# Torque question

hi

A 33.5 kg block (m1) is on a horizontal surface, connected to a 7.10 kg block (m2) by a massless string. The frictionless pulley has a radius R = 0.081 m and a moment of inertia I=0.110 kgm2.
A force F = 244.9 N acts on m1 at an angle θ = 29.3°. There is no friction between m1 and the surface. What is the upward acceleration of m2?
an image is attached

i wrote force equation for both masses.
none of the are in equilibrium so each has its acceleration.
i also wrote one equation for the torque on the pulley. summing
all torques and equaling to I*alfa.

problem is i have too many variables. t1 t2 a1 a2 and not enough equations.
i gotta be missing something.
even if i unite t1 and t2 into one variable of t1-t2,
and decide that alfa is the square root of squares of a1 and a2
i am left with one equation and two variables.

can anyone explain if i am wrong or what am i missing
or give a detailed outline of solution?

thanks

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rl.bhat
Homework Helper
You can write three equations.
F*cosθ - T1 = m1*a ........(1)
T1 - T2 = I*α.......(2)
T2 - m2*g = m2*a.........(1)
Now solve for a.

thanks. it worked