← 2013 Paper 1

UPSC Maths 2013 Paper 1 Q5d — Solution

10 marks · Section B

Question

The base of an inclined plane is 4 metres in length and the height is 3 metres. A force of 8 kg acting parallel to the plane will just prevent a weight of 20 kg from sliding down. Find the coefficient of friction between the plane and the weight.

Technique

Decompose weight into parallel/perpendicular components; equilibrium with limiting friction uphill (since “just prevent” sliding down).

Solution

Step 1 — Geometry of the plane

Base $=4$, height $=3$, hypotenuse $=\sqrt{16+9}=5$. So the inclination $\theta$ satisfies

Step 2 — Force balance on the block

Take the inclined plane with the block on it. Forces:

• Weight $W=20$ kg-force, vertically down. Components:
  • Along the plane (downhill): $W\sin\theta=20\cdot\tfrac{3}{5}=12$ kg.
  • Perpendicular to the plane: $W\cos\theta=20\cdot\tfrac{4}{5}=16$ kg.
• Applied force $F=8$ kg, parallel to the plane, uphill (since it “prevents sliding down”).
• Normal reaction $N$, perpendicular to the plane (uphill perpendicular).
• Friction $f$, along the plane. The body is about to slide downward, so friction acts uphill, with magnitude $\mu N$ at the limiting equilibrium.

Step 3 — Equilibrium equations

Perpendicular to plane: $N=W\cos\theta=16$ kg.

Along the plane (uphill positive): $F+\mu N-W\sin\theta=0$, i.e.

Answer