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Equilibrium of concurrent forces: NEET MCQs with answers

48 questions on this topic of Laws of Motion (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A heavy sphere of weight W rests in equilibrium against a smooth vertical wall, supported by a light inextensible wire attached to its surface. The line of the wire, when extended, passes through the center of the sphere and makes an angle θ with the vertical wall. The force exerted by the sphere on the wall is:

  1. W cos θ
  2. W sin θ
  3. W cot θ
  4. W tan θ
Show answer and explanation

Answer: (D) W tan θ

Let the normal force from the wall on the sphere be N (directed horizontally away from the wall) and tension in the wire be T (directed along the wire at angle θ to the vertical). In equilibrium: T cos θ = W and T sin θ = N. Dividing gives N = W tan θ. By Newton's third law, the sphere exerts an equal force N = W tan θ on the wall.

medium

2. A particle is in static equilibrium under the action of three concurrent forces F₁, F₂, and F₃. The angle between F₁ and F₂ is 150°, and the angle between F₂ and F₃ is 120°. If F₁ = 10 N, what is the magnitude of F₂?

  1. 10√3 N
  2. 20 N
  3. 20/√3 N
  4. 10 N
Show answer and explanation

Answer: (C) 20/√3 N

The angles between the forces are: between F₁ and F₂ is 150°, between F₂ and F₃ is 120°, so between F₃ and F₁ is 360° - (150° + 120°) = 90°. By Lami's theorem: F₁ / sin(120°) = F₂ / sin(90°). Thus 10 / (√3/2) = F₂ / 1, giving F₂ = 20/√3 N.

hard

3. Three concurrent forces F₁, F₂, and F₃ are in equilibrium. The angle between F₁ and F₂ is 90°, and the angle between F₂ and F₃ is 120°. If |F₁| = 6 N, what is the magnitude of F₃?

  1. 6√3 N
  2. 4√3 N
  3. 12 N
  4. 3√3 N
Show answer and explanation

Answer: (B) 4√3 N

The total angle around the point is 360°, so the angle between F₁ and F₃ is 360° - (90° + 120°) = 150°. Applying Lami's theorem: |F₁| / sin(angle between F₂ and F₃) = |F₃| / sin(angle between F₁ and F₂). Thus, 6 / sin 120° = |F₃| / sin 90°. Since sin 120° = √3/2 and sin 90° = 1, |F₃| = 6 / (√3/2) = 12 / √3 = 4√3 N.

medium

4. Two forces of magnitudes 8 N and 6 N act at a right angle (90°) at a point. A third force F is applied so that the point remains in static equilibrium. The magnitude and direction of F are:

  1. 10 N, directed opposite to the resultant of the two forces
  2. 14 N, directed opposite to the resultant of the two forces
  3. 2 N, directed along the larger force
  4. 10 N, directed along the resultant of the two forces
Show answer and explanation

Answer: (A) 10 N, directed opposite to the resultant of the two forces

The resultant of 8 N and 6 N at 90° is R = √(8² + 6²) = √(64 + 36) = 10 N. For static equilibrium, the third force F must be equal in magnitude and exactly opposite in direction to this resultant (equilibriant force).

easy

5. A particle is in equilibrium under the action of three concurrent forces F⃗₁, F⃗₂, and F⃗₃. If the magnitude of F⃗₁ is 15 N and it acts along the positive x-axis, and F⃗₂ acts along the positive y-axis with magnitude 20 N, what is the magnitude of the third force F⃗₃?

  1. 5 N
  2. 25 N
  3. 35 N
  4. 20 N
Show answer and explanation

Answer: (B) 25 N

For equilibrium, the resultant of all concurrent forces must be zero: F⃗₁ + F⃗₂ + F⃗₃ = 0. Therefore, F⃗₃ = -(F⃗₁ + F⃗₂) = -(15î + 20ĵ) N. The magnitude is √(15² + 20²) = √(225 + 400) = √625 = 25 N.

hard

6. A uniform smooth spherical ball of weight W rests wedged between a vertical wall and an inclined plane making an angle of 60° with the horizontal. The normal force exerted by the vertical wall on the ball is:

  1. W / 2
  2. W √3
  3. W / √3
  4. 2 W / √3
Show answer and explanation

Answer: (C) W / √3

Let N₁ be the normal reaction from the vertical wall (horizontal) and N₂ be the normal reaction from the inclined plane (directed perpendicular to the incline, at 30° to the horizontal or 60° to the vertical). Vertically, N₂ cos 30° = W, so N₂ = W / cos 30° = 2W / √3. Horizontally, N₁ = N₂ sin 30° = (2W / √3) × (1/2) = W / √3.

easy

7. A small sphere of mass 2 kg is suspended by a light string. A horizontal force F holds the string displaced at an angle of 45° with the vertical. Taking g = 10 m/s², the magnitude of F and the tension T in the string are:

  1. F = 20 N, T = 20√2 N
  2. F = 10√2 N, T = 20 N
  3. F = 20√2 N, T = 20 N
  4. F = 20 N, T = 20 N
Show answer and explanation

Answer: (A) F = 20 N, T = 20√2 N

Resolving forces at the point of suspension: T cos 45° = mg = 20 N, which gives T = 20 / cos 45° = 20√2 N. Horizontally, F = T sin 45° = (20√2)(1/√2) = 20 N.

medium

8. A light string of length 100 cm has its ends attached to two fixed points A and B at the same horizontal level, separated by 80 cm. A smooth weightless ring carrying a load of weight W is threaded on the string and hangs in equilibrium. The tension in the string is:

  1. (5/8) W
  2. (5/6) W
  3. (4/5) W
  4. W/2
Show answer and explanation

Answer: (B) (5/6) W

Due to symmetry, the smooth ring settles at the midpoint of the string, so each half has length 50 cm. The horizontal half-span is 40 cm. The angle θ with the vertical satisfies sin θ = 40/50 = 4/5, so cos θ = 3/5. Balancing vertical forces gives 2T cos θ = W, hence T = W / (2 cos θ) = W / (2 × 3/5) = 5W/6.

medium

9. A weight W hangs at rest from two strings, each making an angle of 60° with the vertical. The tension in each string is:

  1. W/√3
  2. W
  3. W/2
  4. 2W
Show answer and explanation

Answer: (B) W

Vertical equilibrium: 2T cos 60° = W, so 2T × 1/2 = W and T = W. The horizontal components cancel.

NCERT Class 11

medium

10. Statement I: A particle can be in equilibrium under the action of three non-coplanar concurrent forces. Statement II: Three non-coplanar vectors can never have a zero resultant.

  1. Both Statement I and Statement II are correct
  2. Both Statement I and Statement II are incorrect
  3. Statement I is correct, but Statement II is incorrect
  4. Statement I is incorrect, but Statement II is correct
Show answer and explanation

Answer: (D) Statement I is incorrect, but Statement II is correct

The sum of three non-coplanar vectors cannot be zero because the resultant of any two lies in their common plane, which cannot cancel the third vector lying outside that plane. Therefore, three non-coplanar concurrent forces can never be in equilibrium. Hence Statement I is incorrect and Statement II is correct.

easy

11. Four coplanar concurrent forces acting at the origin are in equilibrium. Three of the forces are F₁ = (3î + 4ĵ) N, F₂ = (-5î + 2ĵ) N, and F₃ = (6î - 8ĵ) N. The fourth force F₄ must be:

  1. (-2î + 3ĵ) N
  2. (4î - 2ĵ) N
  3. (-4î + 2ĵ) N
  4. (2î - 6ĵ) N
Show answer and explanation

Answer: (C) (-4î + 2ĵ) N

For static equilibrium of concurrent forces, ΣF = 0. Therefore, F₄ = -(F₁ + F₂ + F₃) = -[(3 - 5 + 6)î + (4 + 2 - 8)ĵ] = -(4î - 2ĵ) = (-4î + 2ĵ) N.

medium

12. Three coplanar forces of magnitudes 10 N, F, and 10 N act concurrently on a particle at mutual angles of 120° between any adjacent pair. If the particle is in static equilibrium, the magnitude of F must be:

  1. 10 N
  2. 5 N
  3. 10√3 N
  4. 20 N
Show answer and explanation

Answer: (A) 10 N

By Lami's theorem or vector addition, if three coplanar forces act at mutual angles of 120° to each other, their magnitudes must be equal for their resultant to be zero. Hence, F = 10 N.

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