin θ − mg = 0 2T sin22o − (18)(10) = 0 T = 240.3 N

Ans wer © Manhattan Press (H.K.) Ltd.

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To section 5.6

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