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bhangu23
Feb22-09, 04:19 PM
1. The problem statement, all variables and given/known data
A 60 kg crate is being steadily dragged along the floor by a rope, which makes an angle of 70 degress with the horizontal. If the tension in the rope is 20N:
What is the force of friction?
What is the coefficient of friction?

2. Relevant equations

ffr=Ufn
ffr=Umgcos0


3. The attempt at a solution

i know that
fg=mg
= 60(9.8)
=588N

and
Fn= fgcos70
=588cos70
=201N

i dont knwo what to do from there

bhangu23
Feb22-09, 05:15 PM
anyone?

slider142
Feb22-09, 05:53 PM
I'm not sure why your normal force is Fg*cos(70). From my diagram, I get the force of friction |Ff| = |Fg| - |T|sin(70) where T is the tension.