Torque and Equilibrium of Rigid Body: NEET MCQs with Answers

Explanation: α = τ/I = 1/3 rad/s². ω = αt = 1 rad/s.

Explanation: τ = r × F = 2k̂ × 3ĵ = 6(k̂ × ĵ) = –6î N m.

Explanation: Taking moments about B, N_A d = W(d – x), so N_A = W(d – x)/d — the support nearer the centre of mass carries the larger share of the weight.

Explanation: τ = r × F = | î ĵ k̂ ; 7 3 1 ; −3 1 5 | î-component: (3)(5) − (1)(1) = 15 − 1 = 14 ĵ-component: −[(7)(5) − (1)(−3)] = −[35 + 3] = −38 k̂-component: (7)(1) − (3)(−3) = 7 + 9 = 16 τ = 14î − 38ĵ + 16k̂.  

Explanation: A couple is a pair of equal, opposite and parallel forces separated by a distance. Their vector sum is zero, so the net force on the body is zero and the centre of mass has no acceleration — there is no translation. However, the two forces produce torques in the same sense, so the resultant torque is non-zero and the body rotates. Hence a couple produces purely rotational motion.

Explanation: By definition τ = r × F, and a cross product is perpendicular to both of its factors. Therefore τ ⊥ r and τ ⊥ F, so both scalar products vanish: r · τ = 0 and F · τ = 0.

Explanation: Vertical equilibrium: N₁ + N₂ = w. Taking moments about the centre of mass (or equivalently about B), the anticlockwise and clockwise moments must balance: N₁x = N₂(d − x) Substituting N₂ = w − N₁ : N₁x = (w − N₁)(d − x) = wd − wx − N₁d + N₁x ⟹ N₁d = w(d − x) ⟹ N₁ = w(d − x)/d. The support nearer the centre of mass carries the larger share of the weight.

Explanation: Position vector of the point of application relative to the reference point: r = (2, 0, −3) − (2, −2, −2) = (0, 2, −1) = 0î + 2ĵ − k̂. τ = r × F = | î ĵ k̂ ; 0 2 −1 ; 4 5 −6 | î-component: (2)(−6) − (−1)(5) = −12 + 5 = −7 ĵ-component: −[(0)(−6) − (−1)(4)] = −[0 + 4] = −4 k̂-component: (0)(5) − (2)(4) = −8 τ = −7î − 4ĵ − 8k̂.

Explanation: τ = r × F = (2k̂) × (3ĵ) = 6 (k̂ × ĵ). Since k̂ × ĵ = −î, τ = −6î N-m.

Explanation: Position vector of the point of application relative to (2, −2, −2): r = (2−2, 0−(−2), −3−(−2)) = (0, 2, −1). τ = r × F = (2·(−6) − (−1)·5, (−1)·4 − 0·(−6), 0·5 − 2·4) = (−7, −4, −8), i.e. τ = −7i − 4j − 8k.

Explanation: Taking moments about B: N_A · d = W · (d − x) ⇒ N_A = W(d − x)/d.

Explanation: τ = r × F is perpendicular to BOTH r and F by the definition of the cross product. Hence r·τ = 0 and F·τ = 0.

Explanation: In X the forces add to a net force 2F, while their moments about C, each of magnitude Fa, are in opposite senses and cancel. So the rod is in rotational but not translational equilibrium. In Y the net force is zero, but both moments act in the same sense and add to 2Fa, so the forces form a couple. Thus each case is only a partial equilibrium: zero net force and zero net torque are independent conditions.

Explanation: The smooth edge pushes perpendicular to the stick, not horizontally. Taking torques about the lower end, N(0.75) = 20 × 0.5 × 0.8, so N = 32/3 N. The horizontal component of this push is N sin 37° = 6.4 N, and with no other horizontal force, the friction at the floor must balance it.

Explanation: The top of the ladder is 4 m above the floor, and the smooth wall exerts only a horizontal force N. Taking moments about the foot gives N × 4 = 300 × 1.5, so N = 112.5 N. Horizontal balance makes the friction 112.5 N and vertical balance makes the floor's normal reaction 300 N. So the least coefficient is 112.5/300 = 0.375.