← 2020 Paper 1

UPSC 2020 Maths Optional Paper 1 Q5d — Step-by-Step Solution

10 marks · Section B

Equilibrium of a system of particles · Dynamics & Statics · asked 8× in 13 yrs · Read the full method →

Question

A uniform rod, in vertical position, can turn freely about one of its ends and is pulled aside from the vertical by a horizontal force acting at the other end of the rod and equal to half its weight. At what inclination to the vertical will the rod rest?

Technique

Moment (torque) balance about the fixed hinge; lever arm of the horizontal force is the vertical projection, lever arm of weight is the horizontal projection.

Solution

Setup

Let the rod $OA$ of length $2l$ (so $l$ = half-length) and weight $W$ be hinged (turning freely) at $O$. It rests inclined at angle $\theta$ to the upward vertical. Forces:

• Weight $W$ acting vertically downward at the midpoint $G$, where $OG=l$.
• Horizontal force $F=\dfrac{W}{2}$ applied at the far end $A$, where $OA=2l$.
• Reaction at the hinge $O$ (passes through $O$, so it has zero moment about $O$).

Step 1 — Take moments about the hinge $O$

The rod makes angle $\theta$ with the vertical. Position of a point at distance $s$ from $O$ along the rod: horizontal offset $s\sin\theta$, vertical offset $s\cos\theta$.

• Weight $W$ (downward) at $G$ ($s=l$): horizontal distance from $O$ is $l\sin\theta$. Moment (tending to increase $\theta$, i.e. topple) $=W\cdot l\sin\theta$.
• Horizontal force $F$ at $A$ ($s=2l$): perpendicular distance from $O$ to the horizontal line of action is the vertical offset $2l\cos\theta$. Moment (tending to decrease $\theta$, i.e. restore toward… actually pull aside) $=F\cdot 2l\cos\theta$.

These two moments act in opposite senses; equilibrium requires their balance:

Step 2 — Insert $F=W/2$

Divide by $Wl$:

Answer