Mechanics: Moments: Fluency and Exam Drill (Part 2) - Worksheets, Questions and Revision

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A-Level · Mechanics

M3DB Mechanics: Moments: Fluency and Exam Drill (Part 2)

EDEXCEL 9MA0 · Calculator allowed · about 50 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

State the condition that must be satisfied by the moments of the forces acting on a rigid body for the body to be in rotational equilibrium.

(Total for Question 1 is 1 mark)

A force of 32 N acts at right angles to a crowbar at a point 0.75 m from the pivot. Find the moment of the force about the pivot.

(Total for Question 2 is 1 mark)

A mechanic applies a force of 28 N at right angles to a car jack handle, at a point 1.5 m from the pivot. Find the moment of the force about the pivot.

(Total for Question 3 is 1 mark)

A moment is the turning effect of a force about a point. State the two quantities whose product gives the magnitude of a moment.

(Total for Question 4 is 1 mark)

Explain, in terms of the formula for a moment, why the unit of a moment is written as N m and not N/m.

(Total for Question 5 is 1 mark)

A garden gate is opened by a force of 55 N applied at an angle of 70 degrees to the gate, at a point 0.9 m from the hinge. Find the moment of the force about the hinge, giving your answer to 3 significant figures.

(Total for Question 6 is 2 marks)

A decorator uses a screwdriver as a lever to prise open a paint tin lid. She applies a force of 40 N to the screwdriver at an angle of 25 degrees to the screwdriver, at a distance of 0.14 m from the pivot (the rim of the tin). Find the moment of the force about the pivot, giving your answer to 3 significant figures.

(Total for Question 7 is 2 marks)

A uniform plank is pivoted at its centre. A force of 72 N acts vertically downward at a point 1.05 m from the pivot on one side. Find the distance from the pivot at which a force of 84 N must act vertically downward on the other side, for the plank to remain in equilibrium.

(Total for Question 8 is 2 marks)

A force applied at right angles to a bicycle brake lever, at a distance of 0.08 m from the pivot, produces a moment of 3.6 N m. Find the magnitude of the force.

(Total for Question 9 is 2 marks)

A force of 15 N acting at right angles to a door produces a moment of 12 N m about the hinge. Find the distance from the hinge to the point where the force acts.

(Total for Question 10 is 2 marks)

A uniform strut AB rests horizontally on two smooth supports, one at each end. The strut has weight 130 N. Given that the reaction at end A is 55 N, find the reaction at end B.

(Total for Question 11 is 2 marks)

The table below shows three forces acting at right angles to a light rod that is free to rotate about a fixed pivot O.

Force	Perpendicular distance from O	Sense
18 N	0.4 m	clockwise
25 N	0.5 m	anticlockwise
30 N	0.3 m	clockwise

Find the magnitude of the resultant moment about O, and state its sense.

(Total for Question 12 is 2 marks)

A rod is described as non-uniform. State what this tells you about the position of its centre of mass, compared with a uniform rod of the same length.

(Total for Question 13 is 1 mark)

A non-uniform beam AB has length 9 m and weight 240 N, acting at a point 4 m from A. The beam rests horizontally in equilibrium on two smooth supports, one at C, 1 m from A, and one at D, 7 m from A. By taking moments about C, find the reaction at D.

(Total for Question 14 is 3 marks)

A student is asked to find the reaction at support B for a uniform beam AB of length 6 m and weight 150 N, resting horizontally in equilibrium on smooth supports at A and at B. The student takes moments about A and writes: R_B x 6 = 150 x 6. Identify the error in the student's equation, and write down the correct equation and the resulting value of R_B.

(Total for Question 15 is 3 marks)

A non-uniform rod AB has length 3 m and rests horizontally on a support at A. The rod is held horizontal by a light vertical string attached at B. The weight of the rod is 84 N and acts at a point 2 m from A. Show that the tension in the string at B is 56 N, and find the reaction at the support at A.

(Total for Question 16 is 3 marks)

A non-uniform rod AB has length 2.5 m and weight 90 N, acting at a point 1.5 m from A. The rod is smoothly hinged to a vertical wall at A and held in a horizontal position by a light inextensible string attached to the rod at B, making an angle of 40 degrees with the rod.

(a)Find the tension in the string, giving your answer to 3 significant figures.(2)

(b)Find the magnitude of the resultant force exerted on the rod by the hinge at A, giving your answer to 3 significant figures.(2)

(Total for Question 17 is 4 marks)

A uniform footbridge deck AB has length 10 m and weight 400 N. The deck is held horizontally in equilibrium by two vertical cables attached to the deck at C and D, where AC = 2 m and AD = 8 m. Two maintenance workers stand on the deck: one of combined weight (with equipment) 150 N at a point 3 m from A, and another of weight 90 N at a point 7 m from A.

(a)By taking moments about C, find the tension in the cable at D.(2)

(b)Hence find the tension in the cable at C.(2)

(Total for Question 18 is 4 marks)

A uniform ladder AB has length 4 m and weight 150 N. The ladder rests with end A on rough horizontal ground, where the coefficient of friction between the ladder and the ground is 0.3, and end B against a smooth vertical wall, making an angle of 60 degrees with the horizontal ground. A person of weight 700 N climbs up the ladder.

(a)Given that the ladder is on the point of slipping, find the normal reaction exerted on the ladder by the wall.(2)

(b)By taking moments about A, find the distance up the ladder, measured from A, that the person can climb before the ladder is on the point of slipping, giving your answer to 3 significant figures.(3)

(Total for Question 19 is 5 marks)

Mark scheme · M3DB Mechanics: Moments: Fluency and Exam Drill (Part 2)

Question 1

B1 the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point (oe)
Answer: The sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point.

Question 2

B1 24 cao
Answer: 24 N m

Question 3

B1 42 cao
Answer: 42 N m

Question 4

B1 the magnitude of the force and the perpendicular distance from the point to the line of action of the force
Answer: The magnitude of the force and the perpendicular distance from the pivot to the force's line of action.

Question 5

B1 moment = force x perpendicular distance, so the units are multiplied together (newtons x metres = N m), not divided (oe)
Answer: Moment = force x distance, so the units multiply: N x m = N m (not N/m, which would apply if the quantities were divided).

Question 6

M1 moment = 55 x 0.9 x sin70
A1 awrt 46.5 N m
Answer: 46.5 N m (3 s.f.)

Question 7

M1 moment = 40 x 0.14 x sin25
A1 awrt 2.37 N m
Answer: 2.37 N m (3 s.f.)

Question 8

M1 take moments about the pivot: 72 x 1.05 = 84 x d
A1 0.9 m cao
Answer: 0.9 m

Question 9

M1 F = moment / d = 3.6 / 0.08
A1 45 N cao
Answer: 45 N

Question 10

M1 d = moment / force = 12 / 15
A1 0.8 m cao
Answer: 0.8 m

Question 11

M1 resolve vertically: R_A + R_B = weight, so R_B = 130 - 55
A1 75 N cao
Answer: 75 N

Question 12

M1 total clockwise moment = (18 x 0.4) + (30 x 0.3) = 16.2; total anticlockwise moment = 25 x 0.5 = 12.5
A1 3.7 N m clockwise cao
Answer: 3.7 N m, clockwise

Question 13

B1 the centre of mass need not be at the midpoint of the rod (unlike a uniform rod, whose centre of mass is always at its midpoint) oe
Answer: For a non-uniform rod, the centre of mass need not be at the midpoint (unlike a uniform rod, where it always is).

Question 14

M1 identify distances from C: the weight acts 3 m from C; D is 6 m from C
M1 take moments about C: R_D x 6 = 240 x 3
A1 120 N cao
Answer: 120 N

Question 15

B1 identifies the error: the weight of the beam acts at its midpoint (3 m from A), not at the position of B (6 m from A) oe
M1 correct equation: R_B x 6 = 150 x 3
A1 75 N cao
Answer: The student used the wrong distance for the weight's moment arm (6 m, the position of B, instead of 3 m, the midpoint where the weight acts). Correct equation: R_B x 6 = 150 x 3, giving R_B = 75 N.

Question 16

M1 take moments about A: T x 3 = 84 x 2
A1 cso: T = 168/3 = 56 N, as required
B1 resolve vertically: R_A = 84 - 56 = 28 N cao
Answer: T = 56 N (shown); reaction at A = 28 N.

Question 17

(a) M1 take moments about A: T sin40 x 2.5 = 90 x 1.5
(a) A1 awrt 84.0 N
(a) Answer: 84.0 N (3 s.f.)
(b) M1 resolve to find the components of the hinge force: horizontal = T cos40 (awrt 64.4 N ft), vertical = 90 - T sin40 = 36 N; combine using resultant = √horizontal² + vertical²
(b) A1 awrt 73.7 N
(b) Answer: 73.7 N (3 s.f.)

Question 18

(a) M1 take moments about C: T_D x 6 = 400 x 3 + 150 x 1 + 90 x 5 (the deck's weight is 3 m from C, the first worker is 1 m from C, the second worker is 5 m from C, and D is 6 m from C)
(a) A1 300 N cao
(a) Answer: 300 N
(b) M1 resolve vertically: T_C + T_D = 400 + 150 + 90
(b) A1 340 N cao
(b) Answer: 340 N

Question 19

(a) M1 resolve vertically to find the normal reaction from the ground: N = 150 + 700 = 850; on the point of slipping, friction F = 0.3N; resolve horizontally: wall reaction = F
(a) A1 255 N cao
(a) Answer: 255 N
(b) M1 take moments about A: (wall reaction) x 4 sin60 = 150 x 2 cos60 + 700 x x cos60 (the ladder's weight acts at its midpoint, 2 m from A; the person is a distance x from A)
(b) dM1 substitute the wall reaction found in part (a) and rearrange to make x the subject
(b) A1 awrt 2.10 m
(b) Answer: 2.10 m (3 s.f.)

Question 19

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Mechanics: Moments: Fluency and Exam Drill (Part 2) - Worksheets, Questions and Revision

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M3DB Mechanics: Moments: Fluency and Exam Drill (Part 2)

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