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4 Marking scheme: Worksheet (AS) 1

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a

R2 = 7.02 + 5.02 R = 49 + 25 = 8.6 N

[1] [1]

b Force vertically = 40 − 10 = 30 N and force horizontally = 80 − 20 = 60 N R2 = 302 + 602 R = 900 + 3600 ≈ 67 N 7

a

Fx = F cos θ = 10 cos 45° Fx = 7.07 N ≈ 7.1 N

[1]

Fy = F sin θ = 10 sin 45° F y = 7.07 N ≈ 7.1 N

[1]

b Fx = F cos θ = 85 cos 20° Fx = 79.9 N ≈ 80 N

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9

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Fy = F sin θ = 85 sin 20° Fy = 29.1 N ≈ 29 N

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a

moment = force × perpendicular distance from the pivot to the line of action of the force

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b

The net force acting on the object is zero. The sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.

[1]

The centre of gravity of the beam is 0.75 m away from support A. sum of clockwise moments = sum of anticlockwise moments 60 × 0.75 = RB × 1.5 RB = 30 N b Net force in the vertical direction = 0 RA + RB = 60; RA = 60 − 30 = 30 N

[1]

a

AS and A Level Physics

Original material © Cambridge University Press 2010

[1] [1] [1] [1]

1

4 Marking scheme: Worksheet (AS)

10 a

The net force is zero because the seat is in equilibrium.

[1]

b i

Correct diagram

[1]

Weight = mg = 35 × 9.81 ≈ 343 N

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T2 = 1802 + 3432

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T = 180 2 + 343 2 ≈ 390 N

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180 343 −1 θ = tan (0.525) ≈ 28°

ii tan θ =

11 a

[1] [1]

Fx = F cos θ = 300 cos 30º Fx ≈ 260 N

b The net force is zero, because the roller is moving at constant velocity. Resistive force ≈ 260 N to the left. c

Fy = F sin θ = 300 sin 30° = 150 N

[1] [1] [1] [1] [1]

The net vertical force is zero. [1] [1]

150 + contact force = mg contact force = (50 × 9.81) − 150 ≈ 340 N The line of action of the weight of the ladder is at a perpendicular distance of 0.75 m away from the foot of the ladder. Taking moments about the base of the ladder ⇒ sum of clockwise moments = sum of anticlockwise moments (32 × 9.81) × 0.75 = R × 4.0 32 × 9.81 × 0.75 R= ≈ 59 N 4.0 b The force at the base of the ladder creates zero moment about this point.

12 a

AS and A Level Physics

Original material © Cambridge University Press 2010

[1] [1] [1] [1]

2

4 Marking scheme: Worksheet (AS)

13 a

All forces clearly shown on the diagram.

b Taking moments about the brick ⇒ sum of clockwise moments = sum of anticlockwise moments (62 × 9.81)x + (15 × 9.81) × 0.78 = (30 × 9.81) × 1.56 (1.56 × 30) − (0.78 × 15) x= ≈ 0.57 m 62 x ≈ 0.57 m Distance of centre of gravity from the toes = 1.56 − 0.57 = 0.99 m

AS and A Level Physics

Original material © Cambridge University Press 2010

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View more...
D

[1]

2

D

[1]

3

C

[1]

4

C

[1]

5

A

[1]

6

a

R2 = 7.02 + 5.02 R = 49 + 25 = 8.6 N

[1] [1]

b Force vertically = 40 − 10 = 30 N and force horizontally = 80 − 20 = 60 N R2 = 302 + 602 R = 900 + 3600 ≈ 67 N 7

a

Fx = F cos θ = 10 cos 45° Fx = 7.07 N ≈ 7.1 N

[1]

Fy = F sin θ = 10 sin 45° F y = 7.07 N ≈ 7.1 N

[1]

b Fx = F cos θ = 85 cos 20° Fx = 79.9 N ≈ 80 N

8

9

[1] [1] [1]

[1]

Fy = F sin θ = 85 sin 20° Fy = 29.1 N ≈ 29 N

[1]

a

moment = force × perpendicular distance from the pivot to the line of action of the force

[1]

b

The net force acting on the object is zero. The sum of the clockwise moments about any point is equal to the sum of the anticlockwise moments about the same point.

[1]

The centre of gravity of the beam is 0.75 m away from support A. sum of clockwise moments = sum of anticlockwise moments 60 × 0.75 = RB × 1.5 RB = 30 N b Net force in the vertical direction = 0 RA + RB = 60; RA = 60 − 30 = 30 N

[1]

a

AS and A Level Physics

Original material © Cambridge University Press 2010

[1] [1] [1] [1]

1

4 Marking scheme: Worksheet (AS)

10 a

The net force is zero because the seat is in equilibrium.

[1]

b i

Correct diagram

[1]

Weight = mg = 35 × 9.81 ≈ 343 N

[1]

T2 = 1802 + 3432

[1]

T = 180 2 + 343 2 ≈ 390 N

[1]

180 343 −1 θ = tan (0.525) ≈ 28°

ii tan θ =

11 a

[1] [1]

Fx = F cos θ = 300 cos 30º Fx ≈ 260 N

b The net force is zero, because the roller is moving at constant velocity. Resistive force ≈ 260 N to the left. c

Fy = F sin θ = 300 sin 30° = 150 N

[1] [1] [1] [1] [1]

The net vertical force is zero. [1] [1]

150 + contact force = mg contact force = (50 × 9.81) − 150 ≈ 340 N The line of action of the weight of the ladder is at a perpendicular distance of 0.75 m away from the foot of the ladder. Taking moments about the base of the ladder ⇒ sum of clockwise moments = sum of anticlockwise moments (32 × 9.81) × 0.75 = R × 4.0 32 × 9.81 × 0.75 R= ≈ 59 N 4.0 b The force at the base of the ladder creates zero moment about this point.

12 a

AS and A Level Physics

Original material © Cambridge University Press 2010

[1] [1] [1] [1]

2

4 Marking scheme: Worksheet (AS)

13 a

All forces clearly shown on the diagram.

b Taking moments about the brick ⇒ sum of clockwise moments = sum of anticlockwise moments (62 × 9.81)x + (15 × 9.81) × 0.78 = (30 × 9.81) × 1.56 (1.56 × 30) − (0.78 × 15) x= ≈ 0.57 m 62 x ≈ 0.57 m Distance of centre of gravity from the toes = 1.56 − 0.57 = 0.99 m

AS and A Level Physics

Original material © Cambridge University Press 2010

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