Moments and Stability: Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — 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

---

Section A

1. Define the moment of a force and state its unit. [2]

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. Distinguish between stable, unstable and neutral equilibrium, giving an example of each. [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]

---

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].

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. Stable equilibrium — a small tilt causes the object to return to its original position, e.g. a cone resting on its wide base [1] [1]. Unstable equilibrium — a small tilt causes the object to topple further away from its original position, e.g. a cone balanced on its point [1] [1]. Neutral equilibrium — a small tilt or displacement leaves the object in its new position, with no tendency either to return or to topple further, e.g. a ball resting on a flat, horizontal surface [1] [1].

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] [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]; crate A therefore returns to its original position over a wider range of tilt and is more stable, even though the two crates have equal weight and the same base area [1] [1].

---

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.