Question

A frictional force of 2400 N keeps a crate of machine parts from sliding down a ramp. If the coefficient of static friction between the box and the ramp is 0.20 and the incline of the ramp is 30.0°, what is the normal force of the ramp on the box? What is the mass of the crate of machine parts?

Answer 1

To find the normal force of the ramp on the box and the mass of the crate of machine parts, we can use the following information:

1. The coefficient of static friction (μ) = 0.20
2. The frictional force (F_friction) = 2400 N
3. The angle of the ramp (θ) = 30.0°

### Step 1: Calculate the normal force (N)

The maximum static friction force can be calculated using the formula:
\[
F_friction = \mu \cdot N
\]
Rearranging this for the normal force gives:
\[
N = \frac{F_friction}{\mu}
\]
Substituting in the given values:
\[
N = \frac{2400 \, \text{N}}{0.20} = 12000 \, \text{N}
\]

### Step 2: Calculate the mass of the crate

The normal force is also related to the weight of the crate and the angle of the ramp:
\[
N = mg \cos(\theta)
\]
Where:
- \( m \) is mass,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( \theta \) is the incline of the ramp (30.0°).

Using \( N = 12000 \, \text{N} \):
\[
12000 = m \cdot 9.81 \cdot \cos(30°)
\]
Calculating \( \cos(30°) \):
\[
\cos(30°) = \frac{\sqrt{3}}{2} \approx 0.866
\]
Now plugging that into our equation:
\[
12000 = m \cdot 9.81 \cdot 0.866
\]
Calculating \( 9.81 \cdot 0.866 \):
\[
9.81 \cdot 0.866 \approx 8.487
\]
So our equation simplifies to:
\[
12000 = m \cdot 8.487
\]
Now solving for \( m \):
\[
m = \frac{12000}{8.487} \approx 1417.50 \, \text{kg}
\]

### Final Results
- The normal force (N) on the crate is **12000 N**.
- The mass of the crate of machine parts is approximately **1417.50 kg**.