Asked by Anonymous
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.
Answers
Answered by
Henry
Wc = m*g = 1700kg * 9.8N/kg = 16,660 N.
= Wt. of car.
a. Fn=16,660*cos15 = 16.1 N.=Normal force.
b. Fp = 16660*sin15 = 4312 N. = Force
parallel to the road.
Fp-Fs = 0
4312-Fs = 0
Fs = 4312 N. = Force of static friction.
= Wt. of car.
a. Fn=16,660*cos15 = 16.1 N.=Normal force.
b. Fp = 16660*sin15 = 4312 N. = Force
parallel to the road.
Fp-Fs = 0
4312-Fs = 0
Fs = 4312 N. = Force of static friction.
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