(a) Show that in an inclined plane, angle of friction is equal to angle of repose. **OR** (b) Derive an expres

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Q. (a) Show that in an inclined plane, angle of friction is equal to angle of repose. OR

(b) Derive an expression for power and velocity.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2023Subjective· 5mImportance★★★★★

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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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Both the angle of repose and the angle of friction satisfy tan(theta) = mu, so they must be numerically equal.

This question offers a choice between (a) showing the angle of friction equals the angle of repose on an inclined plane, and (b) deriving the relation between power and velocity; part (a) is answered here.

Angle of repose: consider a block of mass m resting on a rough inclined plane whose angle of inclination theta can be increased. The angle of repose is the maximum angle of inclination at which the block just remains on the verge of sliding down (on the point of slipping), under gravity and friction alone, with no external push.

At this angle, resolving the weight mg along and perpendicular to the incline:

Component along the incline (tending to slide the block down): mg sin(theta)
Component perpendicular to the incline (balanced by the normal reaction N): mg cos(theta), so N = mg cos(theta)

At the verge of sliding, the force along the incline is balanced by the limiting (maximum static) friction force f = mu N:

mg sin(theta) = mu N = mu (mg cos(theta))

Dividing both sides by mg cos(theta):

tan(theta) = mu ... (i)

…

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Q.(a) Show that in an inclined plane, angle of friction is equal to angle of repose. OR

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