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April 16, 2014

April 16, 2014

Posted by **COFFEE** on Saturday, February 10, 2007 at 5:56pm.

Break the force F in to vertical and horizontal components.

Friction= (mg + force vertical)mu

net force= ma

horizontal F - force friction = ma

Ok, so this is what I tried...

T = uk*m*g / ((cos theta)+uk(sin theta))

T = ((.25)(3)(9.8)) / ((cos -45)+(.25)(sin -45))

T = 13.9 N

then,

FNet = m*g - T*sin(theta)

FNet = (3)(9.8) - (13.9)*(sin -45)

FNet = 39.2

then,

fk = uk*FNet

fk = .25*39.2

fk = 9.8

Well that was the wrong answer but I don't see anything wrong with my approach. Any suggestions???

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