A 6.5 m long ladder rests against as vertical wall reaching a height of 6.0 m. A 60 kg man stands hlf way up the ladder. A. Find the torque of the force exerted by the man on the ladder bout the upper end of the ladder. b.Assuming the weight of the ladder of be negligible as compared to the man and assuming the wall to be smooth find the force exerted by the ground on the ladder.

Explanation

To solve the problem step-by-step, we will break it down into two parts as specified in the question.

### Part A: Finding the Torque of the Force Exerted by the Man on the Ladder about the Upper End of the Ladder

1. **Identify the Forces Acting on the Ladder:**
- The weight of the man (W) is acting downwards at the midpoint of the ladder. The weight can be calculated as:
\[ W = mg = 60 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 588 \, \text{N} \]

2. **Determine the Geometry of the Ladder:**
- The length of the ladder (L) is 6.5 m, and it reaches a height (h) of 6.0 m against the wall.
- The horizontal distance (d) from the base of the ladder to the wall can be found using the Pythagorean theorem:
\[ d = \sqrt{L^2 - h^2} = \sqrt{(6.5)^2 - (6.0)^2} = \sqrt{42.25 - 36} = \sqrt{6.25} = 2.5 \, \text{m} \]

3. **Calculate the Angle of the Ladder:**
- The angle (θ) can be calculated using:
\[ \sin(\theta) = \frac{h}{L} = \frac{6.0}{6.5} \]
- Thus, \( \theta = \arcsin\left(\frac{6.0}{6.5}\right) \).

4. **Calculate the Torque (τ) about the Upper End of the Ladder:**
- The torque due to the weight of the man about the upper end of the ladder is given by:
\[ \tau = W \cdot \left(\frac{L}{2}\right) \cdot \sin(\theta) \]
- Substituting the values:
\[ \tau = 588 \, \text{N} \cdot \left(\frac{6.5}{2}\right) \cdot \sin(\theta) \]
- Calculate \( \sin(\theta) \):
\[ \sin(\theta) = \frac{h}{L} = \frac{6.0}{6.5} \approx 0.923 \]
- Therefore, substituting this back:
\[ \tau = 588 \cdot 3.25 \cdot 0.923 \approx 1685.1 \, \text{N m} \]

### Part B: Finding the Force Exerted by the Ground on the Ladder

1. **Identify the Forces Acting on the Ladder:**
- The vertical force exerted by the ground (R) must balance the weight of the man. Therefore:
\[ R = W = 588 \, \text{N} \]

2. **Determine the Horizontal Force:**
- Since the wall is smooth, there is no friction at the wall. The horizontal force exerted by the ground (F) must balance the horizontal component of the force exerted by the man.

3. **Calculate the Torque about the Base of the Ladder:**
- The torque due to the weight of the man about the base of the ladder is:
\[ \tau_{\text{man}} = W \cdot \left(\frac{L}{2}\right) \cdot \sin(\theta) \]
- The torque due to the horizontal force (F) about the base of the ladder is:
\[ \tau_F = F \cdot L \cdot \cos(\theta) \]

4. **Set the Torques Equal:**
- For the ladder to be in equilibrium:
\[ \tau_{\text{man}} = \tau_F \]
- Therefore:
\[ 588 \cdot 3.25 \cdot \sin(\theta) = F \cdot 6.5 \cdot \cos(\theta) \]

5. **Solve for F:**
- Rearranging gives:
\[ F = \frac{588 \cdot 3.25 \cdot \sin(\theta)}{6.5 \cdot \cos(\theta)} \]
- Substitute \( \sin(\theta) \) and \( \cos(\theta) \):
\[ F = \frac{588 \cdot 3.25 \cdot \frac{6.0}{6.5}}{6.5 \cdot \frac{2.5}{6.5}} = \frac{588 \cdot 3.25 \cdot 6.0}{2.5 \cdot 6.5} \]
- Calculate F to find the horizontal force exerted by the ground.