Solving a Frictionless Pulley Problem: Find m2's Acceleration

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SUMMARY

The discussion focuses on solving a physics problem involving a frictionless pulley system with two blocks: a 33.5 kg block (m1) and a 7.10 kg block (m2). A force of 244.9 N is applied to m1 at an angle of 29.3°, and the pulley has a moment of inertia of 0.110 kgm² and a radius of 0.081 m. The user initially struggled with too many variables and insufficient equations but was guided to use three key equations to find the upward acceleration of m2. The final equations provided were F*cosθ - T1 = m1*a, T1 - T2 = I*α, and T2 - m2*g = m2*a, leading to a successful solution.

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zhenyazh
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hi
can some one please help me with this for an exam?

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 got to 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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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
 

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