A car is traveling up a hill that is inclined at an angle θ above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) θ = 17o and (b) θ = 30o.
M*g = Wt. of car.
a. Fn = Mg*Cos17 = 0.956Mg.
Fn/Mg = 0.956Mg/Mg = 0.956.
b. Fn = Mg*Cos30 = 0.866Mg.
Fn/Mg = 0.866Mg/Mg = 0.866.
To determine the ratio of the magnitude of the normal force to the weight of the car, we need to consider the forces acting on the car when it is traveling up the hill.
1. Weight (W): The weight of the car is the force exerted by gravity on the car and is equal to the mass of the car (m) multiplied by the acceleration due to gravity (g). The weight acts vertically downward and can be calculated using the formula W = m * g.
2. Normal Force (N): The normal force is the force exerted by a surface perpendicular to the surface. In this case, it is the force exerted by the hill on the car. The direction of the normal force is perpendicular to the surface of the hill and acts in a direction opposite to the weight of the car (up the hill). This force is always equal in magnitude but opposite in direction to the weight of the car when the car is on a level surface.
Now, let's calculate the ratio of the magnitude of the normal force to the weight of the car for the given angles:
(a) When θ = 17°:
- The weight of the car (W) remains the same regardless of the inclination angle.
- The normal force (N) can be determined using trigonometry. In this case, the normal force is the component of the weight acting perpendicular to the incline.
- N = W * cos(θ), where θ is the angle of inclination. Substitute the values:
N = W * cos(17°)
- Finally, calculate the desired ratio: N/W
(b) When θ = 30°:
- Apply the same steps as in case (a) to calculate N and then determine the ratio.
By following these steps and applying the appropriate trigonometric concepts, you should be able to determine the ratio of the magnitude of the normal force to the weight of the car for both cases.