The angle of the slope (\(\theta\)) can be calculated using the formula:
\[ \theta = \cos^{-1} \left( \dfrac{F_f}{\mu \cdot N} \right) \]
Where:
- \(\theta\) is the angle of the slope (in degrees)
- \(F_f\) is the force of friction (in newtons, N)
- \(\mu\) is the coefficient of friction (dimensionless)
- \(N\) is the normal force (in newtons, N)
Example 1: Calculating the Angle of the Slope for a Box
Problem: A box experiences a force of friction of 60 N on a slope with a coefficient of friction of 0.3. The normal force acting on the box is 200 N. What is the angle of the slope?
Calculation:
Given:
- \(F_f = 60 \, \text{N}\)
- \(\mu = 0.3\)
- \(N = 200 \, \text{N}\)
Using the formula:
\[ \theta = \cos^{-1} \left( \dfrac{F_f}{\mu \cdot N} \right) \]
\[ \theta = \cos^{-1} \left( \dfrac{60}{0.3 \cdot 200} \right) \]
\[ \theta = \cos^{-1} \left( \dfrac{60}{60} \right) \]
\[ \theta = \cos^{-1}(1) \]
\[ \theta = 0^\circ \]
Answer: The angle of the slope is 0 degrees (Note: This implies an error in problem setup, as a slope cannot have a 0-degree angle if friction is present).
Example 2: Calculating the Angle of the Slope for a Car
Problem: A car experiences a force of friction of 900 N on a slope with a coefficient of friction of 0.45. The normal force acting on the car is 2500 N. What is the angle of the slope?
Calculation:
Given:
- \(F_f = 900 \, \text{N}\)
- \(\mu = 0.45\)
- \(N = 2500 \, \text{N}\)
Using the formula:
\[ \theta = \cos^{-1} \left( \dfrac{F_f}{\mu \cdot N} \right) \]
\[ \theta = \cos^{-1} \left( \dfrac{900}{0.45 \cdot 2500} \right) \]
\[ \theta = \cos^{-1} \left( \dfrac{900}{1125} \right) \]
\[ \theta = \cos^{-1}(0.8) \]
\[ \theta \approx 36.87^\circ \]
Answer: The angle of the slope is approximately 36.87 degrees.
Example 3: Calculating the Angle of the Slope for a Sled
Problem: A sled experiences a force of friction of 25 N on a slope with a coefficient of friction of 0.35. The normal force acting on the sled is 70 N. What is the angle of the slope?
Calculation:
Given:
- \(F_f = 25 \, \text{N}\)
- \(\mu = 0.35\)
- \(N = 70 \, \text{N}\)
Using the formula:
\[ \theta = \cos^{-1} \left( \dfrac{F_f}{\mu \cdot N} \right) \]
\[ \theta = \cos^{-1} \left( \dfrac{25}{0.35 \cdot 70} \right) \]
\[ \theta = \cos^{-1} \left( \dfrac{25}{24.5} \right) \]
\[ \theta = \cos^{-1}(1.02) \]
(Note: This implies an error in the problem setup, as \(\cos^{-1}\) of a value greater than 1 is not possible.)
