← 2020 Paper 1

UPSC 2020 Maths Optional Paper 1 Q7c-i — Step-by-Step Solution

10 marks · Section B

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

Question

A square framework formed of uniform heavy rods of equal weight WW jointed together, is hung up by one corner. A weight WW is suspended from each of the three lower corners, and the shape of the square is preserved by a light rod along the horizontal diagonal. Find the thrust of the light rod.

Technique

Principle of virtual work with the single DOF θ\theta (half-angle of the rods); T=(dV/dθ)/(dL/dθ)T=(dV/d\theta)/(dL/d\theta).

Solution

Setup

Label the square ABCDABCD hung from the top corner AA, so the diagonal ACAC is vertical and the diagonal BDBD is horizontal. The four uniform rods AB,BC,CD,DAAB, BC, CD, DA each have weight WW. Each rod makes 4545^\circ with the vertical (square geometry). Hung weights WW act at the three lower corners B,C,DB, C, D. A light rod lies along the horizontal diagonal BDBD.

Use the principle of virtual work. Let each upper rod make angle θ\theta with the downward vertical (the square configuration is θ=45\theta=45^\circ), and let each rod have length 2a2a. Take AA as origin, yy measured upward (so the frame is at y<0y<0).

Step 1 — Heights of all weights as functions of θ\theta

Going down the rods at angle θ\theta to the vertical, each rod’s vertical extent is 2acosθ2a\cos\theta.

Step 2 — Total potential energy

Weights: four rod weights WW at the rod midpoints, plus hung weights WW at BB, CC, DD.

V=W[2(acosθ)]+W[2(3acosθ)]+W(2acosθ)2B,D hung+W(4acosθ)C hungV=W\big[2(-a\cos\theta)\big]+W\big[2(-3a\cos\theta)\big]+\underbrace{W(-2a\cos\theta)\cdot2}_{B,D\text{ hung}}+\underbrace{W(-4a\cos\theta)}_{C\text{ hung}} =Wacosθ[2+6+4+4]=16Wacosθ.=-Wa\cos\theta\big[2+6+4+4\big]=-16\,Wa\cos\theta .

Step 3 — The diagonal BDBD and virtual work

Length of the horizontal diagonal:

LBD=2(2asinθ)=4asinθ.L_{BD}=2\,(2a\sin\theta)=4a\sin\theta .

Let the thrust (compressive force) in the light rod be TT; it pushes the corners BB and DD apart, so it does positive work when LBDL_{BD} increases. The principle of virtual work for the single degree of freedom θ\theta:

δV=TδLBDT=dV/dθdLBD/dθ.\delta V = T\,\delta L_{BD}\quad\Rightarrow\quad T=\frac{dV/d\theta}{dL_{BD}/d\theta}. dVdθ=16Wasinθ,dLBDdθ=4acosθ,\frac{dV}{d\theta}=16\,Wa\sin\theta,\qquad \frac{dL_{BD}}{d\theta}=4a\cos\theta, T=16Wasinθ4acosθ=4Wtanθ.T=\frac{16Wa\sin\theta}{4a\cos\theta}=4W\tan\theta .

Step 4 — Evaluate at the square configuration θ=45\theta=45^\circ

T=4Wtan45=4W.T=4W\tan45^\circ=4W .

Answer

 Thrust in the light rod =4W \boxed{\ \text{Thrust in the light rod }=4W\ }
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