Statis frictional force problem.

[neoncrazy101]

1. Homework Statement

A car (m = 1520 kg) is parked on a road that rises 17.3 ° 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?

Coefficient of static friction: rubber on dry concrete: 1.02. Homework Equations

F=ma
Fs max = coefficient of friction x normal force

3. The Attempt at a Solution
I can figure out the normal force just fine (1520cos17.3) then I convert it into N and it comes out to roughly 14200N. My issue is figuring out the static frictional force.

I've tried just doing the \mu(Fn) but its not right. My homework is online and we get 5 chances to get it but it will show us the answer after we click submit per each try and the static frictional answer for this question was 4430N but I cannot for the life of me figure out how they got 4430N. Can anyone help show me how they got it?

EDIT: Never mind. I figured it out. I should of taken, 1520sin17.3. then times 2.205 (convert to lbs) and then times 4.448 (to convert to N) and that gave me an answer of 4433N which is close enough variance in the program my class uses.

 

[tiny-tim]

hi neoncrazy101!

(have a mu: µ )

static force of friction usually has nothing to do with µ …

you find it from equilibrium considerations, just like any other reaction force ​

 

[Doc Al]

I've tried just doing the \mu(Fn) but its not right.

μN gives the maximum possible value of static friction; the actual friction will be less.

Hint: Analyze the forces parallel to the incline.

 

[neoncrazy101]

hi neoncrazy101!

(have a mu: µ )

static force of friction usually has nothing to do with µ …

you find it from equilibrium considerations, just like any other reaction force ​

Really? It doesn't? Man, I love my book lol. It only gives me that equation.

 

[SteamKing]

To convert kg to N, use g = 9.81 m/s^2 [1 kg = 9.81 N]