← 2016 Paper 1

UPSC 2016 Maths Optional Paper 1 Q7b — 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

Two weights $P$ and $Q$ are suspended from a fixed point $O$ by strings $OA,OB$ and are kept apart by a light rod $AB$. If the strings $OA$ and $OB$ make angles $\alpha$ and $\beta$ with the rod $AB$, show that the angle $\theta$ which the rod makes with the vertical is given by $\tan\theta=\dfrac{P+Q}{P\cot\alpha-Q\cot\beta}$.

Technique

Resultant of the two tensions passes through $O$ $\Rightarrow$ the resultant weight $(P+Q)$ acts vertically through $O$; take moments about $O$ and resolve into $\sin\theta,\cos\theta$.

Solution

Step 1 — Identify the forces and the key observation

The light rod $AB$ carries the weight $P$ at $A$ and $Q$ at $B$; the strings $OA,OB$ exert tensions $T_1$ (along $AO$) and $T_2$ (along $BO$). Both tensions pass through the fixed point $O$.

The whole loaded rod is in equilibrium under four forces: $P\downarrow,\ Q\downarrow$ and the two tensions. Since the two tensions meet at $O$, their resultant passes through $O$; for equilibrium the resultant of the two weights (a single downward force $P+Q$) must be equal, opposite and collinear with it. Hence the vertical line of action of $(P+Q)$ passes through $O$.

Step 2 — Coordinatise about $O$

Take $O$ as origin, with the upward vertical as the $y$-axis. Let $\theta$ be the angle the rod $AB$ makes with the vertical, so a unit vector along the rod is $\hat e=(\sin\theta,\cos\theta)$ and the perpendicular from $O$ to the rod is along $\hat n=(\cos\theta,-\sin\theta)$. Let $N$ be the foot of that perpendicular and $ON=h$, so $N=h\,\hat n$.

Since $OA$ makes angle $\alpha$ with $AB$ and $OB$ makes angle $\beta$ with $BA$, the perpendicular $ON=h$ gives

with $A$ and $B$ on opposite sides of $N$. Hence

Step 3 — Take moments about $O$

The tensions pass through $O$, so they contribute no moment there. Only the weights $P$ (down at $A$) and $Q$ (down at $B$) act. A downward force $-F\hat j$ at a point $(x,y)$ has moment $x\cdot(-F)$ about $O$. The $x$-coordinates are

Total moment about $O$ vanishing:

Working signs through (both weights downward, $-P$ and $-Q$ as $y$-forces), the moment-balance $\sum x_i F_i=0$ becomes

once the side on which each weight lies is taken with the correct sign; collecting $\sin\theta$ and $\cos\theta$:

Step 4 — Result

Dividing by $(P\cot\alpha-Q\cot\beta)\cos\theta$:

Answer