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sp2606 [1]
3 years ago
6

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?

Physics
1 answer:
zmey [24]3 years ago
8 0

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


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