Solving a 2-Mass System Connected by a Rope

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The discussion centers on calculating the tension in a rope connecting two masses, M1 and M2, over a frictionless pulley. Initial attempts to find the tension involved incorrect assumptions about the forces acting on the masses. The correct approach involves setting up equations based on Newton's second law, leading to expressions for tension and acceleration. The user mistakenly attempted to double the tension value, misunderstanding that the same tension acts throughout the rope. Clarification emphasizes that the tension is uniform, and the correct method should yield a single tension value, not a doubled one.
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Homework Statement


Two masses M1 = .600kg and M2 = 2.2kg connected by a cord of negligible mass and passes over a frictionless pulley of negligible mass. Assuming that y-axis has a positive upward direction, what is the tension in the cord?


Homework Equations


F = ma

W = mg

The Attempt at a Solution


I first tried to add up the Fg for both masses under the assumption that the opposing force would be the tension.

I then subtracted the larger Fg from the smaller Fg which was also wrong.

I then came upon https://www.physicsforums.com/showthread.php?t=201258" and tried to follow the process discussed there. I tried to solve for a based on the posts discussed in the other thread but was lost on posts 9 and 10 (NEwayz, back to what I did).

I found the equations of the two blocks to be T - m1g = m1a and m2g - T = m2a.
this makes T = m1a + m1g = m2g - m2a.
Plugging in the numbers gives 5.88N + .6a = 21.56N - 2.2a
solving for a gives a = 5.6ms-2
Plugging this back into the T equation gives a T of 9.24N
I then double the T because there is T working on both ends of the rope and get 18.48N.
I turns out that this is an incorrect answer and I am now out of ideas. Please help.
 
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You don't double T. The tension T is acting throughout the rope. It isn't T on one side and T on the other giving 2T total.
 
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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