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?

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.