Find the acceleration of 2 objects given the coefficient of kinetic friction?

Find the acceleration reached by each of the two objects attached by pulleys. One, 7.00 kg, is on an inclined plane of 37 degrees and the other, 12 kg, is hanging freely off the side. The coefficient of kinetic friction between the 7.00 kg object and the plane is 0.300.





I have tried this two different ways.


First:


m2gsinΘ -ukm2gcosΘ - T = m2a


T - m1g = m1a


I found T, and set the two equations equal to eachother and found


a = (m2gsinΘ - ukm2gcosΘ - m1g) / (m1 + m2).


After plugging everything in I found a = -4.882 m/s2 but this is the wrong answer.





Second try:


For 12kg object


m1g - T = m1a


(12)(9.8) - T = 12a


T = 117.6 -12a


For 7kg object


the frictional force


f = μ N = μ mg cos 37 =16.4359 N


T = f + m2g + m2a


T = 16.436 + (7)(9.8) + 7a = 117.6 - 12a


19a = 117.6 -85.036


a = 1.714


This answer is still wrong. Where did I go wrong?

Find the acceleration of 2 objects given the coefficient of kinetic friction?
http://en.wikipedia.org/wiki/Coefficient...