Let's analyze each part of the problem step by step.

Given Data:

• Mass of ladder, $m = 30 \, \text{kg}$
• Length of ladder, $L = 4 \, \text{m}$
• Angle with the ground, $\theta = 65^\circ$
• Gravitational acceleration, $g = 9.81 \, \text{m/s}^2$

Step 1: Normal Force Pushing the Ladder from the Floor

The normal force $N$ counteracts the gravitational force acting on the ladder. The gravitational force $F_g$ can be calculated as: $F_g = m \cdot g = 30 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 294.3 \, \text{N}$ Therefore, the normal force is $N = 294.3 \, \text{N}$, which you have already identified correctly.

Step 2: Torque Exerted by the Ladder’s Weight

Since the ladder's weight acts at its midpoint, the distance from the pivot (floor) to the point where the weight acts is: $d = \frac{L}{2} = \frac{4 \, \text{m}}{2} = 2 \, \text{m}$ The torque $\tau$ exerted by the ladder's weight can be calculated using: $\tau = F_g \cdot d \cdot \sin(\theta)$ $\tau = 294.3 \, \text{N} \cdot 2 \, \text{m} \cdot \sin(65^\circ) = 248.75 \, \text{N·m}$ So, the torque from the weight is approximately $248.75 \, \text{N·m}$.

Step 3: Torque Balance and Normal Force from the Wall

Let's denote the normal force from the wall as $F_{N,\text{wall}}$. For the ladder to be in equilibrium, the torques around the pivot point (floor) must balance.

The torque exerted by the normal force from the wall can be calculated as: $\tau_{\text{wall}} = F_{N,\text{wall}} \cdot L \cdot \cos(\theta)$ Setting this equal to the torque from the ladder's weight for equilibrium: $F_{N,\text{wall}} \cdot 4 \, \text{m} \cdot \cos(65^\circ) = 248.75 \, \text{N·m}$ $F_{N,\text{wall}} = \frac{248.75 \, \text{N·m}}{4 \, \text{m} \cdot \cos(65^\circ)}$ $F_{N,\text{wall}} \approx 146.12 \, \text{N}$

Step 4: Minimum Coefficient of Friction

The frictional force $F_f$ must balance the normal force from the wall to prevent the ladder from slipping: $F_f = F_{N,\text{wall}} = 146.12 \, \text{N}$ The frictional force is also given by: $F_f = \mu \cdot N$ where $\mu$ is the coefficient of friction and $N = 294.3 \, \text{N}$. Thus: $\mu = \frac{F_{N,\text{wall}}}{N} = \frac{146.12 \, \text{N}}{294.3 \, \text{N}}$ $\mu \approx 0.4966$

Summary:

1. Normal Force from the Floor: $N = 294.3 \, \text{N}$
2. Torque from the Weight: $\tau = 248.75 \, \text{N·m}$
3. Normal Force from the Wall: $F_{N,\text{wall}} = 146.12 \, \text{N}$
4. Minimum Coefficient of Friction: $\mu \approx 0.4966$

Do you want any additional details or explanations on these calculations?

Here are 5 related questions to expand your understanding:

1. How would the normal forces change if the ladder were heavier?
2. What would happen to the torque if the ladder was placed at a steeper angle?
3. How would the minimum coefficient of friction change if the ladder was on a smoother surface?
4. Can you calculate the frictional force if the coefficient of friction is known to be 0.6?
5. What would the torque be if the ladder’s length was doubled but the mass remained the same?

Tip: Remember that equilibrium conditions (both translational and rotational) are essential to solving problems involving ladders, beams, or any object in static equilibrium.