When a body slides down from rest along a smooth inclined plane making an angle of $45^\circ$ with the horizontal, it takes time $T$. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time $pT$, where $p$ is some number greater than 1. Calculate the co-efficient of friction between the body and the rough plane.

When a body slides down from rest along a smooth inclined plane making an angle of 45^° with the horizontal, i

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45^\circ

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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 reduces the acceleration down the incline, increasing the time taken. By comparing the accelerations (which scale as $1/t^2$ for motion from rest over a fixed distance), we find that the coefficient of friction is $\mu = 1 - \frac{1}{p^2}$ .

Why this approach works

When a body slides down an incline, gravity pulls it down while friction (if present) opposes the motion. The net force determines the acceleration, and for motion from rest over a fixed distance the time taken depends on that acceleration through the kinematic relation $s = \frac{1}{2}at^2$ . A smaller acceleration means a longer time. By comparing the smooth and rough cases—same angle, same distance, different times—we can extract the coefficient of friction.

The key insight: time squared is inversely proportional to acceleration when distance is held constant.

Step-by-step solution

1. Acceleration on the smooth incline

On a frictionless plane at angle $\theta = 45^\circ$ , only the component of gravity along the slope acts:

a_{\text{smooth}} = g \sin 45^\circ = \frac{g}{\sqrt{2}}.

The body starts from rest and travels distance $s$ in time $T$ :

s = \frac{1}{2} a_{\text{smooth}} T^2 = \frac{1}{2} \cdot \frac{g}{\sqrt{2}} \cdot T^2.

2. Acceleration on the rough incline

Now friction acts up the plane. The normal force is $N = mg \cos 45^\circ = \frac{mg}{\sqrt{2}}$ , so the friction force is

f = \mu N = \mu \cdot \frac{mg}{\sqrt{2}}.

The net force down the plane is

mg \sin 45^\circ - f = \frac{mg}{\sqrt{2}} - \mu \cdot \frac{mg}{\sqrt{2}} = \frac{mg}{\sqrt{2}}(1 - \mu).

Hence the acceleration is

a_{\text{rough}} = g \sin 45^\circ - \mu g \cos 45^\circ = \frac{g}{\sqrt{2}}(1 - \mu).

The same distance $s$ is now covered in time $pT$ : …

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Why this approach works

Step-by-step solution

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