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