A`4 m` long ladder weighing `25 kg` rests with its upper and against a smooth wall and lower and on rought ground.What should be the minimum coefficient of friction between the ground and ladder for it to be inclined at `60^(@)` with the horizontal without slipping? `(Take g =10 m//s^(2))`

Explanation

To solve the problem of determining the minimum coefficient of friction required for a ladder to remain stable against slipping, we can follow these steps:

### Step 1: Identify the forces acting on the ladder
1. **Weight (W)**: The weight of the ladder acts downwards at its center of mass, which is at the midpoint of the ladder. The weight can be calculated as:
\[ W = mg = 25 \, \text{kg} \times 10 \, \text{m/s}^2 = 250 \, \text{N} \]
2. **Normal Force from the Ground (N2)**: This acts upwards at the base of the ladder.
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