whozum said:
If a string has a mass pulling it to the left at 10N and another mass pulling it on the right at 5N, what is the tension in the string?
Both ends of the string have a minimum of 5N acting on them, so that is the tension. The other 5N from the left will result in a net force on the entire system.
Onto your problem
You are given the tension in the third rope, and the three masses. If you read above, you can figure out that the force applied by the hand minus the force from the other three blocks is 65.4N
While I expect you know what you meant to say, there are a couple of things you did say that you need to look at again.
If a string has a mass pulling it to the left at 10N and another mass pulling it on the right at 5N, what is the tension in the string?
The only way this could happen is if there is a tension gradient in the string, and that could only happen if the string had mass, which is not the case in this problem. A massless string experiencing different forces at both ends would have infinite acceleration. In this problem, and in your calculations, the tension in each string is constant and serves as a mechanism for two masses to exert equal and opposite forces upon one another.
Onto your problem
You are given the tension in the third rope, and the three masses. If you read above, you can figure out that the force applied by the hand minus the force from the other three blocks is 65.4N
Your statement suggests that 65.4N is the difference between the force applied by the hand and forces to the left, which fits with your description of tension, but that description is not correct. I don't think you want to talk about 65.4N as being the difference between the force of the hand and other forces. 65.4 N is one force (or more correctly one action-reaction pair of forces), the force applied by the hand (and the reaction force T3), and that is all. If the "object" experiencing the force is the group of three masses, then your statement is correct, but only because the the 65.4N is the only external force acting, so the "force from the other three blocks" means external forces acting on those blocks which is zero. If other forces (friction for example) were acting on the blocks, and the blocks were considered as one object, the 65.4N would still be 65.4N and T3, but the net external force would be reduced and the acceleration would be less. In this frictionless problem, if you are looking at M3, T3 minus the force from the other blocks is not 65.4N; T3 is 65.4N and T3 - T2 is the net force acting on block M3.
I think what the OP needs to see explicitly is something that is contained in your analysis, but not so obviously stated up front, which is that each block is acted upon by one or two forces and that the blocks share a common acceleration
T1 = M1 a
T2 - T1 = M2 a
T3 - T2 = M3 a
(T3 = force applied by gigantic hand = 65.4N)
Add the first three equations to arrive at your starting point.
T3 = (M1 + M2 + M3) a
a = T3/(M1 + M2 + M3)
Then step through the first three equations to find the tensions as you have outlined.
If there were friction forces f1, f2 and f3 involved, the equations would become
T1 - f1 = M1 a
T2 - T1 - f2 = M2 a
T3 - T2 - f3 = M3 a
(T3 = force applied by gigantic hand = 65.4N)
EDIT
This is not correct
\vec{F}_{m2} = m\vec{a} - \vec{F_{m1}} = (24kg)(1m/s^2) - (11.2N) = 12.8N
It should be
T2 - T1 = (24kg)(1m/s^2)
T2 = (24kg)(1m/s^2) + T1
