A Level Physics 4.1 Turning Effects of Forces MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 4.1 Turning Effects of Forces – moments, couples, centre of gravity and stability. Every question is an original practice question with a fully worked answer, written for AS Level revision.
Moments & the Principle of Moments
Q1. The moment of a force about a pivot is defined as
- (A) force × distance moved in the direction of the force
- (B) force × perpendicular distance from the pivot to the line of action of the force
- (C) force ÷ distance from the pivot
- (D) mass × acceleration of the point of application
Show Answer
Answer: B – moment = F × d, where d is the perpendicular distance from the pivot to the line of action of the force. Its unit is the newton metre (N m).
Q2. A force of 40 N is applied perpendicular to a door at a distance of 0.25 m from the hinge. The moment of the force about the hinge is
- (A) 1.6 N m
- (B) 160 N m
- (C) 10 N m
- (D) 65 N m
Show Answer
Answer: C – moment = F × d = 40 × 0.25 = 10 N m.
Q3. A force of 60 N is applied to the end of a horizontal beam at 0.50 m from a pivot. The force acts at 30° above the beam. The moment of the force about the pivot is
- (A) 15 N m
- (B) 26 N m
- (C) 30 N m
- (D) 52 N m
Show Answer
Answer: A – only the component perpendicular to the beam produces a moment: Fperp = 60 sin 30° = 30 N. Moment = 30 × 0.50 = 15 N m. (Using the full 60 N wrongly gives 30 N m.)
Q4. A see-saw is balanced on a central pivot. A child of weight 300 N sits 1.2 m from the pivot on one side. A second child of weight 400 N sits on the other side. The distance of the second child from the pivot is
- (A) 0.30 m
- (B) 1.6 m
- (C) 0.60 m
- (D) 0.90 m
Show Answer
Answer: D – principle of moments: 400 × x = 300 × 1.2 = 360, so x = 360 ÷ 400 = 0.90 m.
Q5. A uniform metre rule is pivoted at its 50 cm mark (its centre of gravity, so its weight causes no moment). A weight of 3.0 N hangs at the 20 cm mark and an unknown weight W hangs at the 95 cm mark. The rule is balanced. W is
- (A) 4.5 N
- (B) 2.0 N
- (C) 3.0 N
- (D) 6.8 N
Show Answer
Answer: B – the 3.0 N weight is 0.30 m from the pivot and W is 0.45 m from the pivot. Moments: W × 0.45 = 3.0 × 0.30 = 0.90, so W = 0.90 ÷ 0.45 = 2.0 N.
Q6. A beam rests on a single pivot and is in equilibrium. Loads of 30 N and 50 N hang from the beam (in addition to its own weight, which acts at the pivot). The upward reaction force at the pivot is
- (A) 20 N
- (B) 40 N
- (C) 80 N
- (D) 1500 N
Show Answer
Answer: C – for equilibrium the resultant force must also be zero, so the upward reaction balances the total downward load: 30 + 50 = 80 N.
Q7. A mechanic pulls a spanner of length 0.25 m with a force of 120 N. The force is applied at 60° to the spanner. The moment applied to the nut is
- (A) 26 N m
- (B) 30 N m
- (C) 15 N m
- (D) 60 N m
Show Answer
Answer: A – perpendicular component = 120 sin 60° = 104 N. Moment = 104 × 0.25 = 26 N m. Maximum moment would need the force perpendicular to the spanner.
Q8. A crane jib is pivoted at its base. A load of weight 500 N hangs 2.0 m horizontally from the pivot. A counterweight on the other side acts 1.25 m from the pivot. For the jib to balance, the counterweight must provide a downward force of
- (A) 200 N
- (B) 313 N
- (C) 625 N
- (D) 800 N
Show Answer
Answer: D – F × 1.25 = 500 × 2.0 = 1000, so F = 1000 ÷ 1.25 = 800 N.
Q9. A uniform rod of length 2.0 m is pivoted at its centre. A force of 8 N acts downward at the left-hand end and a force of 12 N acts downward 0.50 m to the right of the centre. To balance the rod, an additional downward force at the right-hand end must be
- (A) 14 N
- (B) 6.0 N
- (C) 2.0 N
- (D) 4.0 N
Show Answer
Answer: C – anticlockwise moment = 8 × 1.0 = 8 N m; clockwise moment = 12 × 0.50 = 6 N m. An extra 2 N m clockwise is needed at 1.0 m from the pivot, so the force is 2 ÷ 1.0 = 2.0 N.
Q10. The same force F is applied to a door in four ways. In which case is the moment about the hinge greatest?
- (A) F applied at the midpoint of the door, perpendicular to the door
- (B) F applied at the handle (edge furthest from the hinge), perpendicular to the door
- (C) F applied at the handle, at 30° to the door
- (D) F applied close to the hinge, perpendicular to the door
Show Answer
Answer: B – moment = F × perpendicular distance. It is greatest when the force is applied at the largest distance from the pivot and acts fully perpendicular to the door.
Couples & Torque
Q11. A couple consists of
- (A) two equal forces acting in the same direction at the same point
- (B) two unequal, opposite forces with the same line of action
- (C) two equal, opposite, parallel forces whose lines of action do not coincide
- (D) a single force applied away from the centre of gravity
Show Answer
Answer: C – a couple is a pair of equal and opposite parallel forces with different lines of action, so it produces rotation only.
Q12. The resultant force produced by a couple is
- (A) zero
- (B) equal to twice one of the forces
- (C) equal to one of the forces
- (D) dependent on the separation of the forces
Show Answer
Answer: A – the two forces are equal and opposite, so they cancel: resultant force = 0. The body does not accelerate translationally, but the couple still exerts a torque.
Q13. Two parallel forces of 20 N act in opposite directions. The perpendicular distance between their lines of action is 0.40 m. The torque of the couple is
- (A) 0.50 N m
- (B) 50 N m
- (C) 16 N m
- (D) 8.0 N m
Show Answer
Answer: D – torque of a couple = one force × separation of the lines of action = 20 × 0.40 = 8.0 N m. (Do not double the force: 2 × 20 × 0.40 = 16 N m is the common error.)
Q14. A driver grips a steering wheel of diameter 0.38 m with both hands and applies a force of 15 N with each hand, in opposite tangential directions. The torque applied to the steering column is
- (A) 2.9 N m
- (B) 5.7 N m
- (C) 11 N m
- (D) 30 N m
Show Answer
Answer: B – the hands form a couple with separation equal to the diameter: torque = 15 × 0.38 = 5.7 N m.
Q15. A tap is turned by two fingers applying a couple: each force is 6.0 N and the forces are 0.15 m apart. The torque is
- (A) 40 N m
- (B) 1.8 N m
- (C) 0.90 N m
- (D) 0.45 N m
Show Answer
Answer: C – torque = 6.0 × 0.15 = 0.90 N m.
Q16. A free body is acted on by a couple only. The body will
- (A) rotate with no acceleration of its centre of mass
- (B) move in a straight line at constant speed
- (C) accelerate in the direction of one of the forces
- (D) remain completely at rest
Show Answer
Answer: A – the resultant force is zero, so the centre of mass does not accelerate; the non-zero torque produces angular acceleration (rotation).
Q17. The forces of a couple are kept the same but the perpendicular separation of their lines of action is doubled. The torque of the couple
- (A) is halved
- (B) is unchanged
- (C) increases by a factor of four
- (D) is doubled
Show Answer
Answer: D – torque = F × separation, so doubling the separation doubles the torque.
Q18. A couple made from two forces of 25 N each must produce a torque of 15 N m. The required separation of the forces is
- (A) 0.30 m
- (B) 0.60 m
- (C) 1.7 m
- (D) 375 m
Show Answer
Answer: B – separation = torque ÷ force = 15 ÷ 25 = 0.60 m.
Q19. Which statement about the torque of a couple is correct?
- (A) It depends on the pivot chosen for the calculation
- (B) It is zero about the midpoint of the two forces
- (C) It has the same value about every point
- (D) It equals the resultant force multiplied by the separation
Show Answer
Answer: C – the torque of a couple is a free vector: F × separation about any point. The resultant force is zero, so (D) would give zero, which is wrong.
Q20. A body is acted on by two couples: one of torque 12 N m clockwise and one of torque 7 N m anticlockwise. The resultant torque on the body is
- (A) 5 N m clockwise
- (B) 19 N m clockwise
- (C) 5 N m anticlockwise
- (D) zero
Show Answer
Answer: A – taking clockwise as positive: 12 − 7 = 5 N m clockwise.
Centre of Gravity & Applications
Q21. The centre of gravity of a body is
- (A) the point where the mass of the body is greatest
- (B) always at the geometric centre of the body
- (C) the point through which the resultant force on the body acts
- (D) the point from which the whole weight of the body may be considered to act
Show Answer
Answer: D – the centre of gravity is the single point at which the entire weight of the body can be taken to act, for the purpose of calculating moments.
Q22. The centre of gravity of an irregular plane lamina is found experimentally by
- (A) balancing the lamina on a knife edge in one direction only
- (B) suspending the lamina freely from several points in turn, tracing a vertical plumb line each time; the lines cross at the centre of gravity
- (C) measuring its area and dividing by two
- (D) dropping the lamina and noting how it lands
Show Answer
Answer: B – when freely suspended, the lamina hangs with its centre of gravity directly below the pivot, so a plumb line from the pivot passes through it. Two or more such lines intersect at the centre of gravity.
Q23. The centre of gravity of a uniform circular ring is
- (A) at the centre of the ring, where there is no material
- (B) on the rim of the ring
- (C) halfway between the centre and the rim
- (D) undefined, because the ring is hollow
Show Answer
Answer: A – by symmetry the centre of gravity is at the geometric centre. The centre of gravity of a body need not lie within its material.
Q24. Which design change makes a tall vehicle LEAST likely to topple when cornering?
- (A) Raising the heavy engine to roof level
- (B) Narrowing the wheel base
- (C) Placing heavy components low down and widening the wheel base
- (D) Reducing the mass of the wheels
Show Answer
Answer: C – stability increases when the centre of gravity is as low as possible and the base is as wide as possible.
Q25. A box on a slope will topple (rather than slide back) at the point where
- (A) its centre of gravity passes over the centre of the base
- (B) the line of action of its weight falls outside the base
- (C) the friction at the base becomes zero
- (D) its weight acts through the highest corner
Show Answer
Answer: B – as the box tilts, the centre of gravity moves towards the edge of the base. Once the vertical line through the centre of gravity falls outside the base, the weight produces a moment about the edge that tips the box over.
Q26. A rectangular block has a base of width 0.40 m and its centre of gravity 0.60 m above the base. The block is tilted slowly. The angle to the vertical at which it topples is
- (A) 56°
- (B) 34°
- (C) 45°
- (D) 18°
Show Answer
Answer: D – the block topples when the line of action of the weight passes through the corner of the base: tan θ = (half base width) ÷ (height of centre of gravity) = 0.20 ÷ 0.60 = 0.33, so θ = tan−1(0.33) ≈ 18°.
Q27. Before a double-decker bus is approved, it is tilted on a test platform. Carrying passengers on the upper deck makes toppling more likely mainly because
- (A) the passengers raise the centre of gravity of the bus
- (B) the passengers increase the friction of the tyres
- (C) the total weight of the bus decreases
- (D) the wheel base becomes wider
Show Answer
Answer: A – mass added high up raises the combined centre of gravity, so a smaller tilt angle moves the line of action of the weight outside the wheel base.
Q28. A tower crane has a large concrete counterweight on the short arm opposite the load. Its main purpose is to
- (A) increase the total weight so the crane sinks into the ground
- (B) reduce the torque the motor needs to rotate the jib
- (C) produce a moment that balances the moment of the load about the tower, keeping the resultant centre of gravity over the base
- (D) reduce air resistance on the jib
Show Answer
Answer: C – the counterweight’s moment opposes the load’s moment, so the crane does not tip forwards when lifting.
Q29. In stable equilibrium, a body displaced slightly and released returns to its original position. While it is displaced, its centre of gravity
- (A) falls, so its gravitational potential energy decreases
- (B) rises, so its gravitational potential energy increases
- (C) stays at exactly the same height
- (D) moves outside the base
Show Answer
Answer: B – a stable object sits at a minimum of potential energy: tilting it raises the centre of gravity, and the moment of its weight then restores it. (Neutral equilibrium: height unchanged; unstable: height falls.)
Q30. Which example is in neutral equilibrium?
- (A) A cone standing on its base
- (B) A pencil balanced on its tip
- (C) A book standing upright on its spine
- (D) A uniform sphere resting on a flat horizontal surface
Show Answer
Answer: D – displacing the sphere leaves the height of its centre of gravity unchanged, so it has no tendency to roll back or to roll away: neutral equilibrium.
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