Questions on moments

Question 1

Define the moment of a force about a point, state its units, and explain the role of sense.

Accepted Answer

The moment is the force multiplied by the perpendicular distance from the point to the force's line of action, measured in newton metres. Each moment turns either clockwise or anticlockwise about the point, and the two senses oppose one another, so equilibrium of moments means the two totals are equal. B1 for force times perpendicular distance, B1 for newton metres, B1 for the role of sense.

Question 2

An adult of weight 300 N sits 2 m from a seesaw's pivot. Find how far from the pivot a child of weight 200 N must sit on the other side to balance it.

Accepted Answer

Take moments about the pivot. Clockwise 300 × 2 = 600 N m, anticlockwise 200 × x, so 200x = 600 and x = 3 m. M1 for taking moments about the pivot, A1 for 200x = 600, A1 for 3 m. Taking moments about the pivot also removes the reaction there from the equation, since its perpendicular distance is zero.

Question 3

A force of 15 N acts at a point 0.4 m from a hinge, in a direction making 30° with the line from the hinge to that point. Find the moment of the force about the hinge.

Accepted Answer

The perpendicular distance from the hinge to the line of action is 0.4 sin 30° = 0.2 m, so the moment is 15 × 0.2 = 3 N m. Equivalently, the component of the force perpendicular to the line from the hinge is 15 sin 30° = 7.5 N, and 7.5 × 0.4 = 3 N m. M1 for using the perpendicular distance or the perpendicular component, A1 for 3 N m. Multiplying 15 by 0.4 gives 6 N m, the moment the force would have if it acted at right angles to the line, which it does not.

Question 4

A uniform rod of length 3 m and weight 60 N is pivoted at one end and held horizontal by a vertical force F applied at the other end. Explain why taking moments about the pivot removes the reaction at the pivot from the equation, and find F.

Accepted Answer

The reaction at the pivot acts through the pivot, so its perpendicular distance from that point is zero and its moment about it is zero, whatever its size or direction. B1 for that reason. Taking moments about the pivot: the rod is uniform, so its weight acts at the midpoint, 1.5 m away, and F × 3 = 60 × 1.5 gives F = 30 N. M1 for the moments equation with the weight at 1.5 m, A1 for 30 N. The pivot then carries the remaining 30 N upwards, found by resolving vertically rather than by moments.

Question 5

A uniform 6 m beam of weight 90 N rests on a support at its centre. A person of weight 450 N sits 1 m from the centre. Find where a 300 N weight must be placed on the other side to keep the beam horizontal.

Accepted Answer

Uniform means the beam's weight acts at its midpoint, which here is the support itself, so its perpendicular distance from the pivot is zero and it contributes no moment. Balancing the rest, 450 × 1 = 300 × x gives x = 1.5 m from the centre. B1 for the beam's weight giving no moment, M1 for the moments equation, A1 for 1.5 m.

Question 6

A uniform 8 m beam of weight 200 N rests horizontally on supports at 1 m and 6 m from one end. Find the reaction at each support.

Accepted Answer

The beam is uniform, so its weight acts at the 4 m mark. Take moments about the support at 1 m, which removes R₁ from the equation: 200 × 3 = R₂ × 5, so R₂ = 120 N. Vertical equilibrium then gives R₁ = 200 − 120 = 80 N. B1 for the weight acting at the 4 m mark, M1 for taking moments about a support, A1 for the second reaction, M1 for resolving vertically, A1 for the first reaction. Check by taking moments about the end instead: 80 × 1 + 120 × 6 = 800 = 200 × 4.

Question 7

A non-uniform 10 m beam of weight 500 N rests on supports at its two ends, which carry reactions of 300 N and 200 N. Find how far the centre of mass lies from the end with the larger reaction.

Accepted Answer

Take moments about the 300 N end, which removes that reaction: 500 × x = 200 × 10, so x = 4 m. M1 for taking moments about the 300 N end, A1 for 500x = 2000, A1 for 4 m. The reactions already sum to 500 N, so vertical equilibrium is satisfied and carries no new information. The centre of mass sits nearer the end carrying more, which is the sense check.

Question 8

A uniform ladder AB of mass 20 kg and length 6 m rests with its foot A on rough horizontal ground and its top B against a smooth vertical wall, inclined at 60° to the horizontal. Find the magnitude of the force the wall exerts on the ladder.

Accepted Answer

The wall is smooth, so the only force it can exert is a normal reaction perpendicular to the wall, which is horizontal. Call it S. The ladder is uniform, so its weight 20g = 196 N acts at the midpoint, 3 m along the ladder from the foot. Take moments about the foot, which removes both the normal reaction and the friction there. The weight acts vertically, so its perpendicular distance from the foot is the horizontal distance 3 cos 60° = 1.5 m. The wall's push acts horizontally, so its perpendicular distance from the foot is the height of the top, 6 sin 60° = 5.196 m. Balancing the two: S × 6 sin 60° = 196 × 3 cos 60°, so S = 294/5.196 = 56.6 N. B1 for the weight acting 3 m along the ladder, M1 for taking moments about the foot, A1 for the moments equation with both perpendicular distances, M1 for rearranging to S = 98/tan 60°, A1 for 56.6 N. Resolving horizontally then gives the friction at the foot as 56.6 N as well.

Question 9

A uniform rod AB of mass 5 kg and length 2 m is freely hinged to a wall at A and held horizontal by a light inextensible string joining B to a point of the wall directly above A. The string makes an angle of 40° with the rod. Find the tension in the string.

Accepted Answer

The rod is uniform, so its weight 5g = 49 N acts at the midpoint, 1 m from the hinge. The tension is not perpendicular to the rod, so it cannot be multiplied by 2 m. Either use the perpendicular distance from the hinge to the line of the string, which is 2 sin 40°, or resolve the tension into T sin 40° perpendicular to the rod and T cos 40° along it; the part along the rod has its line of action through the hinge and so turns the rod not at all. Both readings give the same moment. Taking moments about the hinge also removes the force the hinge itself exerts: T × 2 sin 40° = 49 × 1, so T = 49/1.286 = 38.1 N. B1 for the weight acting at the midpoint, M1 for taking moments about the hinge with the perpendicular part of the tension, A1 for 2T sin 40° = 49, A1 for 38.1 N.

Question 10

A uniform rectangular lamina ABCD of weight 40 N stands in a vertical plane on rough horizontal ground, with AB = 1.2 m along the ground and BC = 0.8 m vertical. A horizontal force P is applied at D, along the top edge, pushing towards C, and the ground is rough enough to prevent sliding. State where the weight of the lamina acts, and find the value of P at which the lamina is on the point of toppling.

Accepted Answer

The lamina is uniform, so its weight acts at its centre of symmetry, the centre of the rectangle, 0.6 m horizontally from B and 0.4 m above the ground. On the point of toppling about B the lamina is about to lose contact everywhere except at B, so the normal reaction and friction act at B and have no moment about it. Taking moments about B: P acts 0.8 m above B, and the weight acts 0.6 m from B on the other side, so 0.8P = 40 × 0.6 = 24 and P = 30 N. B1 for the weight at the centre, 0.6 m from B, M1 for taking moments about B with the reaction moved to B, A1 for 0.8P = 24, A1 for 30 N. Pushing at the top edge gives P the longest lever arm available; the same push lower down would need a larger force.

Question 11

A uniform beam AB of length 6 m and weight 150 N rests horizontally on supports at C and D, where AC = 1 m and AD = 4 m. A person of weight 600 N walks slowly from C towards B. Find the distance of the person from A at the instant the beam is about to tilt, and state which support it tilts about.

Accepted Answer

The beam tilts about D: once the person is beyond D, the far support C is the one that can lose contact, and on the point of tilting the reaction at C is zero. Take moments about D with R_C = 0. The weight acts at the midpoint, 3 m from A, which is 1 m on the A side of D; the person is x − 4 m beyond D. So 150 × 1 = 600(x − 4), giving x − 4 = 0.25 and x = 4.25 m from A. B1 for tilting about D with R_C = 0, M1 for taking moments about D, A1 for 150 = 600(x − 4), A1 for 4.25 m. The answer lies between D and B, as it must for tilting about D; while the person is between C and D both reactions are positive and nothing tilts.

Question 12

A uniform plank AB of weight 400 N rests horizontally on two supports P and Q, and a load hangs from the plank, as shown in the diagram. The plank's own weight is not marked on the diagram. Use the diagram to find the magnitude of the reaction at Q, and then the magnitude of the reaction at P.

Accepted Answer

The plank is uniform, so its whole weight of 400 N acts at the midpoint of AB, and the scale puts that midpoint at the 4 m mark. B1 for placing the 400 N at 4 m. The rest is read off the same scale: P sits at 1.5 m and Q at 6.5 m, so PQ = 5 m; the weight is 4 − 1.5 = 2.5 m from P; and the 250 N load hangs at 7.5 m, which is 7.5 − 1.5 = 6 m from P and 1 m beyond Q.
Take moments about P, which removes the reaction there from the equation because its line of action passes through that point. M1 for taking moments about P. Then R_Q × 5 = 400 × 2.5 + 250 × 6 = 1000 + 1500 = 2500. A1 for the moments equation with all three distances read correctly. So R_Q = 2500/5 = 500 N. A1 for 500 N.
The plank is also in vertical equilibrium, so the two reactions carry everything pressing down: R_P + R_Q = 400 + 250 = 650. M1 for resolving vertically. Hence R_P = 650 − 500 = 150 N. A1 for 150 N.
Check by taking moments about Q instead: 150 × 5 + 250 × 1 = 750 + 250 = 1000 and 400 × 2.5 = 1000, which agree. The load hangs outside the supports, so its moment about P is the largest term in the equation and Q carries more than three times what P does; moving the load out to B itself, 1.5 m beyond Q, would only bring R_P down to 200 − 50 × 1.5 = 125 N, so this plank cannot be tipped by this load anywhere along it; it would take a heavier load, or a longer overhang, to bring R_P to zero.

Question 13

A uniform rod AB of length 4 m and mass 12 kg rests horizontally on supports at A and at C, where AC = 2.5 m. A particle of mass m kg is placed at B. Find the greatest value of m for which the rod remains in equilibrium, and state the reaction at C at that instant.

Accepted Answer

The rod is uniform, so its weight 12g acts at the midpoint, 2 m from A and therefore 0.5 m from C on the A side. The particle at B is 1.5 m from C on the other side. As m grows the rod threatens to tip about C, so the condition at the point of tipping is R_A = 0. Taking moments about C with R_A = 0: 12g × 0.5 = mg × 1.5, so m = 6/1.5 = 4 kg. With A carrying nothing, C supports everything: R_C = (12 + 4)g = 156.8 N. B1 for the weight acting at the midpoint, M1 for setting the reaction at A to zero at the point of tipping, M1 for taking moments about C, A1 for m = 4 kg, M1 for resolving vertically, A1 for 156.8 N. g cancels from the moments equation, so working in masses is fine as long as you restore g for the reaction.

Question 14

A non-uniform plank AB of length 6 m rests horizontally on supports at its two ends. When a load of weight 100 N is placed on the plank at A, the reaction at B is 150 N. When the same load is placed at B instead, the reaction at A is 90 N. Find the weight of the plank and the distance of its centre of mass from A.

Accepted Answer

The plank is not uniform, so its weight cannot be assumed to act at the midpoint. Let that weight be W newtons and let its line of action be x metres from the end labelled A. Nothing given fixes either quantity, so two equations are needed, and each of the two loadings supplies one. In the first loading take moments about A: the load sits at the pivot and contributes nothing, and the reaction of 150 N acts 6 m away, so W × x = 150 × 6 = 900. In the second loading take moments about B, where again the load contributes nothing, and the weight is now 6 − x from the pivot: W(6 − x) = 90 × 6 = 540. Adding the two equations kills x, because Wx + W(6 − x) = 6W, so 6W = 1440 and W = 240 N. Substituting back, x = 900/240 = 3.75 m. M1 for taking moments about the loaded end in the first situation, A1 for Wx = 900, M1 for taking moments about the loaded end in the second situation, A1 for W(6 − x) = 540, dM1 for adding the two to eliminate x, A1 for the weight of 240 N, A1 for 3.75 m. As a check, vertical equilibrium in the first situation gives the other reaction as 240 + 100 − 150 = 190 N, and moments about the far end confirm it: 190 × 6 = 1140 = 240 × 2.25 + 100 × 6.

Question 15

A uniform ladder of mass 15 kg and length 5 m rests with its foot on rough horizontal ground and its top against a smooth vertical wall, inclined at 60° to the horizontal. A person of mass 60 kg stands on the ladder 4 m from its foot. Taking g = 9.8 m s⁻², find the force exerted by the wall on the ladder, and hence find the least value of the coefficient of friction between the ladder and the ground for which the ladder does not slip.

Accepted Answer

The wall is smooth, so its force S is horizontal. Take moments about the foot, which removes the ground's reaction and friction from the equation. The ladder's weight 15g = 147 N acts 2.5 m along it, at horizontal distance 2.5 cos 60° = 1.25 m from the foot; the person's weight 60g = 588 N acts at 4 cos 60° = 2 m; S acts at the height of the top, 5 sin 60° = 4.33 m. So 4.33S = 147 × 1.25 + 588 × 2 = 183.75 + 1176 = 1359.75, giving S = 314 N. Vertically the wall supplies nothing, so R = 75g = 735 N. Horizontally friction at the foot balances the wall: F = S = 314 N, and F ≤ μR requires μ ≥ 0.427. M1 for taking moments about the foot with both weights, A1 for the equation with all three perpendicular distances, A1 for S = 314 N, B1 for R = 735 N, M1 for F = S with F ≤ μR, A1 for 0.427. The person's moment is more than six times the ladder's, so where the person stands matters more than the ladder's own weight; moving the person to the top would raise S further and demand more friction.

Question 16

A uniform rectangular lamina ABCD of weight 40 N stands in a vertical plane on rough horizontal ground, with AB = 1.2 m along the ground and BC = 0.8 m vertical. A horizontal force P, applied at D along the top edge and pushing towards C, is slowly increased from zero. It can be shown that the lamina would topple about B when P reaches 30 N. The coefficient of friction between the lamina and the ground is 0.6. Find the value of P at which the lamina would begin to slide, and determine whether it slides or topples first. Find also the set of values of the coefficient of friction for which it would topple before sliding, and describe what happens at the boundary of that set.

Accepted Answer

The toppling value is given as 30 N, so only the sliding threshold has to be found. Resolving vertically, R = 40 N, since P is horizontal and adds nothing to the reaction. The limiting friction is 0.6 × 40 = 24 N, and sliding begins when P reaches 24 N. M1 for R = 40 N with limiting friction μR, A1 for 24 N. Since 24 < 30, the lamina slides first, at P = 24 N, and never reaches the toppling value. A1 for the comparison and the conclusion. For toppling to come first, the lamina must still be held by friction when P reaches 30 N, so the limiting friction must exceed 30 N: 40μ > 30, giving μ > 0.75. M1 for 40μ > 30, A1 for μ > 0.75 with the boundary case stated. At μ = 0.75 exactly the limiting friction is 30 N, so the two thresholds coincide: the lamina is on the point of sliding and on the point of toppling at the same value P = 30 N, and neither happens before the other. A taller, narrower lamina lowers the toppling value and makes toppling the likelier outcome.