Posted by Shelby on .
A car (m = 1540 kg) is parked on a road that rises 18.9 ° 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?
Wc = mg = 1540kg * 9.8N/kg = 15,092 N. = Wt of car.
Fc = 15092N. @ 18.9 Deg.
Fp = 15092*sin18.9 Deg. = 4889 N. = Force parallel to road.
a. Fv = 15092*cos18.9 = 14,278 N. = Force perpendicular to road = The normal.
b. Fn = Fp - Fs = 0,
4889 - Fs = 0,
Fs = 4889 N. = Force of static friction.