Coefficient of Kinetic Friction for crate

AI Thread Summary
To find the coefficient of kinetic friction for a 1500 N crate being pushed with a 700 N force at a 20° angle below the horizontal, the correct approach involves calculating the normal force accurately. The normal force should include the weight of the crate plus the vertical component of the applied force, which is 700sin20, not 1500sin20. The horizontal component of the applied force is 700cos20, which is used to determine friction. The initial calculations were incorrect due to miscalculating the normal force. Accurate calculations will yield the correct coefficient of kinetic friction.
Tylersaurus
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Homework Statement



A 1500 N crate is being pushed across a level floor at a constant speed by a force F of 700 N at an angle of 20° below the horizontal.

What is the coefficient of kinetic friction between the crate and the floor?

Homework Equations


f=mu(n)

The Attempt at a Solution



I tried 700cos20/1500 + 1500sin20 using the horizontal component of the applied force divided by the normal force, which was mg plus the vertical component of the applied force, but it didn't work, obviously. Any suggestions?
 
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Your description seems right, but your numbers don't look right. The vertical component of the applied force is 700sin20... not 1500sin20
 
Tylersaurus said:

Homework Statement



A 1500 N crate is being pushed across a level floor at a constant speed by a force F of 700 N at an angle of 20° below the horizontal.

What is the coefficient of kinetic friction between the crate and the floor?

Homework Equations


f=mu(n)


The Attempt at a Solution



I tried 700cos20/1500 + 1500sin20 using the horizontal component of the applied force divided by the normal force, which was mg plus the vertical component of the applied force, but it didn't work, obviously. Any suggestions?
I'm not sure of your equations, but the force is applied below the horizontal; you seem to have assumed when calculating the normal force that the applied force is above the horizontal.
 
PhanthomJay said:
I'm not sure of your equations, but the force is applied below the horizontal; you seem to have assumed when calculating the normal force that the applied force is above the horizontal.

no, he assumed it was below the horizontal. He just used 1500 + 1500sin20 for the normal force, instead of 1500 + 700sin20.
 
Yeah, I got that later. Thanks, though.
 
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