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A 32.3kg block (m1) is on a horizontal surface, connected to a 6.50kg block (m2) by a massless string as shown in the Figure below. The frictionless pulley has a radius R = 0.097m and a moment of inertia I=0.060 kg*m2. A force F = 237.7N acts on m1 at an angle theta = 32.7. There is no friction between m1 and the surface. 
What is the upward acceleration of m2?
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Added Sun, 21 Jun '15
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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
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Added Sun, 21 Jun '15
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