Frictional Force & Normal forces

Homework Statement

A car (m = 1840 kg) is parked on a road that rises 18.7 ° 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?

F=ma Fg=mg Ff=uN

The Attempt at a Solution

errr.... what forces cancels out? is it FN=Fg=Ff or is it FN+Fg+Ff=0? i don't remember...

The Attempt at a Solution

rock.freak667
Homework Helper
On your free body diagram, at the angle of θ=18.7°, what are the forces acting? (After you split the weight of the car into components parallel to the road and perpendicular to the road)

frictional force, force of gravity, and normal force.

rock.freak667
Homework Helper
frictional force, force of gravity, and normal force.

So if there is no net movement perpendicular to the road, that is the magnitude of the normal force?

So if there is no net movement perpendicular to the road, that is the magnitude of the normal force?

the same as the gravitational force? 1 840 * 9.8 = 18 032 N?

rock.freak667
Homework Helper
the same as the gravitational force? 1 840 * 9.8 = 18 032 N?

No, the component of the weight perpendicular to the plane is equal to the normal force.

so that's equal to the y-component of the gravitational force?

18032 cos(18.7deg ) = 17080.10 N

and so frictional force is 18032sin18.7=5781.29 N ?

rock.freak667
Homework Helper
so that's equal to the y-component of the gravitational force?

18032 cos(18.7deg ) = 17080.10 N

and so frictional force is 18032sin18.7=5781.29 N ?

Yes but only because the net force in the x and y directions are zero.