Recent content by Travis

With F_x = -10 g + \mu_k N - 6 g \sin 37^o = -10 g + \mu_k 6 g \cos 37^o - 6 g \sin 37^o

I got -10 -10+6sin(37)-6sin(37) So m=6kg Theta=37Degrees Mu=0.33 a=1.2

Ok I understand that m is mue and yes I got most of what you said.

(10 \cos 60^o, 10 \sin 60^o)

Yes I am starting to understand!

Yes, isn't is also true that the ball will never be perfectly horizontal even if you pull the ropes as hard as possible. It will always hang somewhat, I think I remember that from some pervious classes.

Sorry, I don't understand it.:frown: I know how to plug everything into a formula if I have all the information but If its missing information then I don't have a clue on what to do.

It has a diagram on the test with the block and a pully with a weight hanging from it.

My teacher doesn't really teach out of a book, but just out of his head. I have no clue on how to do this stuff. I have to redo a test and one of the questions is... A 6-kg mass is placed on a plane inclined at 37degrees. It is attached to a 10-kg suspended mass. If the coefficient of kinetic...