Question
A car (m = 1670 kg) is parked on a road that rises 10.3 ° 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?
Answers
Wc = mg = 1670kg * 9.8N/kg = 16,366N. =
Weight of car.
Fc = (16,366N.,10.3 deg.),
Fp = 16,366sin10.3 = 2826.3N. = Force
parallel to plane.
a. Fv = 16,366cos10.3 = 16,102.3N. = Force perpendicular to plane = Normal force.
b. Ff = Force due to friction.
Fn = ma = 0 = Net force, a = 0.
Fn =Fp - Ff = 0,
2826.3 - Ff = 0,
Ff = 2826.3N. = Force of friction.
Weight of car.
Fc = (16,366N.,10.3 deg.),
Fp = 16,366sin10.3 = 2826.3N. = Force
parallel to plane.
a. Fv = 16,366cos10.3 = 16,102.3N. = Force perpendicular to plane = Normal force.
b. Ff = Force due to friction.
Fn = ma = 0 = Net force, a = 0.
Fn =Fp - Ff = 0,
2826.3 - Ff = 0,
Ff = 2826.3N. = Force of friction.