Force-Pulley problem solving for accel

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Homework Statement


A 30.7-kg block (m1) is on a horizontal surface, connected to a 6.3-kg block (m2) hanging vertically by a massless string. The pulley is massless and frictionless. A force of 224.9 N acts on m1 at an angle of 30.5degrees. The coefficient of kinetic friction (muk between m1 and the surface is 0.209. Determine the upward acceleration of m2.


Homework Equations


Fnetx=m*ax, Fnety=m*ay, fk=muk*Fn,


The Attempt at a Solution


I drew the free-body diagram with normal force, Fn, and gravity, m*g, and the y component of the applied Force, Fy=F*sin30.5, on the y-axis with the Fnety= Fn+Fy-m*g=0. On the x-axis I have tension from m2, T, and fk=muk*Fn, and the x component, Fx=F*cos30.5. The Fnetx=T+muk*Fn-F*cos30.5, and I don't have a numerical solution for this yet. I know from Fnety that Fn=m*g-F*sin30.5 and that I'm looking to solve for T from m2 because that will be the same T on m2 on the other side of the pulley. From there I can solve for the acceleration easily but I'm having a terrible time with eliminating variables and isolating T. Am I looking over a trig identity or a simple substitution somewhere? Any suggestions would be great, thanks.
 
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I've haven't looked at all your working to make sure that it's correct, but I will say that you shouldn't worry about having trigonometric functions in your equations, remember that sin(30.5) and cos(30.5) are just numbers. Just solves these two equations as you would any pair of simultaneous equations.