State the formula for the moment of a force about a point, in terms of the force and the perpendicular distance from the point to the line of action of the force.
(Total for Question 1 is 1 mark)
2
A force of 40 N acts at right angles to a lever at a point 0.5 m from the pivot. Find the moment of the force about the pivot.
(Total for Question 2 is 1 mark)
3
A force of 25 N acts at right angles to a rod at a point 1.6 m from a pivot. Find the moment of the force about the pivot.
(Total for Question 3 is 1 mark)
4
A force of 15 N acts at right angles to a bar at a point 3 m from a pivot. Find the moment of the force about the pivot.
(Total for Question 4 is 1 mark)
5
A force of 80 N acts on a rod at an angle of 60 degrees to the rod, at a distance of 0.5 m from a pivot. Find the moment of the force about the pivot, giving your answer to 3 significant figures.
(Total for Question 5 is 2 marks)
6
A force of 45 N acts on a bar at an angle of 40 degrees to the bar, at a distance of 0.8 m from a pivot. Find the moment of the force about the pivot, giving your answer to 3 significant figures.
(Total for Question 6 is 2 marks)
7
A uniform rod is pivoted at its centre. A force of 60 N acts vertically downward at a distance of 0.9 m from the pivot on one side. Find the distance from the pivot at which a force of 45 N must act vertically downward on the other side, for the rod to remain in equilibrium.
(Total for Question 7 is 2 marks)
8
A uniform beam AB of length 6 m and weight 180 N rests horizontally in equilibrium on two smooth supports, one at each end. Find the reaction at each support.
(Total for Question 8 is 2 marks)
9
A non-uniform rod AB of length 2 m is held horizontal in equilibrium by two vertical strings, one attached at each end. The tension in the string at A is 30 N and the tension in the string at B is 50 N. Find the weight of the rod.
(Total for Question 9 is 2 marks)
10
State the SI unit of a moment.
(Total for Question 10 is 1 mark)
11
A force acting at right angles to a lever at a pivot produces a moment of 54 N m. Given that the force has magnitude 90 N, find the distance from the pivot to the point where the force acts.
(Total for Question 11 is 2 marks)
12
A uniform rod AB has length 4 m and weight 100 N. The rod rests horizontally in equilibrium on two smooth supports, one at A and one at a point 2.5 m from A. By taking moments about A, find the reaction at the support 2.5 m from A.
(Total for Question 12 is 2 marks)
13
A moment of 45 N m is produced by a force acting at right angles to a lever, at a distance of 0.75 m from the pivot. Find the magnitude of the force.
(Total for Question 13 is 2 marks)
14
A rod is described as "uniform". State what this tells you about the position of its centre of mass.
(Total for Question 14 is 1 mark)
15
A uniform beam AB has length 5 m and weight 200 N. The beam rests horizontally in equilibrium on two smooth supports, one at C, 1 m from A, and one at D, 1 m from B. By taking moments about C, find the reaction at D.
(Total for Question 15 is 3 marks)
16
A mechanic applies a force of 70 N to the end of a wrench of length 0.3 m, at an angle of 55 degrees to the wrench, to loosen a bolt modelled as a fixed pivot at the other end of the wrench. Find the moment of the force about the pivot, giving your answer to 3 significant figures.
(Total for Question 16 is 3 marks)
17
A uniform rod AB has length 2 m and weight 70 N. 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 50 degrees with the rod. Find the tension in the string, giving your answer to 3 significant figures, and find the magnitude of the vertical component of the force exerted on the rod by the hinge at A.
(Total for Question 17 is 4 marks)
18
A uniform beam AB has length 8 m and weight 200 N. The beam is held horizontally in equilibrium by two vertical cables attached to the beam at points C and D, where AC = 2 m and AD = 7 m. A load of weight 120 N is attached to the beam at point E, where AE = 4 m. By taking moments about C, find the tension in the cable at D, and hence find the tension in the cable at C.
(Total for Question 18 is 4 marks)
19
A uniform ladder AB has length 4 m and weight 160 N. The ladder rests with end A on rough horizontal ground and end B against a smooth vertical wall, making an angle of 65 degrees with the horizontal ground. The ladder is in equilibrium and on the point of slipping. By taking moments about A, find the normal reaction at the wall, NB, and hence find the coefficient of friction between the ladder and the ground. Give both answers to 3 significant figures.
(Total for Question 19 is 4 marks)
Mark scheme · M3D Mechanics: Moments: Fluency and Exam Drill
Question 1
B1 moment = force x perpendicular distance
Answer: Moment = Force x perpendicular distance
Question 2
B1 20 cao
Answer: 20 N m
Question 3
B1 40 cao
Answer: 40 N m
Question 4
B1 45 cao
Answer: 45 N m
Question 5
M1 moment = 80 x 0.5 x sin60
A1 awrt 34.6 N m
Answer: 34.6 N m (3 s.f.)
Question 6
M1 moment = 45 x 0.8 x sin40
A1 awrt 23.1 N m
Answer: 23.1 N m (3 s.f.)
Question 7
M1 take moments about the pivot: 60 x 0.9 = 45 x d
A1 1.2 m cao
Answer: 1.2 m
Question 8
M1 by symmetry, or by taking moments, each reaction supports half the weight
A1 90 N at each support
Answer: 90 N at each support
Question 9
M1 resolve vertically: weight = sum of the two tensions
A1 80 N cao
Answer: 80 N
Question 10
B1 newton metre (N m)
Answer: newton metre (N m)
Question 11
M1 d = moment / force = 54/90
A1 0.6 m cao
Answer: 0.6 m
Question 12
M1 take moments about A: R x 2.5 = 100 x 2 (weight acts at the midpoint, 2 m from A)
A1 80 N cao
Answer: 80 N
Question 13
M1 F = moment / d = 45/0.75
A1 60 N cao
Answer: 60 N
Question 14
B1 the centre of mass acts at the midpoint of the rod
Answer: The centre of mass is at the midpoint of the rod.
Question 15
M1 identify distances from C: weight (midpoint, 2.5 m from A) is 1.5 m from C; D (4 m from A) is 3 m from C
M1 take moments about C: RD x 3 = 200 x 1.5
A1 100 N cao
Answer: 100 N
Question 16
M1 perpendicular distance = 0.3 sin55
M1 moment = 70 x (0.3 sin55)
A1 awrt 17.2 N m
Answer: 17.2 N m (3 s.f.)
Question 17
M1 take moments about A: T sin50 x 2 = 70 x 1 (weight at the midpoint, 1 m from A)
A1 awrt 45.7 N
M1 resolve vertically: hinge vertical component + T sin50 = 70
A1 35 N cao
Answer: T = 45.7 N (3 s.f.); vertical component of hinge force = 35 N
Question 18
M1 take moments about C: TD x 5 = 200 x 2 + 120 x 2 (beam's weight at 4 m from A is 2 m from C; load at E, 4 m from A, is also 2 m from C; D is 5 m from C)
A1 128 N cao
M1 resolve vertically: TC + TD = 200 + 120
A1 192 N cao
Answer: TD = 128 N, TC = 192 N
Question 19
M1 take moments about A: NB x 4 sin65 = 160 x 2 cos65 (weight at the midpoint)
A1 awrt 37.3 N
M1 resolve vertically to find NA = 160 N, resolve horizontally to find F = NB (ft), then μ = F/NA