The ladder in the picture has a mass of 30 kilograms and a length 4 meters. What is the normal force pushing the ladder up from the floor?

 
 = 
294.3
Correct N

Assume that the ladder's weight is evenly distributed, so it can be treated as a single force through the middle. If the ladder is at a 65° angle from the ground, what is the torque exerted by the weight (using the floor as the pivot point)?

τ = 
248.74
Correct248.75310886058 N·m

The torque from the ladder must be balanced by the torque caused by the normal force on the wall. Calculate this force.

 = 
 N

The normal force from the wall must be balanced by the friction force from the floor. Determine the minimum coefficient of friction to keep the ladder from slipping.

μ = 


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.

The ladder in the picture has a mass of 30 kilograms and a length of 4 meters. What is the normal force pushing the ladder up from the floor? Assume that the ladder's weight is evenly distributed, so it can be treated as a single force through the middle. If the ladder is at a 65° angle from the ground, what is the torque exerted by the weight (using the floor as the pivot point)? The torque from the ladder must be balanced by the torque caused by the normal force on the wall. Calculate this force. The normal force from the wall must be balanced by the friction force from the floor. Determine the minimum coefficient of friction to keep the ladder from slipping.

Solve a ladder statics problem involving a 30 kg ladder at a 65° angle. Learn how to calculate the normal force, torque exerted by the ladder's weight, normal force from the wall, and the minimum coefficient of friction required to prevent slipping. Suitable for students in Grades 11-12.

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