The normal force (\(N\)) on an object on a slope can be calculated using the formula:
\[ N = \dfrac{F_f}{\mu \cdot \cos(\theta)} \]
Where:
- \(N\) is the normal force (in newtons, N)
- \(F_f\) is the force of friction (in newtons, N)
- \(\mu\) is the coefficient of friction (dimensionless)
- \(\theta\) is the angle of the slope (in degrees)
Example 1: Calculating the Normal Force for a Box on a Slope
Problem: A box on a slope experiences a force of friction of 50 N. The coefficient of friction between the box and the slope is 0.4, and the slope is inclined at 20 degrees. What is the normal force acting on the box?
Calculation:
Given:
- \(F_f = 50 \, \text{N}\)
- \(\mu = 0.4\)
- \(\theta = 20^\circ\)
Using the formula:
\[ N = \dfrac{F_f}{\mu \cdot \cos(\theta)} \]
\[ N = \dfrac{50}{0.4 \cdot \cos(20^\circ)} \]
\[ N = \dfrac{50}{0.4 \cdot 0.94} \]
\[ N = \dfrac{50}{0.376} \]
\[ N = 133.2 \, \text{N} \]
Answer: The normal force acting on the box is 133.2 N.
Example 2: Calculating the Normal Force for a Car on a Slope
Problem: A car on a slope experiences a force of friction of 800 N. The coefficient of friction between the car's tires and the slope is 0.5, and the slope is inclined at 10 degrees. What is the normal force acting on the car?
Calculation:
Given:
- \(F_f = 800 \, \text{N}\)
- \(\mu = 0.5\)
- \(\theta = 10^\circ\)
Using the formula:
\[ N = \dfrac{F_f}{\mu \cdot \cos(\theta)} \]
\[ N = \dfrac{800}{0.5 \cdot \cos(10^\circ)} \]
\[ N = \dfrac{800}{0.5 \cdot 0.985} \]
\[ N = \dfrac{800}{0.4925} \]
\[ N = 1624.4 \, \text{N} \]
Answer: The normal force acting on the car is 1624.4 N.
Example 3: Calculating the Normal Force for a Sled on a Hill
Problem: A sled on a hill experiences a force of friction of 20 N. The coefficient of friction between the sled and the hill is 0.2, and the hill is inclined at 30 degrees. What is the normal force acting on the sled?
Calculation:
Given:
- \(F_f = 20 \, \text{N}\)
- \(\mu = 0.2\)
- \(\theta = 30^\circ\)
Using the formula:
\[ N = \dfrac{F_f}{\mu \cdot \cos(\theta)} \]
\[ N = \dfrac{20}{0.2 \cdot \cos(30^\circ)} \]
\[ N = \dfrac{20}{0.2 \cdot 0.866} \]
\[ N = \dfrac{20}{0.1732} \]
\[ N = 115.5 \, \text{N} \]
Answer: The normal force acting on the sled is 115.5 N.
