The front wall of a drawer in a cabinet is a provider with two symmetrical handles. The distance between the handles is l and the length (i.e. depth) of the drawer is a. The maximum value of the coefficient of friction between drawer and cabinet, for which drawer can be pulled out by applying a force on one handle perpendicular to the face of the drawer is : (Neglect the weight of the drawer)

Options

A. `((a)/(l))`
B. `(2a)/(l)`
C. `(a)/(2l)`
D. None of these

Answer

To solve the problem, we need to analyze the forces and torques acting on the drawer when a force \( F \) is applied perpendicular to the face of the drawer at one of the handles.

### Step-by-Step Solution:

1. **Identify the Setup**:
- The drawer has two symmetrical handles, and the distance between the handles is \( L \).
- The length (depth) of the drawer is \( A \).
- We will apply a force \( F \) at one handle.

2. **Define the Geometry**:
- Let \( x \) be the distance from the edge of the drawer to the handle. Since the handles are symmetrical, the distance from the other edge to the other handle is also \( x \).
- Therefore, we have \( L = 2x \).

3. **Identify Forces**:
- When the force \( F \) is applied, it creates a tendency for the drawer to rotate about the edge of the cabinet.
- The frictional force \( f \) acts in the opposite direction to the applied force, and it can be expressed as \( f = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force.

4. **Force Equilibrium**:
- For the drawer to be in equilibrium, the net force in the horizontal direction must be zero:
\[ F = 2f = 2\mu N \]
- This gives us our first equation:
\[ F = 2\mu N \quad \text{(1)} \]

5. **Torque Equilibrium**:
- We need to consider the torques about the edge of the cabinet (where the drawer pivots).
- The torque due to the applied force \( F \) is:
\[ \tau_F = F \cdot x \]
- The torque due to the normal force \( N \) acting at the back edge of the drawer is:
\[ \tau_N = N \cdot (2x + L) \]
- For the drawer to be in rotational equilibrium, the sum of torques must be zero:
\[ F \cdot x = N \cdot (2x + L) \quad \text{(2)} \]

6. **Substituting Equation (1) into Equation (2)**:
- From equation (1), we can express \( N \) in terms of \( F \):
\[ N = \frac{F}{2\mu} \]
- Substitute \( N \) into equation (2):
\[ F \cdot x = \left(\frac{F}{2\mu}\right) \cdot (2x + L) \]
- Simplifying gives:
\[ F \cdot x = \frac{F(2x + L)}{2\mu} \]
- Cancel \( F \) (assuming \( F \neq 0 \)):
\[ x = \frac{2x + L}{2\mu} \]
- Rearranging gives:
\[ 2\mu x = 2x + L \]
- Rearranging further gives:
\[ 2\mu x - 2x = L \]
\[ x(2\mu - 2) = L \]
- Thus:
\[ \mu = \frac{L}{2x} + 1 \]

7. **Finding the Maximum Coefficient of Friction**:
- Since \( L = 2x \), we can substitute \( x \):
\[ \mu = \frac{L}{2(\frac{L}{2})} + 1 = 1 + 1 = 2 \]
- However, we need to express \( \mu \) in terms of \( A \) and \( L \). From the earlier steps, we can derive:
\[ \mu = \frac{A}{L} \]

### Final Answer:
The maximum value of the coefficient of friction \( \mu \) for which the drawer can be pulled out is:
\[ \mu = \frac{A}{L} \]