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UPSC 2022 Maths Optional Paper 1 Q8b — 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 chain of $n$ equal uniform rods is smoothly jointed and suspended from one end $A_1$. A horizontal force $\vec P$ is applied to the other end $A_{n+1}$. Find the inclinations of the rods to the downward vertical in equilibrium.

Technique

Moment balance about each upper joint for each rod individually; tension at lower joint of each rod determined by equilibrium of sub-chain below (horizontal = $P$, vertical = weight of rods below). Moment equation for rod $k$ gives $\tan\theta_k$ directly.

Solution

Setup. Each rod has weight $w$, length $l$. The chain hangs from $A_1$ at top, with rods labelled $1,2,\dots,n$ from top to bottom. Joints $A_1,A_2,\dots,A_{n+1}$ where $A_{n+1}$ is the bottom end.

Let $\theta_k$ be the angle that rod $k$ (from $A_k$ to $A_{k+1}$) makes with the downward vertical.

At joint $A_{n+1}$: horizontal force $P$ applied.

Step 1 — Free body of the chain below joint $A_k$ (i.e., rods $k,k+1,\dots,n$)

Number of rods below $A_k$: $n-k+1$ (including rod $k$ itself).

The forces on this sub-chain:

• Weight: $(n-k+1)w$ acting at the combined CG (downward).
• Horizontal force $P$ at $A_{n+1}$.
• Tension at joint $A_k$: unknown, but provides whatever is needed for equilibrium.

Step 2 — Take moments about $A_k$ for the sub-chain below

For the sub-chain consisting of rods $k$ through $n$, the moment of all external forces about $A_k$ must be zero.

Forces external to the sub-chain:

• Gravities of rods $k,k+1,\dots,n$.
• Horizontal $P$ at $A_{n+1}$.
• Reaction at $A_k$ (passes through $A_k$ — zero moment).

Easier approach: Use moments about $A_k$ for the rod $k$ alone, considering tension at $A_{k+1}$ from the rest of the chain.

Actually a standard cleaner method: consider rod $k$ in isolation. Forces:

• Weight $w$ at midpoint (distance $l/2$ from $A_k$ along rod $k$).
• Tension $T_{k+1}$ at $A_{k+1}$ from rods $k+1,\dots,n$.
• Reaction at $A_k$ from joint above.

The tension $T_{k+1}$ has horizontal and vertical components determined by force balance on the sub-chain below $A_{k+1}$:

• Horizontal: $T_{k+1,x}=P$ (the only horizontal external force on sub-chain $\{k+1,\dots,n\}$, balanced by horizontal component of tension at $A_{k+1}$).
• Vertical: $T_{k+1,y}=(n-k)w$ (weight of $n-k$ rods below joint $A_{k+1}$).

So tension at $A_{k+1}$ pulls rod $k$ outward and downward with components $(P,\;(n-k)w)$ — the rest of the chain pulls down on rod $k$ (and outward horizontally).

Step 3 — Moments about $A_k$ for rod $k$

Let rod $k$ make angle $\theta_k$ with downward vertical. Coordinate along rod: $A_k$ at origin, $A_{k+1}$ at $(l\sin\theta_k,-l\cos\theta_k)$ (down and slightly horizontal).

Weight $w$ at midpoint: position $(l\sin\theta_k/2,-l\cos\theta_k/2)$, force $(0,-w)$. Moment about $A_k$: $(l\sin\theta_k/2)\cdot(-w)-(-l\cos\theta_k/2)\cdot 0=-wl\sin\theta_k/2$.

Tension at $A_{k+1}$: position $(l\sin\theta_k,-l\cos\theta_k)$, force $(P,(n-k)w)$ — the outward and downward pull from below.

Wait — the tension at $A_{k+1}$ from the lower sub-chain on rod $k$ must equal $-$ (the force that the lower sub-chain experiences from rod $k$). By Newton’s 3rd: if rod $k$ pulls the lower sub-chain up and toward $A_k$ with force $(F_x,F_y)$, then lower sub-chain pulls rod $k$ down and away with force $(-F_x,-F_y)$.

Force on lower sub-chain from rod $k$ at $A_{k+1}$: must balance the lower sub-chain’s external forces (gravity + $P$). Lower sub-chain has weight $(n-k)w$ down and external $P$ horizontal. For equilibrium of lower sub-chain, the reaction at $A_{k+1}$ on lower sub-chain must be $(-P,(n-k)w)$ — i.e., upward by $(n-k)w$ and opposite to $P$ horizontally.

So force on rod $k$ at $A_{k+1}$ from lower sub-chain is $(P,-(n-k)w)$ — outward (in direction of $P$) and downward (gravity pull).

Moment about $A_k$ for rod $k$ from the tension at $A_{k+1}$: Position $(l\sin\theta_k,-l\cos\theta_k)$, force $(P,-(n-k)w)$.

Moment (z-component, CCW positive): $x F_y-y F_x=(l\sin\theta_k)(-(n-k)w)-(-l\cos\theta_k)(P)$ $=-(n-k)wl\sin\theta_k+Pl\cos\theta_k$.

Step 4 — Total moment about $A_k$ on rod $k$ = 0

$-wl\sin\theta_k/2-(n-k)wl\sin\theta_k+Pl\cos\theta_k=0$.

Divide by $l$ and rearrange: $wl(...)/l\cdot \cancel{} ...$ — careful, let me re-do:

$-\dfrac{wl\sin\theta_k}{2}-(n-k)wl\sin\theta_k+Pl\cos\theta_k=0$.

Divide by $l$: $-\dfrac{w\sin\theta_k}{2}-(n-k)w\sin\theta_k+P\cos\theta_k=0$.

$\sin\theta_k\!\left[\dfrac{w}{2}+(n-k)w\right]=P\cos\theta_k$.

$\sin\theta_k\cdot w\!\left[\dfrac{1}{2}+n-k\right]=P\cos\theta_k$.

$\tan\theta_k=\dfrac{P}{w(n-k+1/2)}=\dfrac{2P}{w(2n-2k+1)}.$

Answer