Q.Two inclined planes meet at a common apex where a frictionless pulley is mounted. On the left plane, which makes an angle $\theta_1$ with the horizontal, a body A of mass $m$ rests; the coefficient of friction between A and this plane is $\mu$. On the right plane, which makes an angle $\theta_2$ with the horizontal and is frictionless, a body B of mass $m$ rests. A light string joins A and B, running up the left slope, over the pulley at the apex, and down the right slope. Which of the following statements are true? (Note: more than one of the given options may be correct.)

Friction on an Inclined Plane

Imagine a block placed on a ramp. If the ramp is flat, the block stays put. As you tilt the ramp, something changes. At a small tilt, the block still doesn't move — something is holding it in place. That something is static friction, acting up the slope, exactly balancing the component of gravity that tries to pull the block down.

Now tilt the ramp a little more. At some critical angle, the block just begins to slide. At that instant, static friction has reached its maximum possible value. Tilt it further, and the block accelerates downward — now kinetic friction acts up the slope, but it is weaker than the maximum static friction was.

That is the entire story in a nutshell. Let's make it precise.

The forces on an inclined plane

Take a block of mass $m$ on a plane inclined at angle $\theta$ to the horizontal. The weight $mg$ points straight down. Resolve it into two components:

• Perpendicular to the plane: $mg\cos\theta$ — this pushes the block into the surface. The normal reaction $N$ balances it: $N = mg\cos\theta$.
• Parallel to the plane (down the slope): $mg\sin\theta$ — this is what tries to make the block slide.

Friction $f$ acts parallel to the plane, opposing the relative motion (or the tendency to move). So if the block is at rest or sliding down, friction acts up the slope. If you push the block up the slope, friction acts down the slope.

Three cases

1. Block at rest (static friction)

The block does not move. So the net force along the slope is zero:

Here $f_s$ is static friction. It is not a fixed value — it adjusts itself to whatever is needed, up to a maximum. The maximum possible static friction is:

The block stays at rest as long as:

The angle $\theta_s = \tan^{-1}(\mu_s)$ is called the angle of repose. At this angle, the block is just about to slide.

2. Block sliding down (kinetic friction)

Once $\theta > \theta_s$, the block accelerates down. Kinetic friction acts up the slope, with magnitude:

The net force down the slope is:

So the acceleration down the plane is:

3. Block pushed up the slope …