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

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SUMMARY

The discussion clarifies the calculation of the frictional force on a crate sliding up a ramp inclined at 30 degrees. The kinetic coefficient of friction (μk) is 0.30, and the mass of the crate is 10.0 kg. The correct approach involves using cos(30) to determine the normal force, which is essential for calculating the frictional force. The misunderstanding arises from confusing the weight component (mgsin(30)) with the frictional force, which is a distinct force represented separately in the free body diagram.

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  • Understanding of free body diagrams
  • Knowledge of friction coefficients (static and kinetic)
  • Basic trigonometry, specifically sine and cosine functions
  • Newton's laws of motion
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  • Learn about the role of normal force in friction calculations
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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
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.

 

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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.
 
jfk313 said:
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.
 

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