Frictional Force & Normal forces

[onyxorca]

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?

Homework Equations

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...

 

[rock.freak667]

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)

 

[onyxorca]

frictional force, force of gravity, and normal force.

 

[rock.freak667]

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?

 

[onyxorca]

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]

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.

 

[onyxorca]

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]

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.