Force of Friction on object on an incline

tascja

1. The problem statement, all variables and given/known data
A 3812.0 kg truck is parked on a 18.1° slope. What is the friction force on the truck?


2. Relevant equations

3. The attempt at a solution
So i set up a triangle with and then used:
weight of truck= (9.8 m/s^2 *3812 Kg) = 37367.6 N

sin(18.1) = (37367.6 N) /(-Ff)
Ff = -120246 N

Could someone explain why this is wrong.. im thinking maybe i set up my triangle incorrectly, but i dont know how else i would draw it.. thanks in advance

LowlyPion

Friction is based on the force normal to the incline, not normal to the horizontal.

tascja

I thought friction went parallel to the incline?

LowlyPion

It acts in a direction || to the surface but is determined by the force ⊥ to the surface through μ which is the ratio of the || force (required to overcome) by the ⊥ force.

http://hyperphysics.phy-astr.gsu.edu...frict.html#fri

http://hyperphysics.phy-astr.gsu.edu...frict.html#nor

tascja

sorry but im not following what your saying.. i dont know the materials so i cant find μ...

LowlyPion

You don't need μ in this case. The Frictional Force is what it balancing the truck's component of gravity down the incline. They just want you to find it. Not calculate it from the Normal force.

tascja

ok so Ff = mgcos(theta) .. and then is 18.1 my angle?

LowlyPion

Is M*g*cosθ the component of gravity || to the incline?

tascja

oh no then i would be mgsinθ = (3812 kg)(9.81 m/s^3)(sin18.1) = 1.162e4 N