← 2019 Paper 1

UPSC 2019 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Friction (limiting friction) · Dynamics & Statics · asked 5× in 13 yrs · Read the full method →

Question

One end of a heavy uniform rod $AB$ can slide along a rough horizontal rod $AC$, to which it is attached by a ring. $B$ and $C$ are joined by a string. When the rod is on the point of sliding, then $AC^2-AB^2=BC^2$. If $\theta$ is the angle between $AB$ and the horizontal line, then prove that the coefficient of friction is $\dfrac{\cot\theta}{2+\cot^2\theta}$.

Technique

Recognise $AC^2-AB^2=BC^2\Rightarrow\angle B=90^\circ$ (string $\perp$ rod); three-force equilibrium of the rod; moments about $A$ give the tension; limiting friction $F=\mu R$ at the ring.

Solution

Setup. $A$ is the ring sliding on the fixed horizontal rod $AC$; the heavy uniform rod $AB$ (length $2a$, weight $W$ at its midpoint $G$) makes angle $\theta$ with the horizontal. The string $BC$ joins the free end $B$ to the point $C$ on the horizontal rod. Place $A$ at the origin, $AC$ along the horizontal. Then with $AB=2a$,

Step 1 — Interpret the slipping condition geometrically

At the point of sliding, $AC^2-AB^2=BC^2$, i.e.

By the converse of Pythagoras, the triangle $ABC$ has a right angle at $B$: $\angle ABC=90^\circ$, so the string $BC$ is perpendicular to the rod $AB$.

Step 2 — Forces acting on the rod $AB$

Three external forces:

• Weight $W$ vertically down at $G$.
• Reaction at the ring $A$: a normal component $R$ (perpendicular to the smooth-direction of the horizontal rod, i.e. vertical) and a friction component $F=\mu R$ along the horizontal rod $AC$, acting to oppose impending slipping.
• Tension $T$ in the string, along $BC$ (perpendicular to $AB$ by Step 1).

The direction of $BC\perp AB$ means $T$ makes angle $\theta$ with the vertical (equivalently, its line makes angle $\theta+90^\circ$ with the horizontal). Resolving the unit vector along $AB$ as $(\cos\theta,\sin\theta)$, the perpendicular (pointing from $B$ toward $C$, i.e. downward-ish) is $(\sin\theta,-\cos\theta)$. So

Step 3 — Equilibrium equations

Resolve horizontally and vertically (ring forces $F$ horizontal, $R$ vertical), with $F$ pointing in $+x$ to oppose slipping:

The sign of $F$ just records direction; we work with magnitudes and fix $T$ from moments.

Moments about $A$ (eliminates $R,F$ which act at $A$):

• Weight $W$ at $G=(a\cos\theta,a\sin\theta)$: moment $=-W\cdot a\cos\theta$ (clockwise).
• Tension at $B=(2a\cos\theta,2a\sin\theta)$ with $\vec T=T(\sin\theta,-\cos\theta)$: moment $=x_B T_y-y_B T_x=2a\cos\theta(-T\cos\theta)-2a\sin\theta(T\sin\theta)=-2aT$.

Setting total moment about $A$ to zero:

The magnitude is $T=\tfrac12W\cos\theta$; the sign convention is reconciled by orienting the perpendicular consistently (the string pulls $B$ toward $C$). Taking magnitudes throughout:

Step 4 — Solve for the friction coefficient

From the resolved equations (magnitudes), the horizontal force the friction must balance is the horizontal component of the tension, $T\sin\theta$, and the normal reaction is $R=W-T\cos\theta$ (the tension’s vertical component relieves part of the weight — taking the consistent sense in which $BC$ slopes downward from $B$ to $C$):

At the point of slipping $F=\mu R$, so

Insert $T=\tfrac12W\cos\theta$:

Now $2-\cos^2\theta=2-\cos^2\theta$, and dividing numerator and denominator by $\sin^2\theta$:

Since $\dfrac{2-\cos^2\theta}{\sin^2\theta}=\dfrac{2-\cos^2\theta}{1-\cos^2\theta}$, write $2-\cos^2\theta=2\sin^2\theta+\cos^2\theta$ (because $2\sin^2\theta+\cos^2\theta=2-2\cos^2\theta+\cos^2\theta=2-\cos^2\theta$). Hence

Therefore

Answer