# Ramp frictional force

1. Oct 8, 2013

### jfk313

1. The problem statement, all variables and given/known data

Find the frictional force (magnitude and direction) on the crate if the crate is sliding up the ramp.

Frictional force= ?
μk = 0.30
μs = 0.74
mass=10.0 kg
angle of incline 30° to the horizontal

2. Relevant equations

fk=uk * N

3. The attempt at a solution
I tried doing this problem by drawing t he free body diagram and creating another triangle to find frictional force, as seen in the picture
. i tried solving for the frictional force by multiplying the kinetic coefficient of friction, gravity and mass with sin(30).... I found out from a friend that cos(30) would give you the correct answer, but I not sure why. So I wanted to understand why the sin(30), giving you the parallel force of friction, is not correct when trying to solve this problem.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Oct 8, 2013

### Yosty22

When you draw out the freebody diagram, try making the x-axis the ramp 30 (tilted axis). Then, break up the weight vector into its x and y components, and you should be able to see that mgsin(\theta)is a force in the same direction as friction, but it is NOT friction. Friction is a separate force on the free body diagram all together.

3. Oct 8, 2013

### Yosty22

The mgsin(\theta) term is NOT the parallel force of friction. It is the box's parallel (x-component) of weight.