Determining the angle of the slope (\(\theta\)) using the normal force and the mass of the object is a key concept in physics. The angle can be calculated using the rearranged formula:
\[ \theta = \cos^{-1}\left(\dfrac{F_n}{m \cdot g}\right) \]
Where:
- \(\theta\) is the angle of the slope (in degrees)
- \(F_n\) is the normal force (in newtons, N)
- \(m\) is the mass of the object (in kilograms, kg)
- \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\))
Example 1: Calculating the Angle of a Slope for a Car
Problem: A car with a mass of 1500 kg experiences a normal force of 12000 N. What is the angle of the slope?
Calculation:
Given:
- \(F_n = 12000 \, \text{N}\)
- \(m = 1500 \, \text{kg}\)
- \(g = 9.8 \, \text{m/s}^2\)
Using the formula:
\[ \theta = \cos^{-1}\left(\dfrac{F_n}{m \cdot g}\right) \]
\[ \theta = \cos^{-1}\left(\dfrac{12000}{1500 \cdot 9.8}\right) \]
\[ \theta = \cos^{-1}\left(\dfrac{12000}{14700}\right) \]
\[ \theta \approx \cos^{-1}(0.8163) \]
\[ \theta \approx 35^\circ \]
Answer: The angle of the slope is approximately 35 degrees.
Example 2: Calculating the Angle of a Slope for a Crate
Problem: A crate with a mass of 200 kg experiences a normal force of 1500 N. What is the angle of the slope?
Calculation:
Given:
- \(F_n = 1500 \, \text{N}\)
- \(m = 200 \, \text{kg}\)
- \(g = 9.8 \, \text{m/s}^2\)
Using the formula:
\[ \theta = \cos^{-1}\left(\dfrac{F_n}{m \cdot g}\right) \]
\[ \theta = \cos^{-1}\left(\dfrac{1500}{200 \cdot 9.8}\right) \]
\[ \theta = \cos^{-1}\left(\dfrac{1500}{1960}\right) \]
\[ \theta \approx \cos^{-1}(0.7653) \]
\[ \theta \approx 40^\circ \]
Answer: The angle of the slope is approximately 40 degrees.
Example 3: Calculating the Angle of a Slope for a Barrel
Problem: A barrel with a mass of 100 kg experiences a normal force of 500 N. What is the angle of the slope?
Calculation:
Given:
- \(F_n = 500 \, \text{N}\)
- \(m = 100 \, \text{kg}\)
- \(g = 9.8 \, \text{m/s}^2\)
Using the formula:
\[ \theta = \cos^{-1}\left(\dfrac{F_n}{m \cdot g}\right) \]
\[ \theta = \cos^{-1}\left(\dfrac{500}{100 \cdot 9.8}\right) \]
\[ \theta = \cos^{-1}\left(\dfrac{500}{980}\right) \]
\[ \theta \approx \cos^{-1}(0.5102) \]
\[ \theta \approx 59^\circ \]
Answer: The angle of the slope is approximately 59 degrees.
