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Moments questions
A rigid body can turn as well as move, so equilibrium needs a second condition. The moment of a force is force times perpendicular distance from the pivot, and a beam in equilibrium has both zero resultant force and zero resultant moment. Taking moments about a well-chosen point is usually what supplies the missing equation.
6 original questions · 23 marks · the moments notes · Mechanics
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Define the moment of a force about a point, state its units, and explain the role of sense.
Worked 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. Perpendicular is the word that earns the mark; a distance measured along the force does no turning at all.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.
Worked 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. Less weight needs more distance in exact proportion. Taking moments about the pivot also removes the reaction there from the equation, since its perpendicular distance is zero.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.
Worked 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. The 90 N is not irrelevant to the reaction at the support, but it is irrelevant to this moments equation, and saying why is worth a mark.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.
Worked 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. Choosing a pivot that sits under an unknown force is the labour-saving move of the whole topic, and full marks need both a moments equation and a resolving equation.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.
Worked 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. For a non-uniform beam you must never assume the weight acts at the midpoint.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.
Worked 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 RA = 0. Taking moments about C with RA = 0: 12g × 0.5 = mg × 1.5, so m = 6/1.5 = 4 kg. With A carrying nothing, C supports everything: RC = (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. Setting the far reaction to zero is how every on-the-point-of-tipping question is solved, and g cancels from the moments equation, so working in masses is fine as long as you restore g for the reaction.
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