UPSC 2022 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Friction (limiting friction) · Dynamics & Statics · asked 5× in 13 yrs · Read the full method →

Question

A body of weight $w$ rests on a rough inclined plane of inclination $\theta$ , coefficient of friction $\mu>\tan\theta$ . Find the work done in slowly dragging the body distance $b$ up the plane and then dragging it back, the applied force in each case parallel to the plane.

Technique

Two-direction force balance for slow (quasi-static) motion; identify friction direction (opposes motion); sum work over round trip.

Solution

Setup. Inclined plane at angle $\theta$ . Normal reaction $N=w\cos\theta$ . Friction $f=\mu N=\mu w\cos\theta$ .

The condition $\mu>\tan\theta$ means $\mu w\cos\theta>w\sin\theta$ — friction exceeds the component of gravity along the slope. So the body would not slide on its own; it requires a push to move up and a push to move down.

Step 1 — Dragging up

Forces along the slope (taking up the slope as positive):

Applied force $F_{up}$ (up).
Gravity component $-w\sin\theta$ .
Friction $-\mu w\cos\theta$ (opposing motion, so directed down when body moves up).

Slow motion (quasi-static): net force = 0.

$F_{up}=w\sin\theta+\mu w\cos\theta=w(\sin\theta+\mu\cos\theta)$ .

Work done in dragging up distance $b$ :

W_{up}=F_{up}\cdot b=w b(\sin\theta+\mu\cos\theta).

Step 2 — Dragging back down

When pulling the body back down, the body still moves slowly. Friction now opposes the motion, so friction is directed up the slope.

Applied force $F_{down}$ (now pointing down the slope to drag it down, but the body’s weight component already wants to pull it down). However, since $\mu>\tan\theta$ , gravity alone is not enough — needs additional force.

Forces along the slope (taking down the slope as positive):

Applied force $F_{down}$ (down).
Gravity component $+w\sin\theta$ .
Friction $-\mu w\cos\theta$ (opposing motion; body moves down, so friction is up the slope).

Slow motion: net = 0.

$F_{down}+w\sin\theta=\mu w\cos\theta$ , $F_{down}=\mu w\cos\theta-w\sin\theta=w(\mu\cos\theta-\sin\theta)$ .

(Positive since $\mu>\tan\theta\Rightarrow\mu\cos\theta>\sin\theta$ .)

Work done in dragging down distance $b$ :

W_{down}=F_{down}\cdot b=w b(\mu\cos\theta-\sin\theta).

Step 3 — Total work

W=W_{up}+W_{down}=wb(\sin\theta+\mu\cos\theta)+wb(\mu\cos\theta-\sin\theta)=2\mu w b\cos\theta.

Answer

\boxed{\;W=2\mu w b\cos\theta.\;}