normal force = 1700*9.81*cos 15
gravity force down hill = 1700*9.81 * sin 15
Third law, Friction force up hill equal in magnitude and opposite in sign to gravity force down hill.
A Mercedes-Benz 300SL (m = 1700 kg) is parked on a road that rises 15 degrees above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires? Important: Assume that the road is higher up to the right and lower down to the left.
1 answer
1 answer