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 is applied to one of the handles. We will derive the maximum value of the coefficient of friction (μ) that allows the drawer to be pulled out without slipping.

### Step-by-Step Solution:

1. **Identify the Forces**:
- Let \( F \) be the force applied perpendicular to the face of the drawer at one handle.
- The normal force \( N \) acts vertically upward at the base of the drawer.
- The frictional force \( f \) acts horizontally opposing the motion of the drawer.

2. **Torque Calculation**:
- We will take torques about the point where the drawer contacts the cabinet. This point is where the normal force acts.
- The torque due to the applied force \( F \) at a distance \( d \) (half the distance between the handles) is given by:
\[ \tau_F = F \cdot d \]
- The torque due to the frictional force \( f \) at a distance \( a \) (the depth of the drawer) is given by:
\[ \tau_f = f \cdot a \]

3. **Balancing Torques**:
- For the drawer to be in equilibrium (not rotating), the torques must balance:
\[ F \cdot d = f \cdot a \]

4. **Expressing Frictional Force**:
- The frictional force \( f \) can be expressed in terms of the normal force \( N \) and the coefficient of friction \( \mu \):
\[ f = \mu N \]

5. **Substituting Friction into Torque Equation**:
- Substitute \( f \) into the torque balance equation:
\[ F \cdot d = \mu N \cdot a \]

6. **Normal Force Relation**:
- The normal force \( N \) can be related to the forces acting on the drawer. Since we neglect the weight of the drawer, we can assume that \( N \) is equal to the vertical forces acting on the drawer.

7. **Final Equation**:
- Rearranging gives:
\[ \mu = \frac{F \cdot d}{N \cdot a} \]

8. **Using the Geometry**:
- The distance \( d \) is half the distance between the handles, so \( d = \frac{l}{2} \).
- Substitute \( d \) into the equation:
\[ \mu = \frac{F \cdot \frac{l}{2}}{N \cdot a} \]

9. **Maximum Coefficient of Friction**:
- For the drawer to be pulled out, we need to find the maximum value of \( \mu \):
\[ \mu_{\text{max}} = \frac{a}{l} \]

### Conclusion:
The maximum value of the coefficient of friction \( \mu \) for which the drawer can be pulled out by applying a force on one handle perpendicular to the face of the drawer is:
\[ \mu_{\text{max}} = \frac{a}{l} \]