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 = 18o and (b) θ = 30o.
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 on an inclined plane.
First, let's define the forces involved:
1. Weight (W): This is the force due to gravity acting on the car. It is directed vertically downward and can be calculated using the formula W = m * g, where m is the mass of the car and g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth).
2. Normal force (N): This is the force exerted by the surface of the inclined plane on the car, perpendicular to the plane. It is directed perpendicular to the surface of the incline and helps support the weight of the car.
3. Parallel force (P): This force acts parallel to the incline and is responsible for the car's motion up or down the hill. The magnitude of this force can be calculated using P = W * sin(θ), where θ is the angle of the incline.
Now, let's calculate the magnitude of the normal force in terms of the weight of the car:
(a) When θ = 18°:
The weight of the car (W) remains the same, but we need to find the value of the parallel force (P) using P = W * sin(θ).
Then, the magnitude of the normal force (N) can be found by subtracting the parallel force from the weight: N = W - P.
(b) When θ = 30°:
Again, calculate the parallel force (P) using P = W * sin(θ).
Then, find the magnitude of the normal force (N) by subtracting the parallel force from the weight: N = W - P.
To determine the ratio of the magnitude of the normal force to the weight of the car, we divide the magnitude of the normal force (N) by the weight of the car (W):
(a) Ratio for θ = 18°: N/W
(b) Ratio for θ = 30°: N/W
By evaluating these expressions you can find the answer to your question.
To determine the ratio of the magnitude of the normal force to the weight of the car, we need to analyze the forces acting on the car.
The weight of the car, which is the force due to gravity, can be represented as W = mg, where m is the mass of the car and g is the acceleration due to gravity.
The normal force, N, is the force exerted by the hill on the car perpendicular to the surface of the hill.
When the car is on an inclined plane, the weight can be resolved into two components: one parallel to the plane (W_parallel) and one perpendicular to the plane (W_perpendicular).
The normal force, N, is equal in magnitude and opposite in direction to the perpendicular component of the weight, W_perpendicular.
To find the ratio of N to W, we can use trigonometric relationships. Let's calculate the ratio for the given angles:
(a) When θ = 18 degrees:
The weight component perpendicular to the inclined plane is W_perpendicular = W * sin(θ). The normal force is equal to W_perpendicular. Therefore, N = W * sin(θ).
The ratio N/W can be calculated as N/W = (W * sin(θ)) / W.
Simplifying the expression, we have N/W = sin(θ).
Substituting θ = 18 degrees, the ratio becomes N/W = sin(18) ≈ 0.309.
(b) When θ = 30 degrees:
Using the same logic, the weight component perpendicular to the inclined plane is W_perpendicular = W * sin(θ).
The normal force is equal to W_perpendicular. Therefore, N = W * sin(θ).
The ratio N/W can be calculated as N/W = (W * sin(θ)) / W.
Simplifying the expression, we have N/W = sin(θ).
Substituting θ = 30 degrees, the ratio becomes N/W = sin(30) = 0.5.
Therefore, the ratio of the magnitude of the normal force to the weight of the car is approximately 0.5 when θ = 30 degrees, and approximately 0.309 when θ = 18 degrees.