A block of base 10 cm x 10 cm and height 15 cm is kept on an inclined plane. The coefficient of fric

Q: Why does the normal force shift to the edge during toppling?

As the incline angle increases, the line of action of gravity shifts downhill. To maintain rotational equilibrium, the normal force must shift downhill as well to provide a counter-torque. When it reaches the edge, it can't shift any further, and toppling begins.

Q: Why don't we consider the mass of the block in our final conditions?

Notice that mass $m$ appears on both sides of the inequalities for both sliding ($mg \sin\theta > \mu mg \cos\theta$) and toppling ($mg \sin\theta \times \frac{h}{2} > mg \cos\theta \times \frac{b}{2}$). It cancels out, meaning the critical angles depend only on the geometry and friction, not the mass!

Q: What if the block was wider, say base 20 cm and height 15 cm?

If the base was wider, the ratio $b/h$ would be $20/15 = 4/3 \approx 1.33$. Since $1.33 < \sqrt{3} (1.732)$, it would still topple first. But if the base was even wider, say 30 cm, $b/h = 2$, which is greater than $1.732$, so it would slide first!

Q: Does the value of friction affect the toppling angle?

No! The toppling condition $\tan\theta > b/h$ is purely geometric. Friction only determines whether the block will slide *before* it reaches that toppling angle. If friction is high enough, it will always topple.

Q: Why do we take torque about the lower edge $O$?

At the verge of toppling, the entire block is about to pivot around that specific lower edge. The normal force and friction both act exactly at that edge, so their torques are zero, making the calculation much simpler!

Question

A block of base $10	ext{ cm} 	imes 10	ext{ cm}$ and height $15	ext{ cm}$ is kept on an inclined plane. The coefficient of friction between them is $\sqrt{3}$. The inclination $	heta$ of this inclined plane from the horizontal plane is gradually increased from $0^{\circ}$. Then,

Options:

(A) at $	heta = 30^{\circ}$, the block will start sliding down the plane

(B) the block will remain at rest on the plane up to certain $	heta$ and then it will topple

(C) at $	heta = 60^{\circ}$, the block will start sliding down the plane and continue to do so at higher angles

(D) at $	heta = 60^{\circ}$, the block will start sliding down the plane and on further increasing $	heta$, it will topple at certain $	heta$

Sigma Percentile · Accepted Answer

### The Battle on the Incline

Imagine placing a block on a wooden plank and slowly lifting one end. What happens first? Does it slide down like a sled, or does it tumble over like a falling tower? This classic physics problem is a beautiful tug-of-war between two phenomena: **sliding** and **toppling**.

To solve this, we must act as referees and evaluate the conditions for both events independently. Whichever condition is satisfied at a smaller angle $	heta$ will be the winner!

### The Condition for Sliding

Let's first analyze the forces trying to make the block slide. Gravity pulls the block straight down with a force $mg$. We can resolve this into two components:
1. $mg \sin	heta$ acting parallel to the incline, urging the block downwards.
2. $mg \cos	heta$ acting perpendicular to the incline, pressing the block against the surface.

The surface pushes back with a normal reaction $N = mg \cos	heta$. Friction, the ultimate defender, opposes the sliding motion. The maximum static friction available is $f_{	ext{max}} = \mu N = \mu mg \cos	heta$.

For the block to break free and start sliding, the downward pull must overpower this maximum friction:

$$mg \sin	heta > \mu mg \cos	heta$$

Dividing both sides by $mg \cos	heta$, we get the elegant sliding condition:

$$	an	heta > \mu$$

Given that $\mu = \sqrt{3}$, the block will slide if $	an	heta > \sqrt{3}$, which corresponds to an angle $	heta > 60^\circ$.

### The Condition for Toppling

Now, what about toppling? As the incline gets steeper, the normal force $N$ shifts downwards to counteract the torque produced by gravity. The critical moment arrives when $N$ reaches the very edge of the block (let's call it point $O$).

If we take torques about this edge $O$, the component $mg \sin	heta$ tries to rotate the block downhill (clockwise), while $mg \cos	heta$ tries to keep it seated (counter-clockwise).

For the block to topple, the overturning torque must exceed the restoring torque:

$$(mg \sin	heta) 	imes \frac{h}{2} > (mg \cos	heta) 	imes \frac{b}{2}$$

Rearranging this gives us the toppling condition:

$$	an	heta > \frac{b}{h}$$

We are given the base $b = 10	ext{ cm}$ and height $h = 15	ext{ cm}$. Plugging these in:

$$	an	heta > \frac{10}{15} = \frac{2}{3}$$

### The Grand Finale

We now have our two thresholds:
- Sliding requires $	an	heta > \sqrt{3} \approx 1.732$
- Toppling requires $	an	heta > \frac{2}{3} \approx 0.667$

As we gradually increase the angle $	heta$ from $0^\circ$, the value of $	an	heta$ increases. It will inevitably reach $0.667$ long before it reaches $1.732$.

Therefore, the block will lose its rotational stability and **topple** before it ever gets the chance to slide!

Select Answer:

Visualized Solution

Analyzing the Setup

Components of Gravity

Condition for Sliding

Calculating Sliding Angle

Condition for Toppling

Calculating Toppling Angle

Conclusion

The Sigma Insight: Dynamics of Rigid Body Rotation

The Battle on the Incline

The Condition for Sliding

The Condition for Toppling

The Grand Finale

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