UPSC 2024 Maths Optional Paper 1 Q6a — Step-by-Step Solution

15 marks · Section B

Principle of virtual work · Dynamics & Statics · asked 6× in 13 yrs · Read the full method →

Question

A regular tetrahedron, formed of six light rods each of length $l$ , rests on a smooth horizontal plane. A ring of weight $W$ and radius $r$ is supported by the slant sides. Using the principle of virtual work, find the stress in any of the horizontal sides.

Technique

Parametrise the configuration by base-edge length $s$ ; compute the ring height $z(s)$ ; differentiate; apply virtual work with three base-rod tensions and the ring’s weight.

Solution

Regular tetrahedron of six light rods (length l) on a smooth horizontal plane. The three base rods (red) are the horizontal sides whose stress T is sought; the three slant rods rise to the apex. A ring of radius r and weight W rests horizontally on the three slant rods. The virtual work is done as the base triangle's side s is varied, changing the apex height and hence the ring's height z(s).

Setup. The base is an equilateral triangle of side $s$ (varied); three slant rods of fixed length $l$ meet at the apex. The apex height is $h(s)=\sqrt{l^2-s^2/3}$ . The circumradius of the base triangle is $s/\sqrt{3}$ .

Step 1 — Height of the ring.

The ring of radius $r$ (centred on the axis) rests on the three slant rods. At depth $d$ below the apex the slant rods are at horizontal distance $(s/\sqrt{3})(d/h)$ from the axis. Setting this equal to $r$ :

d=\frac{r h\sqrt{3}}{s},\qquad z(s)=h(s)\!\left(1-\frac{r\sqrt{3}}{s}\right).

Step 2 — Differentiate $z$ at $s=l$ .

h(l)=\frac{l\sqrt{6}}{3},\qquad\frac{dh}{ds}\bigg|_{s=l}=\frac{-l/3}{l\sqrt{2/3}}=\frac{-1}{\sqrt{6}}.

By the product rule:

\frac{dz}{ds}\bigg|_{s=l}=-\frac{1}{\sqrt{6}}\!\left(1-\frac{r\sqrt{3}}{l}\right)+\frac{l\sqrt{6}}{3}\cdot\frac{r\sqrt{3}}{l^2}=-\frac{1}{\sqrt{6}}+\frac{r}{l\sqrt{2}}+\frac{r\sqrt{2}}{l}=-\frac{1}{\sqrt{6}}+\frac{3r}{l\sqrt{2}}.

Step 3 — Virtual work.

In a virtual displacement $\delta s$ (each base edge lengthens by $\delta s$ ):

Three base rods of tension $T$ each do work $-T\,\delta s$ ; total $-3T\,\delta s$ .
Ring of weight $W$ drops by $\delta z=(dz/ds)\delta s$ ; gravity does work $-W\,\delta z$ .

Setting total virtual work to zero:

-3T\,\delta s-W\frac{dz}{ds}\,\delta s=0\;\Rightarrow\;T=-\frac{W}{3}\cdot\frac{dz}{ds}\bigg|_{s=l}.

T=-\frac{W}{3}\!\left(-\frac{1}{\sqrt{6}}+\frac{3r\sqrt{2}}{2l}\right)=\frac{W}{3\sqrt{6}}-\frac{Wr\sqrt{2}}{2l}=\frac{W\sqrt{6}}{18}-\frac{9Wr\sqrt{2}}{18l}=\frac{W\sqrt{2}(l\sqrt{3}-9r)}{18l}.

$T>0$ (tension) when $r<l\sqrt{3}/9$ ; $T<0$ (compression) otherwise.

Answer

\boxed{\;T=\frac{W\sqrt{2}(l\sqrt{3}-9r)}{18l}.\;}