A uniform ladder which is 5m long has a mass of 25kg , leans with its upper end against a smooth vertical wall and its lower end on rough ground. The bottom of the ladder is `3m` from the wall. Calculate the frictional force between the ladder and the ground . `(g=10ms^(-2)`)

Explanation

To solve the problem, we need to analyze the forces acting on the ladder and apply the principles of equilibrium. Here’s a step-by-step solution:

### Step 1: Draw a Diagram
Draw a diagram of the ladder leaning against the wall. Label the points:
- A: Top of the ladder (against the wall)
- B: Bottom of the ladder (on the ground)
- C: The point where the ladder touches the ground
- D: The wall

### Step 2: Identify Given Values
- Length of the ladder (L) = 5 m
- Distance from the wall to the bottom of the ladder (d) = 3 m
- Mass of the ladder (m) = 25 kg
- Acceleration due to gravity (g) = 10 m/s²

### Step 3: Calculate the Height of the Ladder on the Wall
Using the Pythagorean theorem:
\[ h = \sqrt{L^2 - d^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \text{ m} \]
So, the height of the ladder on the wall is 4 m.

### Step 4: Calculate the Weight of the Ladder
The weight (W) of the ladder can be calculated as:
\[ W = m \cdot g = 25 \text{ kg} \cdot 10 \text{ m/s}^2 = 250 \text{ N} \]

### Step 5: Determine the Normal Force (N)
Since the ladder is in equilibrium, the normal force (N) at point B (the ground) will be equal to the weight of the ladder:
\[ N = W = 250 \text{ N} \]

### Step 6: Analyze the Torque about Point B
Taking moments about point B (the bottom of the ladder), we can set up the torque equation. The torque due to the weight of the ladder acts at its center of mass, which is at a distance of 2.5 m from point B (half the length of the ladder). The height at which the weight acts is 2 m (half of the height of the ladder):
\[ \text{Torque due to weight} = W \cdot \text{horizontal distance to line of action of weight} \]
The horizontal distance from B to the line of action of the weight is:
\[ \text{horizontal distance} = \frac{3}{5} \cdot 2 = 1.2 \text{ m} \]
Thus, the torque due to the weight is:
\[ \text{Torque} = 250 \text{ N} \cdot 1.2 \text{ m} = 300 \text{ Nm} \]

### Step 7: Analyze the Torque due to Frictional Force
Let the frictional force be \( F_s \). The torque due to the frictional force at point B is:
\[ \text{Torque due to friction} = F_s \cdot \text{height of ladder} = F_s \cdot 4 \text{ m} \]

### Step 8: Set Up the Torque Equation
For equilibrium, the total torque about point B must be zero:
\[ F_s \cdot 4 = 300 \]
Solving for \( F_s \):
\[ F_s = \frac{300}{4} = 75 \text{ N} \]

### Step 9: Conclusion
The frictional force between the ladder and the ground is \( 75 \text{ N} \).

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