A car (m = 1510 kg) is parked on a road that rises 12.6 ° 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?

Answer 1

Wc = m*g = 1510kg * 9.8N/kg = 14,798 N.
= Wt. of car.

Fc = 14,798N @ 12.6o = Force of car.
Fp = 14798*sin12.6 = 3228 N. = Force
parallel to hill.
a. Fv = 14,798*cos12.6 = 14,442 N.=Force
perpendicular to hill or Normal.

b. Fp-Fs = ma.
3228-Fs = m*a = m*0 = 0.
Fs = 3228 N. = Force of static friction.