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Alonzo19
 3 years ago
A 26.3 kg block (m1) is on a horizontal surface, connected to a 5.70 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.057 m and a moment of inertia I=0.090 kgm2. A force F = 192.9 N acts on m1 at an angle theta = 30.3°. There is no friction between m1 and the surface. What is the upward acceleration of m2?
Alonzo19
 3 years ago
A 26.3 kg block (m1) is on a horizontal surface, connected to a 5.70 kg block (m2) by a massless string as shown in the figure below. The frictionless pulley has a radius R = 0.057 m and a moment of inertia I=0.090 kgm2. A force F = 192.9 N acts on m1 at an angle theta = 30.3°. There is no friction between m1 and the surface. What is the upward acceleration of m2?

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Qasim_Ahmed
 3 years ago
Best ResponseYou've already chosen the best response.1look buddy i dont know the absolute solution for i dont recall the equation but i'll tell you what to do .. first you should analyse F to calculate the force component that acts on the system.. in this case this component equals \[F * \cos \theta \] then you can calculate the m1's acceleration by newtons 2nd low \[F = m_{1} a\] now comes my erased part of my memory ! you can write down the equation of inertia and put in your variables to find the up going acceleration i know i didn't made lot of help .. so please forgive me .. i have an important exam tomorrow in Plasma Physics and i am so distracted .. Good Luck

Alonzo19
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks Ahmed! Let me try this way and see. Also, good luck on your exam :)
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