A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is appli

Q: Why does the normal reaction shift to the edge?

When you apply a horizontal force at a height, it creates a torque that tries to rotate the cube. To counter this and maintain rotational equilibrium, the ground must exert a greater upward force on the leading edge and less on the trailing edge. At the verge of toppling, the trailing edge lifts off, and the entire normal reaction is concentrated at the leading edge.

Q: Why don't we consider the torque of friction?

We are calculating the torque about the pivot point (the leading edge). Since the frictional force acts along the ground and passes directly through this pivot point, its perpendicular distance from the pivot is zero. Hence, its torque is zero.

Q: What if the force was applied at the top edge instead of $3a/4$?

If the force was applied at the top edge (height $a$), its perpendicular distance from the pivot would be $a$. The torque equation would be $F \times a = mg \times (a/2)$, resulting in a smaller required force $F = mg/2$.

Q: How do we know the cube won't slide first?

The problem explicitly states to 'Assume that the cube does not slide'. In reality, for the cube to topple before sliding, the required toppling force must be less than the maximum static friction ($F \le \mu mg$). This means $\mu$ must be greater than $2/3$.

Q: Does the mass of the cube affect the toppling condition?

Yes, the required force $F$ is directly proportional to the mass $m$. A heavier cube has a larger restoring torque ($mg \times a/2$), so you need a proportionally larger force to topple it.

Question

A uniform cube of side $a$ and mass $m$ rests on a rough horizontal table. A horizontal force $F$ is applied normal to one of the faces at a point that is directly above the centre of the face, at a height $3a/4$ above the base. The minimum value of $F$ for which the cube begins to tip about the edge is ......... (Assume that the cube does not slide).

Sigma Percentile · Accepted Answer

Imagine you are trying to push a heavy, tall refrigerator across the floor. If you push it near the bottom, it slides. But if you push it near the top, it feels like it wants to tip over before it even starts moving. This everyday experience is the heart of rotational equilibrium and the physics of toppling.

In this classic JEE problem, we are asked to find the exact threshold where a uniform cube decides to stop sitting flat and begins to tip over its edge. Let's dive into the beautiful mechanics behind this moment of instability.

### The Anatomy of Toppling

When a cube rests on a horizontal surface, gravity pulls it down from its Center of Mass (CM), and the ground pushes back up with a Normal Reaction ($N$). When no horizontal force is applied, this normal reaction is distributed evenly across the base, effectively acting through the center.

However, the moment you apply a horizontal force $F$ at a certain height, you introduce a **torque**. This torque tries to rotate the cube. To prevent rotation, the ground must push harder on the leading edge (the edge you are pushing towards) and less on the trailing edge.

As you increase $F$, the normal reaction shifts further and further towards the leading edge. **The critical moment of toppling occurs when the entire normal reaction $N$ has shifted to the extreme edge (let's call it point $O$).** At this exact instant, the trailing edge loses contact with the ground.

### Setting up the Master Equation

To find the minimum force required to topple the cube, we analyze the torques acting about our pivot point, $O$. Why point $O$? Because at the verge of toppling, both the normal reaction $N$ and the static friction $f$ pass directly through this point.

Since torque is defined as Force $	imes$ Perpendicular Distance, and the perpendicular distance for both $N$ and $f$ from point $O$ is zero, their torques vanish!

$$ 	au_N = 0 \quad 	ext{and} \quad 	au_f = 0 $$

This brilliant choice of pivot leaves us with a battle between only two torques:
1. **The Toppling Torque:** Created by the applied force $F$. It acts at a height of $\frac{3a}{4}$ from the ground. It wants to rotate the cube clockwise.
   $$ 	au_F = F 	imes \frac{3a}{4} $$
2. **The Restoring Torque:** Created by the cube's weight $mg$. The weight acts downwards from the CM, which is horizontally located at a distance of $\frac{a}{2}$ from the edge $O$. It wants to pull the cube back down anti-clockwise.
   $$ 	au_{mg} = mg 	imes \frac{a}{2} $$

### The Final Calculation

For the cube to just begin to tip, the toppling torque must perfectly balance the restoring torque. We set up our equilibrium equation:

$$ 	au_F = 	au_{mg} $$

Substituting our expressions:

$$ F \left( \frac{3a}{4} ight) = mg \left( \frac{a}{2} ight) $$

We can elegantly cancel the side length $a$ from both sides, revealing that the condition for toppling is independent of the cube's size!

$$ F \left( \frac{3}{4} ight) = mg \left( \frac{1}{2} ight) $$

Solving for $F$, we get our final answer:

$$ F = \frac{2}{3}mg $$

### Beyond the Problem: Topple or Slide?

The problem explicitly asked us to assume the cube does not slide. But what if we didn't make that assumption? For the cube to topple *before* it slides, the required toppling force ($\frac{2}{3}mg$) must be less than the maximum possible static friction ($\mu mg$).

This means toppling only happens if $\mu > \frac{2}{3}$. If the floor was smoother, the cube would simply slide away, no matter how high you pushed it!

Visualized Solution

$Free Body Diagram$

$Condition for Toppling$

$Torque Balance$

$τ_{net} = 0$

$τ_{N} = 0, τ_{f} = 0$

$τ_{F} = F \times \frac{3 a}{4}$

$τ_{m g} = m g \times \frac{a}{2}$

$τ_{F} = τ_{m g}$

$F (\frac{3}{4}) = m g (\frac{1}{2})$

$F = \frac{2}{3} m g$

$What if it slides?$

The Sigma Insight: Dynamics of Rigid Body Rotation

The Anatomy of Toppling

Setting up the Master Equation

The Final Calculation

Beyond the Problem

Similar Questions

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

A rod of weight $W$ is supported by two parallel knife edges $A$ and $B$ and is in equilibrium in a horizontal position. The knives are at a distance $d$ from each other. The centre of mass of the rod is at distance $x$ from $A$ . The normal reaction on $A$ is ......... and on $B$ is ......... .

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Free Body Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Toppling

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Torque Balance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τnet​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τN​=0,τf​=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τF​=F×43a​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τmg​=mg×2a​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } τF​=τmg​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F(43​)=mg(21​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } F=32​mg

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if it slides?