Why Does Cos(30) Calculate the Correct Frictional Force on a Sliding Crate?

[jfk313]

Homework Statement

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

Homework Equations

fk=uk * N

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

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https://www.youtube.com/watch?v=https://www.youtube.com/watch?v=

. 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.

 

[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.

 

[Yosty22]

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.

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