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UPSC 2020 Maths Optional Paper 1 Q6c — Step-by-Step Solution

15 marks · Section B

Equilibrium of a system of particles · Dynamics & Statics · asked 8× in 14 yrs · Read the full method →

Question

A beam ADAD rests on two supports BB and CC, where AB=BC=CDAB=BC=CD. It is found that the beam will tilt when a weight of pp kg is hung from AA or when a weight of qq kg is hung from DD. Find the weight of the beam.

Technique

Limiting equilibrium / tilting: at the verge of tilting the far support reaction is zero; take moments about the near support. Two cases, then eliminate the unknown CG position.

Solution

Setup

Let AB=BC=CD=ℓAB=BC=CD=\ell, so measuring from AA:

A:0,B:ℓ,C:2ℓ,D:3ℓ.A:0,\quad B:\ell,\quad C:2\ell,\quad D:3\ell .

Let the beam’s weight be WW, acting at its centre of gravity GG at distance xx from AA. (We do not assume the beam is uniform; if the result forces x=3ℓ2x=\tfrac{3\ell}{2} it would mean p=qp=q, but generally p≠qp\ne q, so xx is unknown too.)

The beam rests on the two supports BB and CC, which can only push up. When a load is hung, one support’s reaction can drop to zero; the beam is on the verge of tilting about the other support.

Step 1 — Tilt when pp is hung from AA

Hanging pp at AA tends to lift the DD–end, so the beam is about to rotate (tilt) about BB; the reaction at CC just vanishes, RC=0R_C=0.

Take moments about BB (the pivot). The downward load pp at AA is at horizontal distance ℓ\ell on the AA–side; the beam weight WW at GG is at distance (x−ℓ)(x-\ell) on the DD–side. At the verge of tilting these balance:

p⋅ℓ=W (x−ℓ).(1)p\cdot \ell = W\,(x-\ell). \tag{1}

Step 2 — Tilt when qq is hung from DD

Hanging qq at DD tends to lift the AA–end, so the beam is about to tilt about CC; RB=0R_B=0.

Moments about CC: load qq at DD is at distance ℓ\ell; weight WW at GG is at distance (2ℓ−x)(2\ell-x) on the AA–side:

q⋅ℓ=W (2ℓ−x).(2)q\cdot \ell = W\,(2\ell - x). \tag{2}

Step 3 — Eliminate xx

Add (1) and (2):

pℓ+qℓ=W(x−ℓ)+W(2ℓ−x)=W[(x−ℓ)+(2ℓ−x)]=W⋅ℓ.p\ell + q\ell = W(x-\ell)+W(2\ell-x)=W\big[(x-\ell)+(2\ell-x)\big]=W\cdot \ell .

Therefore

(p+q)ℓ=Wℓ⇒W=p+q.(p+q)\ell = W\ell \quad\Rightarrow\quad W=p+q .

Answer

 Weight of the beam W=p+q (kg) \boxed{\ \text{Weight of the beam } W=p+q\ \text{(kg)}\ }
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