Failing and Sliding Masses: Solving for Upward Acceleration in a Pulley System

[leavex]

Homework Statement

Mass m1=25.3 kg is on a horizontal surface, connected to mass m2= 5.10 kg by a light string as shown. The pulley has negligible mass and no friction. A force of 201.7 N acts on m1 at an angle of 30.9o.
The coefficient of kinetic friction between m1 and the surface is 0.225. Determine the upward acceleration of m2.

Homework Equations

F=ma
1) Fcos(theta)-T-Mk(m1g-Fsin(theta))=m1a
2)T-m2g=m2a

The Attempt at a Solution

Since they both have the same acceleration, i was trying to look for a

 

[jbsteele]

way to relate m1a and m2a together.1) Fcos(theta)-T-Mk(m1g-Fsin(theta))=m1a 2)T-m2g=m2aI know T is equal to both equations so I figured I could substitute it in. So I have: 1) Fcos(theta)-m2g-Mk(m1g-Fsin(theta))=m1a 2)m2g-m2g=m2aBut then I'm stuck. I'm not sure what I should do next. Help would be much appreciated. Thank you.