Question

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?

Answer 1

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