Masses and Pulley: Find Accelerations of A, B, C

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The discussion focuses on solving for the accelerations of two masses, A and B, on a frictionless table connected by a rope over a pulley to a hanging mass, C. The original poster struggles with deriving the correct equations of motion for masses A and B, believing they should account for an additional acceleration due to the pulley. Other participants clarify that only the tension in the rope affects the acceleration of masses A and B, emphasizing that Newton's second law applies in an inertial frame without needing to add extra accelerations. The consensus is that the equations should reflect the tension alone as the net force acting on the masses. Properly applying these principles will lead to the correct solution for the accelerations of A, B, and C.
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Homework Statement



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Two masses, A and B, lie on a frictionless table. They are attached to either end of a light rope of length l which passes around a pulley of negligible mass. The pulley is attached to a rope connected to a hanging mass, C.

We are supposed to find the accelerations of A, B and C.

The Attempt at a Solution



I'm sorry that I don't have the time to write out my detailed working. I didn't get the answer required. But I notied that the solution given (page 5 from http://hep.uchicago.edu/cdf/frisch/p141/ps4_solutions.pdf) simply states

M_A a_A = T

and so on. My other equations, like the kinematic constraint (relation between ac, aa, and ab) and the equation of motion for mass C is ok. We need 4 equations to solve. So 2 are ok.

My equations for MA and MB are different from the solution here. I have assumed that M_A is in a frame that is accelerating at a rate equal to a_C, in addition to the tension imparted on mass A. Same for B. Can I do it like I have done here, or is this totally wrong?
 
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bigevil said:
My equations for MA and MB are different from the solution here. I have assumed that M_A is in a frame that is accelerating at a rate equal to a_C, in addition to the tension imparted on mass A. Same for B. Can I do it like I have done here, or is this totally wrong?
It's not clear to me what you've done. The equations for Ma and Mb are just Newton's 2nd law (which applies in an inertial frame). The tension is the only horizontal force on those masses. (The fact that the pulley is accelerating will end up affecting the tension, but that will come out of solving the equations for all three masses.)
 
bigevil said:
...
My equations for MA and MB are different from the solution here. I have assumed that M_A is in a frame that is accelerating at a rate equal to a_C, in addition to the tension imparted on mass A. Same for B. Can I do it like I have done here, or is this totally wrong?

I don't understand what you mean by this. Tension T alone is the net force acting on MA, therefore its acceleration is T/MA. There is no other acceleration to be added to this. MA doesn't care how tension T is generated. All it knows is that, if tension T acts on it, its acceleration will be T/MA.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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