A Level Physics 4.2 Equilibrium of Forces MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 4.2 Equilibrium of Forces. Test yourself on the two conditions for equilibrium, resolving forces on slopes and in cables, force triangles and two-support beam problems – every answer comes with full worked reasoning.
Conditions for Equilibrium
Q1. For an extended body to be in equilibrium, which pair of conditions must both be satisfied?
- (A) The resultant force is zero, and the body is at rest
- (B) The resultant force is zero, and the resultant moment about any point is zero
- (C) The resultant moment is zero, and all forces act through the centre of mass
- (D) All the forces on the body are equal in magnitude
Show Answer
Answer: B – An extended body can rotate, so force balance alone is not enough: the resultant force must be zero (no linear acceleration) AND the resultant moment about any point must be zero (no angular acceleration). A body in equilibrium can still be moving at constant velocity, so being at rest is not required.
Q2. Two equal, opposite, parallel forces act on a body, but their lines of action do not coincide (a couple). What happens to the body?
- (A) It stays in complete equilibrium
- (B) It accelerates in a straight line
- (C) It rotates, because the resultant moment is not zero even though the resultant force is zero
- (D) It moves in a circle at constant speed
Show Answer
Answer: C – The two forces cancel, so the resultant force is zero and there is no linear acceleration. But together they form a couple whose moment is not zero, so the body gains angular acceleration and rotates. Zero resultant force alone does not guarantee equilibrium for an extended body.
Q3. For a particle (point object) acted on by several forces, the condition for equilibrium is that
- (A) the vector sum of the forces is zero
- (B) the sum of the moments about the particle is zero
- (C) the forces are equal in magnitude in pairs
- (D) the particle is at rest
Show Answer
Answer: A – A particle has no size, so rotation and moments do not apply to it. Equilibrium simply requires the forces to balance: the vector sum (resultant force) is zero. The particle may be at rest or moving at constant velocity.
Q4. A lift of weight 12 000 N descends at constant velocity, supported by a single cable. Which statement about the tension T in the cable is correct?
- (A) T = 0
- (B) T is slightly less than 12 000 N
- (C) T is slightly more than 12 000 N
- (D) T = 12 000 N, so the resultant force is zero
Show Answer
Answer: D – Constant velocity means zero acceleration, so by Newton’s first law the resultant force is zero. The only vertical forces are the weight (12 000 N down) and the tension (up), so T = 12 000 N. This is dynamic equilibrium: the lift moves, yet the forces balance.
Q5. A book rests on a horizontal table. It is in equilibrium because
- (A) the resultant force and the resultant moment on it are both zero
- (B) no forces act on it
- (C) its weight acts at its centre of mass
- (D) friction holds it in place
Show Answer
Answer: A – Forces do act: the weight is balanced by the normal reaction from the table, giving zero resultant force, and the moments of these forces about any point also cancel, giving zero resultant moment. Both equilibrium conditions are met, so the book stays at rest (static equilibrium).
Q6. In which situation is the resultant force on the body zero, yet the body is NOT in equilibrium?
- (A) A car cruising at constant velocity on a straight road
- (B) A hanging sign at rest outside a shop
- (C) Two equal and opposite forces applied at opposite ends of a ruler, offset from each other
- (D) A book resting on a shelf
Show Answer
Answer: C – The offset equal-and-opposite forces form a couple: they cancel as forces but produce a resultant moment, so the ruler starts to rotate. In the other three cases both the resultant force and the resultant moment are zero, so they are true equilibria.
Q7. A body is in equilibrium under the action of exactly two forces. The two forces must be
- (A) equal in magnitude and perpendicular to each other
- (B) unequal in magnitude but opposite in direction
- (C) equal in magnitude and in the same direction
- (D) equal in magnitude, opposite in direction, and acting along the same line (collinear)
Show Answer
Answer: D – Zero resultant force requires the forces to be equal and opposite. Zero resultant moment requires them to act along the same line of action; if their lines were offset they would form a couple and the body would rotate.
Q8. A couple consists of two forces, each of magnitude 20 N, whose lines of action are 0.30 m apart. What is the moment of the couple?
- (A) 3.0 N m
- (B) 6.0 N m
- (C) 12 N m
- (D) 67 N m
Show Answer
Answer: B – Moment of a couple = one force × perpendicular distance between the lines of action = 20 × 0.30 = 6.0 N m. The moment of a couple is the same about every point, which is why it produces pure rotation.
Q9. Which statement about a body in equilibrium is FALSE?
- (A) The body must be stationary
- (B) The resultant force on the body is zero
- (C) The resultant moment on the body about any point is zero
- (D) The body may be moving at constant velocity
Show Answer
Answer: A – Equilibrium requires zero resultant force and zero resultant moment, not rest. A body moving at constant velocity with no resultant moment is in dynamic equilibrium, so statement A is false.
Q10. An aircraft flies horizontally at constant speed and at constant height. Which row correctly gives the resultant force and the resultant moment acting on the aircraft?
- (A) Resultant force: non-zero; Resultant moment: non-zero
- (B) Resultant force: zero; Resultant moment: non-zero
- (C) Resultant force: non-zero; Resultant moment: zero
- (D) Resultant force: zero; Resultant moment: zero
Show Answer
Answer: D – Constant velocity means lift balances weight and thrust balances drag (zero resultant force), and flying straight without pitching or rolling means the moments also balance (zero resultant moment). Both equilibrium conditions hold while the aircraft is moving.
Resolving Forces in Equilibrium
Q11. A particle is acted on by a force of 30 N towards the east and a force of 40 N towards the north. A third force keeps the particle in equilibrium. What is the magnitude of the third force?
- (A) 10 N
- (B) 70 N
- (C) 50 N
- (D) 35 N
Show Answer
Answer: C – The resultant of the two perpendicular forces is √(302 + 402) = √(900 + 1600) = √(2500) = 50 N. For equilibrium the third force must be equal and opposite to this resultant, so its magnitude is 50 N (directed south-west).
Q12. A block of weight 60 N rests in equilibrium on a slope inclined at 30° to the horizontal. What is the frictional force acting on the block?
- (A) 30 N
- (B) 52 N
- (C) 60 N
- (D) 120 N
Show Answer
Answer: A – Resolve the weight along the slope: component down the slope = mg sin θ = 60 × sin 30° = 60 × 0.50 = 30 N. In equilibrium friction acts up the slope and balances this component, so friction = 30 N. (The normal reaction is mg cos θ = 60 × 0.866 = 52 N, option B’s trap.)
Q13. A sign of mass 8.0 kg hangs at rest from a single vertical cable. Taking g = 9.8 m s−2, what is the tension in the cable?
- (A) 8.0 N
- (B) 78 N
- (C) 0 N
- (D) 160 N
Show Answer
Answer: B – Weight = mg = 8.0 × 9.8 = 78 N. Vertical equilibrium: tension up = weight down, so T = 78 N.
Q14. A book of weight 10 N is held at rest against a vertical wall by a horizontal push of 25 N. What is the frictional force exerted on the book by the wall?
- (A) 25 N
- (B) 0 N
- (C) 35 N
- (D) 10 N
Show Answer
Answer: D – Vertically, the only forces are the weight (10 N down) and friction from the wall. For equilibrium friction must act upwards with magnitude 10 N. Horizontally, the push is balanced by the normal reaction of 25 N – the push controls how large friction could be, but friction itself only needs to match the weight here.
Q15. A force of 120 N acts at 60° above the horizontal. What is the vertical component of this force?
- (A) 60 N
- (B) 69 N
- (C) 104 N
- (D) 120 N
Show Answer
Answer: C – Vertical component = F sin θ = 120 × sin 60° = 120 × 0.866 = 104 N. (The horizontal component is F cos θ = 120 × 0.50 = 60 N, option A.)
Q16. A load of weight 100 N is held in equilibrium by a string whose tension T acts at 60° above the horizontal, together with a horizontal force H. What is the value of T?
- (A) 115 N
- (B) 100 N
- (C) 50 N
- (D) 200 N
Show Answer
Answer: A – Vertical equilibrium: T sin 60° = 100, so T = 100 / 0.866 = 115 N. Horizontally, H = T cos 60° = 115 × 0.50 = 58 N. Because the string is not vertical, its tension must exceed the weight.
Q17. A block of weight 80 N slides down a slope inclined at 45° at constant velocity. What is the magnitude of the frictional force on the block?
- (A) 80 N
- (B) 57 N
- (C) 40 N
- (D) 0 N
Show Answer
Answer: B – Constant velocity means equilibrium along the slope: friction up the slope = component of weight down the slope = mg sin θ = 80 × sin 45° = 80 × 0.707 = 57 N.
Q18. A particle is in equilibrium under four forces. Three of them are: 10 N east, 5 N west and 12 N north. What is the magnitude of the fourth force?
- (A) 7.0 N
- (B) 17 N
- (C) 27 N
- (D) 13 N
Show Answer
Answer: D – Sum the components of the three known forces: east-west: 10 − 5 = 5 N east; north-south: 12 N north. The fourth force must cancel these: 5 N west and 12 N south. Its magnitude is √(52 + 122) = √(25 + 144) = √(169) = 13 N.
Q19. A crate of weight 200 N rests on a slope inclined at 30° to the horizontal. What is the normal reaction exerted on the crate by the slope?
- (A) 100 N
- (B) 200 N
- (C) 173 N
- (D) 231 N
Show Answer
Answer: C – Perpendicular to the slope there is no acceleration, so the normal reaction balances the component of weight perpendicular to the slope: R = mg cos θ = 200 × cos 30° = 200 × 0.866 = 173 N.
Q20. A sledge of weight 500 N is pulled across horizontal snow by a rope with tension 200 N at 30° above the horizontal. The sledge moves at constant velocity. What is the normal reaction from the snow on the sledge?
- (A) 400 N
- (B) 500 N
- (C) 600 N
- (D) 327 N
Show Answer
Answer: A – The rope pulls partly upwards: vertical component = 200 × sin 30° = 100 N. Vertical equilibrium: R + 100 = 500, so R = 400 N. Pulling at an angle reduces the normal reaction (and hence friction) compared with a horizontal pull.
Three-Force & Two-Support Problems
Q21. Three forces act on a body that is in equilibrium. When the three force vectors are drawn tip-to-tail in turn, they form
- (A) a straight line
- (B) an open polygon
- (C) two parallel lines
- (D) a closed triangle
Show Answer
Answer: D – Equilibrium means the vector sum of the forces is zero. Placing the vectors tip-to-tail, the final tip lands back on the starting point, giving a closed triangle of forces. The angles of this triangle can then be used with the sine and cosine rules to find unknown forces.
Q22. A load of weight 240 N hangs from two symmetric ropes. Each rope makes an angle of 60° with the horizontal, and the tensions in the two ropes are equal. What is the tension in each rope?
- (A) 120 N
- (B) 139 N
- (C) 240 N
- (D) 480 N
Show Answer
Answer: B – Vertical equilibrium: the vertical components of the two tensions support the weight. 2T sin 60° = 240, so T = 240 / (2 × 0.866) = 240 / 1.732 = 139 N. The horizontal components cancel by symmetry.
Q23. A heavy load is supported by two symmetric ropes. Keeping the load the same, the ropes are re-arranged so that each makes a smaller angle with the horizontal (the ropes are nearer to horizontal). What happens to the tension in each rope?
- (A) It increases
- (B) It decreases
- (C) It stays equal to half the weight
- (D) It falls to zero
Show Answer
Answer: A – The vertical components must still add up to the weight: 2T sin θ = W. As θ gets smaller, sin θ falls, so T must grow to compensate. This is why a nearly horizontal rope can snap under a load it would easily hold when steep.
Q24. A picture of weight 36 N hangs in equilibrium from two wires of equal length attached symmetrically. Each wire makes an angle of 30° with the horizontal. What is the tension in each wire?
- (A) 18 N
- (B) 72 N
- (C) 36 N
- (D) 21 N
Show Answer
Answer: C – Vertical equilibrium: 2T sin 30° = 36, so 2T × 0.50 = 36, giving T = 36 N in each wire.
Q25. A uniform beam of length 4.0 m and weight 120 N rests on supports at its two ends. A load of 80 N is placed on the beam 1.0 m from the left-hand support. What is the reaction at the right-hand support?
- (A) 120 N
- (B) 100 N
- (C) 60 N
- (D) 80 N
Show Answer
Answer: D – Take moments about the left-hand support (so its reaction has no moment). Clockwise moments: beam weight acts at its centre, 2.0 m along: 120 × 2.0 = 240 N m; load: 80 × 1.0 = 80 N m. Total = 320 N m. Anticlockwise moment: Rright × 4.0. Equilibrium: Rright = 320 / 4.0 = 80 N. (Check: Rleft = 120 + 80 − 80 = 120 N.)
Q26. A uniform plank of length 6.0 m and weight 180 N rests on supports at its ends. A person of weight 720 N stands on the plank 1.5 m from the left-hand support. What is the reaction at the right-hand support?
- (A) 450 N
- (B) 270 N
- (C) 630 N
- (D) 900 N
Show Answer
Answer: B – Moments about the left-hand support: plank weight acts at 3.0 m: 180 × 3.0 = 540 N m; person: 720 × 1.5 = 1080 N m. Total clockwise = 1620 N m. Rright × 6.0 = 1620, so Rright = 270 N. (Check: Rleft = 180 + 720 − 270 = 630 N, and the left support carries more because the person stands nearer to it.)
Q27. A load is held in equilibrium by a horizontal string and a second, slanted string. The closed force triangle for the three forces has sides in the ratio 3 : 4 : 5, where the side of 4 units represents the weight of 40 N. What is the tension in the slanted string, represented by the side of 5 units?
- (A) 50 N
- (B) 40 N
- (C) 30 N
- (D) 70 N
Show Answer
Answer: A – In a force triangle every side uses the same scale. The weight, 40 N, corresponds to 4 units, so 1 unit = 10 N. The slanted string is the 5-unit side: 5 × 10 = 50 N. (The horizontal string is the 3-unit side, 30 N.)
Q28. A particle is in equilibrium under three forces that form a right-angled force triangle. The largest force (the hypotenuse of the triangle) is 50 N and one of the other forces is 30 N. What is the magnitude of the third force?
- (A) 80 N
- (B) 20 N
- (C) 40 N
- (D) 58 N
Show Answer
Answer: C – Use Pythagoras: third force = √(502 − 302) = √(2500 − 900) = √(1600) = 40 N. The three forces form a 3-4-5 triangle scaled by 10.
Q29. A sign of weight 60 N hangs from the far end of a horizontal bracket beam of length 1.2 m pivoted at a wall. A tie cable runs from the far end of the beam up to the wall, making an angle of 45° with the beam. Taking moments about the pivot (and neglecting the beam’s weight), what is the tension in the tie cable?
- (A) 60 N
- (B) 42 N
- (C) 120 N
- (D) 85 N
Show Answer
Answer: D – Only the vertical component of the tension, T sin 45°, has a moment about the pivot (the horizontal component acts along the beam through the pivot line). Moment balance: T sin 45° × 1.2 = 60 × 1.2, so T = 60 / 0.707 = 85 N.
Q30. A child of weight 320 N sits 1.2 m from the pivot of a seesaw. Where must an adult of weight 480 N sit, on the other side of the pivot, for the seesaw to balance?
- (A) 1.8 m from the pivot
- (B) 0.80 m from the pivot
- (C) 1.2 m from the pivot
- (D) 0.53 m from the pivot
Show Answer
Answer: B – Balance requires equal and opposite moments about the pivot (resultant moment zero): 480 × d = 320 × 1.2 = 384 N m, so d = 384 / 480 = 0.80 m. The heavier adult sits closer to the pivot.
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