UPSC 2016 Maths Optional Paper 1 Q7c — Step-by-Step Solution

20 marks · Section B

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Question

A square $ABCD$ , the length of whose sides is $a$ , is fixed in a vertical plane with two of its sides horizontal. An endless string of length $l\ (>4a)$ passes over four pegs at the angles of the board and through a ring of weight $W$ which is hanging vertically. Show that the tension of the string is $\dfrac{W(l-3a)}{2\sqrt{l^2-6la+8a^2}}$ .

Technique

Smooth string $\Rightarrow$ uniform tension $T$ ; length bookkeeping gives each slant $L=\tfrac{l-3a}{2}$ ; vertical equilibrium of the ring $2T\cos\phi=W$ with $\cos\phi=\sqrt{L^2-(a/2)^2}/L$ .

Solution

Step 1 — Set up the geometry

Place the square’s corners (pegs) at

\text{top: }\left(-\tfrac a2,\tfrac a2\right),\ \left(\tfrac a2,\tfrac a2\right);\qquad \text{bottom: }\left(-\tfrac a2,-\tfrac a2\right),\ \left(\tfrac a2,-\tfrac a2\right).

The endless string is a single closed loop threaded through the ring and over the four pegs. By the left–right symmetry the ring $R$ hangs on the central vertical axis, at $R=(0,-y)$ below the square. The two segments leaving the ring rise symmetrically to the two upper pegs; the remainder of the loop runs down the two vertical sides and along the bottom edge.

Step 2 — Account for the string length

The closed loop consists of:

two equal slant segments from the ring $R$ to each top peg, each of length $L$ ;
the two vertical sides (each length $a$ ): from each top peg down to the corresponding bottom peg;
the bottom edge (length $a$ ) joining the two bottom pegs.

Total:

2L+2a+a=l\quad\Longrightarrow\quad L=\frac{l-3a}{2}.

Step 3 — Tension is uniform; resolve at the ring

The string is smooth over the pegs and through the ring, so the tension $T$ is the same throughout. At the ring, the only string segments pulling are the two slant pieces $RA',RB'$ (to the two top pegs), symmetric about the vertical, plus the weight $W$ down.

Let each slant make angle $\phi$ with the vertical. The horizontal offset from $R$ to a top peg is $\tfrac a2$ and the slant length is $L$ , so

\sin\phi=\frac{a/2}{L},\qquad \cos\phi=\frac{\sqrt{L^2-(a/2)^2}}{L}.

Vertical equilibrium of the ring (two tensions up, weight down):

2T\cos\phi=W\quad\Longrightarrow\quad T=\frac{W}{2\cos\phi}=\frac{WL}{2\sqrt{L^2-(a/2)^2}}.

Step 4 — Substitute $L=\dfrac{l-3a}{2}$

L^2-\left(\frac a2\right)^2=\frac{(l-3a)^2-a^2}{4}=\frac{l^2-6la+9a^2-a^2}{4}=\frac{l^2-6la+8a^2}{4}.

Therefore

T=\frac{W\cdot\frac{l-3a}{2}}{2\cdot\frac{\sqrt{l^2-6la+8a^2}}{2}}=\frac{W(l-3a)}{2\sqrt{l^2-6la+8a^2}}.

Answer

\boxed{\;T=\frac{W(l-3a)}{2\sqrt{l^2-6la+8a^2}}.\;}

UPSC 2016 Maths Optional Paper 1 Q7c — Step-by-Step Solution

Question

Technique

Solution

Step 1 — Set up the geometry

Step 2 — Account for the string length

Step 3 — Tension is uniform; resolve at the ring

Step 4 — Substitute L=l−3a2L=\dfrac{l-3a}{2}L=2l−3a​

Answer

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.

Step 4 — Substitute $L=\dfrac{l-3a}{2}$