When an object is on a slope, the force of friction (\(F_f\)) can be calculated using the formula:
\[ F_f = \mu \cdot N \cdot \cos(\theta) \]
Where:
- \(F_f\) is the force of friction (in newtons, N)
- \(\mu\) is the coefficient of friction (dimensionless)
- \(N\) is the normal force (in newtons, N)
- \(\theta\) is the angle of the slope (in degrees)
Example 1: Calculating the Force of Friction for a Box on an Inclined Plane
Problem: A box with a mass of 10 kg is on a slope inclined at 30 degrees. The coefficient of friction between the box and the slope is 0.5. What is the force of friction acting on the box?
Calculation:
Given:
- \(m = 10 \, \text{kg}\)
- \(\mu = 0.5\)
- \(\theta = 30^\circ\)
- \(g = 9.8 \, \text{m/s}^2\)
- \(N = m \cdot g \cdot \cos(\theta) = 10 \cdot 9.8 \cdot \cos(30^\circ) = 10 \cdot 9.8 \cdot 0.866 = 84.868 \, \text{N}\)
Using the formula:
\[ F_f = \mu \cdot N \cdot \cos(\theta) \]
\[ F_f = 0.5 \cdot 84.868 \cdot \cos(30^\circ) \]
\[ F_f = 0.5 \cdot 84.868 \cdot 0.866 \]
\[ F_f = 36.75 \, \text{N} \]
Answer: The force of friction acting on the box is 36.75 N.
Example 2: Calculating the Force of Friction for a Car on an Inclined Road
Problem: A car with a mass of 1200 kg is on a road inclined at 15 degrees. The coefficient of friction between the car's tires and the road is 0.6. What is the force of friction acting on the car?
Calculation:
Given:
- \(m = 1200 \, \text{kg}\)
- \(\mu = 0.6\)
- \(\theta = 15^\circ\)
- \(g = 9.8 \, \text{m/s}^2\)
- \(N = m \cdot g \cdot \cos(\theta) = 1200 \cdot 9.8 \cdot \cos(15^\circ) = 1200 \cdot 9.8 \cdot 0.966 = 11372.064 \, \text{N}\)
Using the formula:
\[ F_f = \mu \cdot N \cdot \cos(\theta) \]
\[ F_f = 0.6 \cdot 11372.064 \cdot \cos(15^\circ) \]
\[ F_f = 0.6 \cdot 11372.064 \cdot 0.966 \]
\[ F_f = 6605.4 \, \text{N} \]
Answer: The force of friction acting on the car is 6605.4 N.
Example 3: Calculating the Force of Friction for a Sled on a Snowy Hill
Problem: A sled with a mass of 30 kg is on a hill inclined at 25 degrees. The coefficient of friction between the sled and the hill is 0.3. What is the force of friction acting on the sled?
Calculation:
Given:
- \(m = 30 \, \text{kg}\)
- \(\mu = 0.3\)
- \(\theta = 25^\circ\)
- \(g = 9.8 \, \text{m/s}^2\)
- \(N = m \cdot g \cdot \cos(\theta) = 30 \cdot 9.8 \cdot \cos(25^\circ) = 30 \cdot 9.8 \cdot 0.906 = 266.508 \, \text{N}\)
Using the formula:
\[ F_f = \mu \cdot N \cdot \cos(\theta) \]
\[ F_f = 0.3 \cdot 266.508 \cdot \cos(25^\circ) \]
\[ F_f = 0.3 \cdot 266.508 \cdot 0.906 \]
\[ F_f = 72.4 \, \text{N} \]
Answer: The force of friction acting on the sled is 72.4 N.
