## Force of Friction on object on an incline

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

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 Recognitions: Homework Help Friction is based on the force normal to the incline, not normal to the horizontal.
 I thought friction went parallel to the incline?

Recognitions:
Homework Help

## Force of Friction on object on an incline

 Quote by tascja I thought friction went parallel to the incline?
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

 sorry but im not following what your saying.. i dont know the materials so i cant find μ...
 Recognitions: Homework Help 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.
 ok so Ff = mgcos(theta) .. and then is 18.1 my angle?

Recognitions:
Homework Help
 Quote by tascja ok so Ff = mgcos(theta) .. and then is 18.1 my angle?
Is M*g*cosθ the component of gravity || to the incline?

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