Moments and Stability: Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Elastic Deformation, Moments and Centre of Gravity revision notes

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Section A

1. Define the moment of a force, state its unit, and give one everyday example of a moment in action, naming the pivot. [3]

2. State the principle of moments. [2]

Section B

3. A uniform beam 4.0 m long is pivoted at its centre. A 60 N force acts downwards 1.5 m to the left of the pivot.

(a) Calculate the moment of this force about the pivot. [2] (b) Calculate the force needed 1.2 m to the right of the pivot to balance the beam. [3] (c) Explain why the weight of the beam can be ignored here. [2]

4. (Extension beyond this resource’s guide, which covers only single-pivot moment problems — finding the reaction forces at two separate supports is a more advanced application of the same principle of moments.) A uniform plank of weight 200 N and length 5.0 m rests on two supports, one at each end. A 400 N load sits 1.5 m from the left support.

(a) Calculate the upward force from the right support, by taking moments about the left support. [4] (b) Hence calculate the force from the left support. [2]

5. Explain what is meant by centre of gravity, and describe how you would find the centre of gravity of an irregular flat sheet of card. [5]

6. Explain, in terms of centre of gravity and base area, why:

(a) a racing car is stable at speed [3] (b) a double-decker bus tips when tilted beyond a certain angle [3]

7. Describe an experiment, using a metre rule pivoted at its centre, to verify the principle of moments. [5]

Section C

8. A tall, narrow display stand and a low, wide display stand have the same weight. Describe and explain, in terms of the position of the centre of gravity, which stand is more stable, and describe two changes that would make the tall stand more stable. [6]

9. A spring has an unstretched length of 10.0 cm. When a load of 2.4 N is hung from it, within the limit of proportionality, its length becomes 14.0 cm.

(a) Calculate the extension of the spring. [1] (b) Calculate the spring constant. [2] (c) Sketch or describe the shape of the load–extension graph for this spring for loads up to and beyond the limit of proportionality. [3]

10. Two crates, A and B, have equal weight and the same base area, but crate A has a significantly lower centre of gravity than crate B. Explain why crate A is more stable than crate B, even though the two crates weigh the same. [4]

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Answers

1. The moment is force × perpendicular distance from the pivot to the line of action of the force [1]; the unit is the newton metre (N m) [1]. Everyday examples include turning a spanner on a nut, pushing open a door, or prising up a paving slab with a lever such as a crowbar — in each case the pivot is the point about which the object turns (the nut, the hinge, or the point where the lever rests on the ground) [1].

2. When a body is in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that point [1] [1].

3. (a) Moment = 60 × 1.5 [1] = 90 N m anticlockwise [1]. (b) For balance, F × 1.2 = 90 [1] [1]; F = 75 N [1]. (c) The beam is uniform and pivoted at its centre, so its weight acts through the pivot [1]; the perpendicular distance is therefore zero, and it exerts no moment [1].

4. (a) Taking moments about the left support: clockwise = (400 × 1.5) + (200 × 2.5) [1] [1] = 600 + 500 = 1100 N m [1]. Anticlockwise = R × 5.0, so R = 1100 ÷ 5.0 = 220 N [1]. (b) Total upward force = total downward force = 600 N [1]; left support = 600 − 220 = 380 N [1].

5. The centre of gravity is the single point through which the entire weight of an object can be considered to act [1]. Method: suspend the card freely from a point near its edge and let it settle [1]; hang a plumb line from the same point and mark the vertical line on the card [1]. Repeat from a second, different point [1]; the centre of gravity is where the two lines intersect — a third line is a useful check [1].

6. (a) The racing car has a very low centre of gravity and a wide wheelbase [1] [1]; when it tilts on a corner, the line of action of the weight stays inside the base, so the moment returns it to the road rather than tipping it [1]. (b) The bus has a high centre of gravity relative to the width of its base [1]; when tilted, the line of action of the weight passes outside the edge of the base [1], so the weight now produces a moment that continues the rotation rather than restoring it, and the bus topples [1].

7. Pivot a metre rule at its centre, so it balances with no load [1]. Hang known weights at measured distances from the pivot on each side of the rule [1]. Adjust the weights and/or their distances until the rule balances (is horizontal) [1]. Calculate the sum of the clockwise moments and the sum of the anticlockwise moments about the pivot from the weights and distances used [1]. Show that, at balance, these two sums are equal, confirming the principle of moments [1].

8. The low, wide stand is more stable [1]. Its centre of gravity is lower and its base is wider [1], so it must be tilted through a larger angle before the vertical line through its centre of gravity falls outside the base [1]; once that line falls outside the base, the weight produces a moment that topples the stand [1]. Changes to the tall stand, any two: add weight low down / a heavy base, lowering the centre of gravity [1]; widen the base [1]; reduce its height or move heavy items lower [1]. (Equal weight does not make the two equally stable: stability depends on the position of the centre of gravity relative to the base.)

9. (a) Extension = 14.0 − 10.0 = 4.0 cm (0.040 m) [1]. (b) k = F/x = 2.4 ÷ 0.040 = 60 N/m [1] [1]. (c) Up to the limit of proportionality, the load–extension graph is a straight line through the origin, since extension is directly proportional to load [1]. Beyond the limit of proportionality, the graph curves away from the straight line [1], showing that extension is no longer proportional to load [1].

10. Stability depends on the position of the centre of gravity relative to the base, not on weight [1]. Because crate A has a lower centre of gravity, it must be tilted through a larger angle before the line of action of its weight moves outside its base than crate B [1] [1]. Until that happens, the weight produces a moment that returns the crate to its original position, so crate A recovers over a wider range of tilt and is more stable [1].

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Where marks are usually lost

• Using the distance along the beam rather than the perpendicular distance.
• Taking moments about a point where an unknown force still has a moment.
• Saying stability depends on weight rather than on the centre of gravity and base.
• Applying k = F/x to data taken beyond the limit of proportionality, where extension is no longer proportional to load.