Look at how the forces break down. For block 1
SFx = m1*a = F*cos(q) - T1
where SFx is the sum of the forces in the x-direction, a is the acceleration, and T1 is the tension from the string. For block 2
SFy = m2*a = T2 - m2*g
where SFy is the sum of the forces in the y-direction, a is the acceleration, and T2 is the tension from the string. Considering the pulley now
St = I*a/r = r*T1 - r*T2
where St is the sum of the torque, I is the moment of inertia of the pulley, r is the radius of the pulley. Combining these three equations will let you solve for the acceleration.
I*a/r^2 = T1 - T2
T1 = F*cos(q) - m1*a
T2 = m2*a + m2*g
so we have
I*a/r^2 = F*cos(q) - m1*a - m2*a - m2*g
which rearranges to
(m1 + m2 + I/r^2)*a = F*cos(q) - m2*g
Solving for a yields
a = (F*cos(q) - m2*g)/(m1 + m2 + I/r^2)
a = (237.7*cos(32.7) - 6.5*9.8) / (6.5+32.3+(0.060/0.097^2))
a = 136.32711 / 45.1768
a = 3.0176
.gif)
.gif)
