A man runs along a straight road. Naturally he leans a little in the forward direction. The angle with the horizontal line and the line joining the man's centre of mass with the point of support is `alpha` ( assume mass per unit length of the man is uniform ) . The man will not slip if the coefficient of friction `mu` satisfies the conditon.

Options

A. `mu gt cot alpha`
B. `mu lt cot alpha`
C. `mu gt tan alpha`
D. `mu lt tan alpha`

Answer

To solve the problem, we need to analyze the forces acting on the man as he runs and leans forward. The goal is to find the condition under which the man will not slip, in terms of the coefficient of friction (μ) and the angle (α).

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Man:**
- The gravitational force (weight) acting downwards: \( mg \)
- The normal force (N) acting upwards from the ground
- The frictional force (F_f) acting horizontally, preventing slipping

2. **Balance the Vertical Forces:**
- In the vertical direction, the forces must balance. Thus, we have:
\[ N = mg \]
- This equation states that the normal force is equal to the weight of the man.

3. **Determine the Frictional Force:**
- The maximum frictional force can be expressed as:
\[ F_f = \mu N \]
- Substituting the expression for normal force, we get:
\[ F_f = \mu mg \]

4. **Analyze the Torque about the Center of Mass:**
- The torque due to the frictional force about the center of mass must be balanced by the torque due to the normal force.
- The torque due to friction (which tries to rotate the man anti-clockwise) is:
\[ \tau_f = F_f \cdot L \sin(\alpha) \]
- The torque due to the normal force (which tries to rotate the man clockwise) is:
\[ \tau_N = N \cdot L \cos(\alpha) \]

5. **Set Up the Torque Balance Equation:**
- For the man to not slip, the torques must balance:
\[ F_f \cdot L \sin(\alpha) = N \cdot L \cos(\alpha) \]
- Canceling \( L \) from both sides gives:
\[ F_f \sin(\alpha) = N \cos(\alpha) \]

6. **Substituting the Expressions for Forces:**
- Substitute \( F_f = \mu mg \) and \( N = mg \):
\[ \mu mg \sin(\alpha) = mg \cos(\alpha) \]
- Dividing both sides by \( mg \) (assuming \( mg \neq 0 \)):
\[ \mu \sin(\alpha) = \cos(\alpha) \]

7. **Rearranging the Equation:**
- Rearranging gives:
\[ \mu = \frac{\cos(\alpha)}{\sin(\alpha)} = \cot(\alpha) \]

8. **Final Condition for No Slipping:**
- Therefore, the condition for the man not to slip is:
\[ \cot(\alpha) \leq \mu \]

### Conclusion:
The man will not slip if the coefficient of friction \( \mu \) satisfies the condition:
\[ \cot(\alpha) \leq \mu \]