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A 1,800 kg car is parked on a road that has an elevation level of 7°. Suppose the coefficient of static friction is .65. What is the force of static friction?

User Vicusbass
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1 Answer

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m = mass of the car = 1800 kg

g = acceleration due to gravity = 9.8 m/s²


F_(g) = force of gravity on the car

force of gravity on the car is given as


F_(g) = mg

inserting the values


F_(g)
= (1800) (9.8) = 17640 N


f_(s) = static frictional force acting on the car

from the force diagram of the car , the static frictional force is opposite to the parallel component of force of gravity on the car . hence static frictional force mus balance the component of force of gravity parallel to incline surface.

hence


f_(s) =
F_(g) Sin7


f_(s) = (17640) Sin7


f_(s) = 2149.8 N


A 1,800 kg car is parked on a road that has an elevation level of 7°. Suppose the-example-1
User Elian Ebbing
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