Block ontop of another causing static friction?

In Fig, blocks A and B have weights of 52 N and 22 N, respectively. (a) Determine the minimum weight of block C to keep A from sliding if μs between A and the table is 0.23. (b) Block C suddenly is lifted off A. What is the acceleration of block A if μk between A and the table is 0.14?





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Here is what I have so far.





A. Fx = Na-(m1g+m2g) = 0


Fy = T-f = 0 -%26gt; f = .23(Fn1)


B. Fy = (Nb+T)-m2g = 0


C. Fy = Nc-m3g = 0





So given this if I have T I can solve for the rest of the equation but I see no easy way to find T here, I am assuming I am denoting the forces incorrectly. But since C is on top of A it increases A's weight correct? But this doesn't mean the normal force for C is equal to A's normal force does it?

Block ontop of another causing static friction?
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Minimum weight of block C=w





Friction = f =(w+52)*0.14





22N balances fricion f





(w+52)*0.23 = 22





w = 22/0.23 - 52





w =95.65 - 52 =43.65 N


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Mass of A = m =52/g





Friction = 0.14*52= 7.28 N





(52/g)a = T- 7.28





(22/g)a=22 - T





Adding,





a(52+22)/g =22-7.28





a(52+22)/g =14.72





a = 14.72*9.8 /74





a =1.949 m/s^2





(b) Block C suddenly is lifted off A, the acceleration of block A is 1.949 m/s^2


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