# Thread: Do I have this correct in the FBD

1. ## Do I have this correct in the FBD

A car thats m = 1700kg is parked on a road that rises 15 degrees 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??

For my FBD I put M at 270 degrees of 1700 kg, then for the 15 degrees angle up from the 0degree line and draw a line, then on the other side DO I PUT another line 15 degrees from the 180 line, to show 165 degree angle.

I am not sure what to do next.?? Do I have the right set up above??

A car thats m = 1700kg is parked on a road that rises 15 degrees 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??

For my FBD I put M at 270 degrees of 1700 kg, then for the 15 degrees angle up from the 0degree line and draw a line, then on the other side DO I PUT another line 15 degrees from the 180 line, to show 165 degree angle.

I am not sure what to do next.?? Do I have the right set up above??

3. Why is the Ff going that way?? and not towards the other way of N? I have them reversed. Why Ff that way, I don't have a clue why.

So working it how I have it
EFx= Fncos 15 + UFncos 105 +0=0

Works out to be U= -(cos15)/(cos 105)
Then sub that into the EFy equation.

SO it's Fnsin 15 + U FROM ABOVE(Fnsin105) - 16660N

Fn = 16660N / (3.8637N) founded to 2 sig figs of 4300N which is the answer.

But if I went by your diagram of Ff in that direction And do the same as above, the 15 and 105 would be switched around.
Can you explain, because I am a bit confused.
Thanks.

$F_f = mg\sin{\theta}$ ... the force of static friction acts opposite to the direction of the weight component parallel to the incline.
$N = mg\cos{\theta}$ ... the Normal force acts opposite to the direction of the weight component perpendicular to the incline.