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UPSC 2016 Maths Optional Paper 1 Q7a — Step-by-Step Solution

15 marks · Section B

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

Question

A uniform rod $AB$ of length $2a$ movable about a hinge at $A$ rests with other end against a smooth vertical wall. If $\alpha$ is the inclination of the rod to the vertical, prove that the magnitude of reaction of the hinge is $\dfrac12 W\sqrt{4+\tan^2\alpha}$ where $W$ is the weight of the rod.

Technique

Three-force rigid-body statics; moments about the hinge to find the wall reaction $S=\tfrac{W}{2}\tan\alpha$; resolve horizontally/vertically for the hinge components.

Solution

Step 1 — Set up forces

The rod $AB$ (length $2a$, weight $W$ at its midpoint $G$) is hinged at $A$ and leans with $B$ on a smooth vertical wall. Three forces act:

• weight $W$ vertically down at $G$;
• wall reaction $S$ at $B$, horizontal (perpendicular to the vertical wall, smooth $\Rightarrow$ no friction);
• hinge reaction at $A$ with components $X$ (horizontal) and $Y$ (vertical).

Place $A$ at the origin; the rod makes angle $\alpha$ with the vertical, so

Step 2 — Moments about $A$

Taking moments about $A$ removes the hinge reaction. The horizontal wall reaction $S$ (at height $2a\cos\alpha$) and the weight $W$ (horizontal offset $a\sin\alpha$):

Step 3 — Resolve for the hinge components

Horizontal: the only other horizontal force is $S$ (pushing the rod away from the wall), so

Vertical: the weight $W$ is balanced solely by the hinge (the wall reaction is horizontal), so

Step 4 — Magnitude of the hinge reaction

Answer