A cubical box of side a sitting on a rough tabletop is pushed horizontally with a gradually increasing force until the box moves. If the force is applied at a height from the table top which is greater than a critical height H , the box topples first. If it is applied at a height less than H , the box starts sliding first. Then the coefficient of friction between the box and the table top is

Options

A. `(a)/(2H)`
B. `(2H)/(a)`
C. `(a)/(H)`
D. `(H)/(a)`

Explanation

To solve the problem, we need to analyze the conditions under which the cubical box either slides or topples when a horizontal force is applied at different heights. Here's a step-by-step breakdown of the solution:

### Step 1: Understanding the Problem
We have a cubical box of side length \( a \) placed on a rough table. A horizontal force \( F \) is applied at a height \( h \) from the table. There are two scenarios:
- If \( h > H \) (critical height), the box will topple.
- If \( h < H \), the box will slide.

### Step 2: Establishing Forces and Conditions
1. **Weight of the Box**: The weight \( W \) of the box acts downwards at its center of mass, which is at a height of \( \frac{a}{2} \) from the base.
2. **Normal Force**: The normal force \( N \) acts upwards from the table and balances the weight of the box, so \( N = mg \).
3. **Frictional Force**: The frictional force \( f \) opposes the motion and is given by \( f = \mu N = \mu mg \), where \( \mu \) is the coefficient of friction.

### Step 3: Translational Equilibrium
When the box is at the critical height \( H \) and is in equilibrium:
- The applied force \( F \) must equal the frictional force \( f \):
\[ F = f = \mu mg \]

### Step 4: Rotational Equilibrium
To analyze the toppling condition, we consider torques about the edge of the box that is in contact with the table (let's call this point A):
1. The torque due to the weight of the box about point A is:
\[ \tau_{mg} = mg \cdot \frac{a}{2} \]
(This is counterclockwise.)

2. The torque due to the applied force \( F \) at height \( h \) is:
\[ \tau_{F} = F \cdot h \]
(This is clockwise.)

For the box to be in rotational equilibrium (neither toppling nor sliding), the net torque about point A must be zero:
\[ F \cdot h = mg \cdot \frac{a}{2} \]

### Step 5: Substitute \( F \) from Translational Equilibrium
From the translational equilibrium, we have \( F = \mu mg \). Substituting this into the torque equation gives:
\[ \mu mg \cdot h = mg \cdot \frac{a}{2} \]

### Step 6: Simplifying the Equation
We can cancel \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \mu h = \frac{a}{2} \]

### Step 7: Solving for the Coefficient of Friction
Rearranging the equation to solve for \( \mu \):
\[ \mu = \frac{a}{2h} \]

### Conclusion
The coefficient of friction between the box and the tabletop is:
\[ \mu = \frac{a}{2H} \]