A rod l m long is acted upon by a couple as shown in the figure. The moment of couple is τ N m. If the force acts at each end of the rod, then magnitude of each force is (sin 30° = cos 60° = 0.5) (a) τ/l (b) (l)/(2τ) (c) (2τ)/(l) (d) 2/τ

Answer: (c). Moment of couple = force × perpendicular distance between the two lines of action. The forces make 30° with the rod, so the perpendicular distance is lsin 30° = l/2. τ = F/2 ⇒ F = (2τ)/(l) So the answer is (c) (2τ)/(l).

A rod l m long is acted upon by a couple as shown in the figure. The… | MHT CET 2019 (3 May, Shift 1)

Q32MHT CET 2019 · 3 MAY · SHIFT 1Easy

Torque, Couple & Equilibrium of Rigid Bodies

A rod $l\text{ m}$ long is acted upon by a couple as shown in the figure. The moment of couple is $\tau\text{ N m}$ . If the force acts at each end of the rod, then magnitude of each force is

$(\sin 30^\circ = \cos 60^\circ = 0.5)$

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Answer: (c)

Moment of couple = force $\times$ perpendicular distance between the two lines of action.
The forces make $30^\circ$ with the rod, so the perpendicular distance is $l\sin 30^\circ = \frac{l}{2}$ .
$\tau = F\cdot\frac{l}{2} \Rightarrow F = \frac{2\tau}{l}$
So the answer is (c) $\frac{2\tau}{l}$ .

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Torque, Couple & Equilibrium of Rigid Bodies question from MHT CET 2019 (3 May, Shift 1)

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