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
weight = w = m g
normal force = m g cos theta
ratio = cos theta
so cos 17
and cos 30
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 on an inclined plane.
The two main forces involved are the weight force (W) acting vertically downwards and the normal force (N) acting perpendicular to the incline.
(a) When θ = 17o:
To find the ratio of the magnitude of N to W, we can express these forces in terms of their component vectors parallel and perpendicular to the incline.
The weight force can be broken down into two components: the force acting parallel to the incline (W_parallel) and the force acting perpendicular to the incline (W_perpendicular).
W_parallel = W * sin(θ) = W * sin(17o)
W_perpendicular = W * cos(θ) = W * cos(17o)
Since the car is traveling up the hill, the normal force will act in the opposite direction to the weight force. Therefore, the magnitude of the normal force is equal to W_perpendicular, which is W * cos(17o).
So, the ratio of the magnitude of the normal force to the weight of the car in this case is:
N/W = W * cos(17o) / W
Simplifying it further, we get:
N/W = cos(17o)
(b) When θ = 30o:
Following a similar approach, we can find the ratio of the magnitude of the normal force to the weight of the car when θ = 30o.
W_parallel = W * sin(θ) = W * sin(30o)
W_perpendicular = W * cos(θ) = W * cos(30o)
N/W = W * cos(30o) / W
Simplifying it further:
N/W = cos(30o)
So, in this case, the ratio of the magnitude of the normal force to the weight of the car is given by the cosine of the angle (30o in this example).