A block of mass $m_1$ rests on an incline that makes an angle $\theta$ with the horizontal. A light string tied to $m_1$ runs up along the slope, passes over a frictionless pulley fixed at the top of the incline, and then hangs vertically, supporting a block of mass $m_2$. The coefficient of friction between $m_1$ and the sloping surface is $\mu$. Which of the following statements are true? (Note: more than one of the given options may be correct.) (a) If $m_2 > m_1 \sin\theta$, the body will move up the plane. (b) If $m_2 > m_1(\sin\theta + \mu\cos\theta)$, the body will move up the plane. (c) If $m_2 < m_1(\sin\theta + \mu\cos\theta)$, the body will move up the plane. (d) If $m_2 < m_1(\sin\theta - \mu\cos\theta)$, the body will move down the plane.

A block of mass m₁ rests on an incline that makes an angle θ with the horizontal. A light string tied to m₁ ru

NCERT Exemplar · Q12

m_1

(a) If

m_2 > m_1 \sin\theta

, the body will move up the plane.

(b) If

m_2 > m_1(\sin\theta + \mu\cos\theta)

, the body will move up the plane.

m_2 < m_1(\sin\theta + \mu\cos\theta)

, the body will move up the plane.

(d) If

m_2 < m_1(\sin\theta - \mu\cos\theta)

, the body will move down the plane.

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Concept understanding — Friction on an Inclined Plane

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:

f_s = mg\sin\theta

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:

f_{s,\text{max}} = \mu_s N = \mu_s mg\cos\theta

The block stays at rest as long as:

mg\sin\theta \le \mu_s mg\cos\theta \quad\Rightarrow\quad \tan\theta \le \mu_s

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

Important

The angle of repose $\theta_s$ satisfies $\tan\theta_s = \mu_s$ . This is a direct link between friction and geometry — you can measure $\mu_s$ by finding the tilt at which sliding begins.

2. Block sliding down (kinetic friction)

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

f_k = \mu_k N = \mu_k mg\cos\theta

The net force down the slope is:

F_{\text{net}} = mg\sin\theta - \mu_k mg\cos\theta

So the acceleration down the plane is:

a = g(\sin\theta - \mu_k\cos\theta)

Note

| Condition | What happens |

|-----------|--------------|

| $\tan\theta < \mu_s$ | Block stays at rest |

| $\tan\theta = \mu_s$ | Block just about to slide |

| $\tan\theta > \mu_s$ | Block slides down with acceleration $g(\sin\theta - \mu_k\cos\theta)$ |

3. Block pushed up the slope …

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Friction always opposes the impending motion, so it flips direction depending on whether $m_1$ tends to go up or down. Including it correctly gives (B) as the up-slope condition and (D) as the down-slope condition; (A) and (C) are wrong.

Concept

Along the incline, $m_1$ feels: tension $T=m_2g$ (up-slope), weight component $m_1g\sin\theta$ (down-slope), and friction $f\le\mu m_1g\cos\theta$ (direction set by the impending slide).

Condition to move UP

If $m_1$ is about to move up, friction acts down the slope. Motion up begins when

$m_2g>m_1g\sin\theta+\mu m_1g\cos\theta\;\Rightarrow\; m_2>m_1(\sin\theta+\mu\cos\theta).$ …

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Concept

Condition to move UP

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