A solid cylinder of mass m is wrapped with an inextensible light string and, is placed on a rough inclined plane as shown in the figure. The frictional force acting between the cylinder and the inclined plane is :
[The coefficient of static friction, μₛ' is 0.4]

Solution

Physics sir · 4 steps

Find the friction needed to hold the cylinder still, then check whether the incline can supply that much.

1. If it stayed still. The string pulls up the slope with tension T at the top of the cylinder, and friction f acts up the slope at the contact point. Both act at distance R from the centre and turn the cylinder in opposite senses. For no rotation, TR = fR, so T = f.

   T + f = mg sin 60° ⇒ f_(needed) = (mg sin 60°)/2 = √3/4mg ≈ 0.43 mg

2. Check the limit. The normal reaction is N = mg cos 60° = mg/2. The largest friction available is μₛ N = 0.4 × mg/2 = 0.2 mg. This is less than the 0.43 mg needed, so the cylinder cannot stay at rest.
3. It slips. The string end is fixed, so the cylinder unwinds about its top point. The contact point then slides down the slope, so friction is limiting and acts up the incline with magnitude μ N.
4. So. f = 0.4 × mg/2 = mg/5, which is option (d).

The idea to keep. Friction is never assumed. First find what equilibrium demands, then compare it with μ N. If the demand is bigger, the body slips and friction is just μ N.

Answer from the NTA final answer key. Solution by PrepWiser.