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UPSC 2015 Maths Optional Paper 1 Q5d — 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 rod of 8 kg is movable in a vertical plane about a hinge at one end; to the other end is fastened a weight equal to half of the rod; this end is fastened by a string of length $l$ to a point at a height $b$ above the hinge vertically. Obtain the tension in the string.

Technique

Moment balance about the hinge; sine rule in triangle $OPA$ to express the angle between rod and string; rod length and tilt angle cancel.

Solution

Setup. Let $O$ be the hinge, $A$ the other end of the rod. Rod $OA$ has weight $W=8$ kg-wt (treating “kg” as kg-weight for the statics, as is conventional in UPSC). Length of rod: not given explicitly — interpret as $l_r$. At $A$, an additional weight $w=W/2=4$ kg-wt is fastened. From $A$, a string of length $l$ runs to a fixed point $P$ situated vertically above $O$ at height $b$ (so $OP=b$).

Geometry. Let $\theta$ be the angle $\angle POA$ (between rod and the vertical upward direction $OP$). In triangle $OPA$:

• $OP=b$ (vertical).
• $OA=l_r$ (rod length).
• $PA=l$ (string).

By law of cosines applied at $O$: $l^2=b^2+l_r^2-2 b\,l_r\cos\theta$.

But the rod length $l_r$ is not given separately — so a likely reading is that the rod has length equal to $l$ (string length) — but that’s also not stated. Re-reading the problem:

A rod of 8 kg is movable in a vertical plane about a hinge at one end, another end is fastened a weight equal to half of the rod, this end is fastened by a string of length $l$ to a point at a height $b$ above the hinge vertically.

Standard textbook problem: typically the rod has length $2a$ or simply $L$, with the string going from the free end $A$ to a point at height $b$ above the hinge. In many versions the rod length equals $b$ or equals $l$, but here both $l$ and $b$ appear separately as given parameters.

Reading. Let me interpret with the rod length as $2a$ (unstated, takes a symbol), or more cleanly, take the rod length to be $L$. Working out: only the moments matter for tension, and we can use virtual work / equilibrium of moments about the hinge.

Step 1 — Moments about the hinge $O$

Forces on the rod:

• Weight of rod $W=8$ kg-wt acting at midpoint $G$ (distance $L/2$ from $O$), vertically downward.
• Weight $w=4$ kg-wt hanging at $A$, vertically downward (distance $L$ from $O$ along rod).
• Tension $T$ in the string $PA$, acting at $A$ along $A\to P$.
• Reaction at $O$ — passes through $O$, no moment.

Let the rod make angle $\theta$ with the vertical (upward) — so $A$ is at horizontal distance $L\sin\theta$ from $O$ and vertical distance $L\cos\theta$ below… wait, the rod swings, the angle convention matters.

Let me set angle $\theta=\angle POA$ between rod and the upward vertical $OP$. Then $A$ is at position $(L\sin\theta,L\cos\theta)$ if we measure $\theta$ from $OP$ — but if $\theta<\pi/2$, $A$ is above $O$, which doesn’t fit a rod hanging down. Let’s take $\theta$ as the angle the rod makes with $OP$, with $\theta$ obtuse (rod hangs down on the same side as the string).

Actually for a hinged-rod equilibrium with a string above, the rod is held up by the string (otherwise gravity drops it to vertical). So $A$ is above the horizontal through $O$, meaning $\theta<\pi/2$.

Position of $A$: $A=(L\sin\theta,L\cos\theta)$ with $O$ at origin and $P=(0,b)$.

Take moments about $O$ (anticlockwise positive). Set up: weight $W$ at $G=(L\sin\theta/2,L\cos\theta/2)$, force $-W\hat j$; moment $=$ position $\times$ force $=L\sin\theta/2\cdot(-W)\cdot(-1)$ — sign tedious. Better: magnitude approach.

Gravity moments (tend to swing the rod toward vertical-down). With $A$ at horizontal distance $L\sin\theta$ from $O$:

• Rod weight at midpoint: moment about $O=W\cdot(L\sin\theta/2)=8\cdot L\sin\theta/2=4L\sin\theta$ (kg-wt $\cdot$ length), clockwise (tending to drop $A$).
• Hanging weight at $A$: moment = $w\cdot L\sin\theta=4L\sin\theta$, clockwise.

Total clockwise gravity moment about $O$: $(4L+4L)\sin\theta=8L\sin\theta$.

Tension moment. Tension $T$ acts at $A$ along $A\to P$. Component perpendicular to the rod (at $A$) gives moment $=T\cdot L\cdot\sin\phi$, where $\phi$ is the angle between the rod $OA$ and the string $AP$ at $A$.

Use sine rule in triangle $OPA$:

So tension moment about $O$ (anticlockwise, opposing gravity) = $T\cdot L\cdot\sin\phi=T\cdot L\cdot\dfrac{b\sin\theta}{l}=\dfrac{TLb\sin\theta}{l}$.

Step 2 — Equilibrium

Set net moment about $O$ to zero:

The rod length $L$ cancels, and $\sin\theta$ cancels — the answer is independent of the rod length and of $\theta$.

Answer