How do you find the acceleration of two objects if the coefficient of kinetic friction between the 7kg object and the plane is 0.250? Given: m1=7kg, m2=12kg, and there is an angle of 37.0 degrees.

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- anonymous

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- anonymous

I need more information. What does the set-up look like?

- anonymous

Okay. There is a triangle, with a 37 degree angle in the left bottom corner, holding up a block that weighs 7kg. This block is attached to a levy on the right upper corner of the triangle which is holding up a 12 kg block hanging downward on the right.

- anonymous

First draw a free body diagram. For the 7kg object you should get the two equations: Fx=T-Ff-mgsin37=m1a and Fy=N-mgcos37=0. So N=mgcos37=54.79. The friction force(Ff)=.25N=13.7. Substitute this into Fx and you will get Fx=T-54.98=m1a. This is the same as T=m1a+54.98. Then from the 12kg object, you should get the equation: Fy=m2g-T=m2a. Which is the same as T=m2g-m2a. Substitute this in for the T in the Fx equation and solve for a. I got 2.76, but I may have done some math wrong somewhere, so you should probably check everything :)

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- anonymous

Thank you so much!

- anonymous

I'll try it our right now.

- anonymous

No problem :)

- anonymous

How did you get 13.7 from the friction force?

- anonymous

The normal force is mgcos37 which is 54.79. Friction force is normal force time the coeffecient of friction. So it is 54.79*.25= 13.7

- anonymous

Oh. lol Thanks.

- anonymous

I got up to the last step. m2g-m2a= m1a + 55.0N
118N-12kg(a)= 7kg+55.0N
How would I solve this? :?

- anonymous

Oh, I got it. Thank you again!

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