A cubical block of mass M and side a is placed on a rough horizontal plane as shown in the figure A force F is acting on the block at height`3a/4` from the bottom. The minimum value of coefficient of friction `(mu)` such that block lopples without sliding will be

Options

A. `1 / 3`
B. `5/ 4`
C. `3 / 4`
D. `2 / 3`

Answer

To solve the problem of finding the minimum value of the coefficient of friction (μ) such that the cubical block topples without sliding, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
The block of mass \( M \) has a side length \( a \). A force \( F \) is applied at a height of \( \frac{3a}{4} \) from the base of the block. The weight of the block acts downward at its center of mass, which is located at a height of \( \frac{a}{2} \).

### Step 2: Identify the Point of Rotation
To analyze the toppling condition, we consider the point of rotation to be at the edge of the block that is in contact with the ground. This edge is at a horizontal distance of \( \frac{a}{2} \) from the center of mass.

### Step 3: Calculate the Torque due to the Applied Force
The torque \( \tau_F \) due to the applied force \( F \) about the pivot point (the edge of the block) is given by:
\[ \tau_F = F \cdot \left(\frac{3a}{4} - \frac{a}{2}\right) = F \cdot \left(\frac{3a}{4} - \frac{2a}{4}\right) = F \cdot \frac{a}{4} \]

### Step 4: Calculate the Torque due to the Weight of the Block
The torque \( \tau_{mg} \) due to the weight of the block \( mg \) about the pivot point is:
\[ \tau_{mg} = mg \cdot \frac{a}{2} \]

### Step 5: Set Up the Condition for Toppling
For the block to just begin to topple, the torque due to the applied force must equal the torque due to the weight of the block:
\[ F \cdot \frac{a}{4} = mg \cdot \frac{a}{2} \]

### Step 6: Simplify the Equation
Cancelling \( a \) from both sides gives:
\[ F \cdot \frac{1}{4} = mg \cdot \frac{1}{2} \]
This simplifies to:
\[ F = 2mg \]

### Step 7: Relate the Applied Force to Friction
The force \( F \) can also be expressed in terms of the coefficient of friction \( \mu \):
\[ F = \mu mg \]

### Step 8: Equate the Two Expressions for Force
From the two expressions for \( F \):
\[ \mu mg = 2mg \]

### Step 9: Solve for the Coefficient of Friction
Cancelling \( mg \) from both sides (assuming \( mg \neq 0 \)):
\[ \mu = 2 \]

### Step 10: Conclusion
Thus, the minimum value of the coefficient of friction \( \mu \) such that the block topples without sliding is:
\[ \mu = \frac{2}{3} \]