A light rod pivots about a point O. A force of 200 N acts perpendicular to the rod at a distance of 0.4 m from O. Find the moment of the force about O.
(Total for Question 1 is 1 mark)
2
A moment of 96 N m is required to turn a lever about its pivot. A force of 160 N is applied perpendicular to the lever. Find the minimum distance from the pivot at which the force must act.
(Total for Question 2 is 1 mark)
3
A uniform rod AB has weight 90 N and rests horizontally in equilibrium on a single smooth pivot at its midpoint. State the reaction at the pivot.
(Total for Question 3 is 1 mark)
4
Two weights of 20 N and 30 N hang from a light rod at distances 0.6 m and 0.4 m from a pivot O, both on the same side of the pivot. Find the total moment of these two weights about O.
(Total for Question 4 is 1 mark)
5
Two point masses of 3 kg and 5 kg are placed on a light rod at x = 0 and x = 8 m respectively. Find the x-coordinate of the centre of mass of the system.
(Total for Question 5 is 1 mark)
6
State the condition, in terms of moments, that a rigid body must satisfy (in addition to zero resultant force) to be in equilibrium under a system of coplanar forces.
(Total for Question 6 is 1 mark)
7
A light rod AB rests horizontally on a pivot at C. A weight of 25 N hangs at a distance 0.8 m from C on one side. Find the distance from C at which a weight of 40 N must hang on the other side, for the rod to be in equilibrium.
(Total for Question 7 is 2 marks)
8
A light rod AB rests horizontally in equilibrium on a pivot at C. Weights of 15 N and 25 N hang from the rod at distances 0.3 m and 0.5 m from C, on the same side of the pivot. A weight W hangs at a distance 0.4 m from C on the other side. By taking moments about C, find W.
(Total for Question 8 is 2 marks)
9
A light rod AB rests horizontally in equilibrium on a pivot at C, under three weights of 15 N, 25 N and 42.5 N hanging from it (and the pivot reaction). Find the reaction at the pivot C.
(Total for Question 9 is 2 marks)
10
A non-uniform rod AB of length 3 m is held horizontal by two vertical strings, one at A and one at B. The tension in the string at A is twice the tension in the string at B, and the weight of the rod is 90 N. Find the tension in the string at B.
(Total for Question 10 is 2 marks)
11
A non-uniform rod is held horizontal by two vertical strings, with tensions TA and TB such that TA = 2TB and TA + TB = 90 N. Find TA.
(Total for Question 11 is 2 marks)
12
A uniform rectangular lamina ABCD has AB = 4 cm and BC = 10 cm. Point masses of 2 kg are attached at each of the four corners, and the lamina itself has negligible mass. Taking A as the origin, with AB along the x-axis and AD along the y-axis, find the coordinates of the centre of mass of the system of four point masses.
(Total for Question 12 is 2 marks)
13
A bent rod OAB has two straight arms, OA and AB, rigidly joined at right angles at A. When the rod is laid flat with OA horizontal and AB vertical, its centre of mass is a horizontal distance 0.3 m and a vertical distance 0.4 m from O. The rod is suspended freely from O and hangs in equilibrium. Find the angle OA makes with the vertical, giving your answer to 3 significant figures.
(Total for Question 13 is 2 marks)
14
A uniform cube of side 0.5 m and weight 40 N rests on rough horizontal ground. A horizontal force is applied to the top edge, increasing until the cube is on the point of toppling about the opposite bottom edge. Find the moment of the weight about this edge at the point of toppling.
(Total for Question 14 is 2 marks)
15
A light rigid rod AB has length 1.2 m and rests horizontally in equilibrium on a smooth pivot at point C on the rod. Weights of 30 N and 50 N hang from the rod at distances 0.2 m and 0.5 m from C, on one side of the pivot. A weight of magnitude W hangs at a distance 0.8 m from C, on the other side. By taking moments about C, find W, and hence find the reaction at the pivot.
(Total for Question 15 is 3 marks)
16
A uniform beam AB has length 6 m and weight 240 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, and hence find the reaction at C.
(Total for Question 16 is 3 marks)
17
A non-uniform rod AB has length 5 m and weight 150 N. The rod is held in equilibrium in a horizontal position by two vertical strings, one at A and one at B. The tension in the string at A is twice the tension in the string at B. Find the tension in each string, and hence, by taking moments about A, find the distance of the centre of mass of the rod from A, giving your answer to 3 significant figures.
(Total for Question 17 is 4 marks)
18
A particle of weight 60 N hangs in equilibrium, attached to two light inextensible strings whose other ends are fixed to a horizontal ceiling, on opposite sides of the vertical through the particle. String 1 makes an angle of 55 degrees with the ceiling, and string 2 makes an angle of 40 degrees with the ceiling. Find the tension in each string, giving your answers to 3 significant figures.
(Total for Question 18 is 4 marks)
19
A uniform cube of side 0.8 m rests on a rough plane inclined at angle θ to the horizontal, which is slowly increased from 0. The coefficient of friction between the cube and the plane is 0.5. Find the angle θ at which the cube would be on the point of toppling about its lower edge (assuming it does not slide first), and find the angle at which it would be on the point of sliding (assuming it does not topple first). Hence determine, with justification, whether the cube slides or topples first as θ increases from 0.
(Total for Question 19 is 4 marks)
Mark scheme · M7D Mechanics: Moments and Statics Depth: Fluency and Exam Drill
Question 1
B1 80 cao
Answer: 80 N m
Question 2
B1 0.6 cao
Answer: 0.6 m
Question 3
B1 90 cao
Answer: 90 N
Question 4
B1 24 cao
Answer: 24 N m
Question 5
B1 5 cao
Answer: x = 5 m
Question 6
B1 the sum of the moments about any point must be zero
Answer: The sum of the moments about any point must be zero.
Question 7
M1 take moments about C: 25 x 0.8 = 40 x d
A1 0.5 m cao
Answer: 0.5 m
Question 8
M1 take moments about C: 15(0.3) + 25(0.5) = 0.4W
A1 42.5 N cao
Answer: 42.5 N
Question 9
M1 resolve vertically: reaction = sum of the three weights (the rod is light)
A1 82.5 N cao
Answer: 82.5 N
Question 10
M1 resolve vertically: TA + TB = 90, with TA = 2TB, so 3TB = 90
A1 30 N cao
Answer: 30 N
Question 11
M1 substitute TA = 2TB into TA + TB = 90 to find TB = 30, then TA = 2(30)
A1 60 N cao
Answer: 60 N
Question 12
M1 recognise that four equal masses at the corners of a rectangle have their centre of mass at the geometric centre
A1 (2, 5) cao
Answer: (2, 5)
Question 13
M1 the rod hangs with its centre of mass directly below O, so the angle OA makes with the vertical satisfies tan(angle) = horizontal distance / vertical distance = 0.3/0.4
A1 awrt 36.9 degrees
Answer: 36.9 degrees (3 s.f.)
Question 14
M1 at the point of toppling, the weight still acts through the centre of the cube, a horizontal distance of half the side (0.25 m) from the toppling edge
A1 10 N m cao
Answer: 10 N m
Question 15
M1 take moments about C: 30(0.2) + 50(0.5) = 0.8W
A1 W = 38.75 N cao
A1 reaction = 30 + 50 + 38.75 = 118.75 N cao (ft)
Answer: W = 38.75 N; reaction at pivot = 118.75 N
Question 16
M1 the weight acts at the midpoint (3 m from A, so 2 m from C); D is 5 m from A, so 4 m from C; take moments about C: RD(4) = 240(2)
A1 RD = 120 N cao
A1 resolve vertically: RC = 240 - 120 = 120 N cao (ft)
Answer: RD = 120 N, RC = 120 N
Question 17
M1 TA + TB = 150 with TA = 2TB, so 3TB = 150
A1 TB = 50 N, TA = 100 N
M1 take moments about A: TB(5) = 150(d), where d is the distance of the centre of mass from A
A1 awrt 1.67 m
Answer: TA = 100 N, TB = 50 N; centre of mass is 1.67 m from A (3 s.f.)
Question 18
M1 resolve horizontally: T1 cos55 = T2 cos40
M1 resolve vertically: T1 sin55 + T2 sin40 = 60, and eliminate T1 (or T2) using the horizontal equation
A1 awrt T1 = 46.1 N
A1 awrt T2 = 34.5 N
Answer: T1 = 46.1 N, T2 = 34.5 N (both 3 s.f.)
Question 19
M1 for a cube, toppling occurs when the line of action of the weight passes through the lower edge, i.e. when tan(θ) = (half the side)/(half the side) = 1
A1 toppling angle = 45 degrees
M1 sliding occurs when tan(θ) = μ = 0.5
A1 sliding angle = awrt 26.6 degrees; since 26.6 < 45, the cube slides first
Answer: Toppling angle = 45 degrees; sliding angle = 26.6 degrees (3 s.f.); the cube slides before it would topple