# An Atwoods Machine hanging from the ceiling

1. The problem statement, all variables and given/known data
This isn’t a homework problem, but I thought it would fit best in this section because it is an actual problem. Therefore, I don’t know what the solution is, but it interests me.

First let me start off by saying I’ve solved Atwoods machines before but now that we’ve introduced a rope A that the Atwoods machine is hanging from, it gets very confusing for me.

The question that I’d like to figure out is:
What is the tension in rope A if the Atwoods machine is at rest on the ceiling, but the masses are not? m2 is heavier than m1. The rope is mass less and has no friction. 2. Relevant equations
Fnet = ma
w = mg
g = |9.8 m/s^2|

3. The attempt at a solution

T1 = T2 = T for m1 and m2

The force acting on m1
Fnet = T – m1a1

The force acting on the mass m2
Fnet = T – m2a2

Up is the positive direction, and m2 is going down, so our new equations are

Fnet = T – m1a
Fnet = m2a – T

But this is sort of where I get confused, because the Atwoods machine is hanging from the ceiling.

I know that m2 has a tension force acting on it in the up direction, and the rope has a resultant force from this tension which is pulling the string down- this causes mass m1 to have a Force in the up direction and its resultant force is what causes m2’s force in the up direction.

So the rope has no net force acting on it, because both of the tensions from m1 and m2 on the rope cancel out.

Now if we have a rope A that is applying a tension in the up direction to our rope.

So our rope has a force pulling it upwards.

Since the rope is mass less we have to "skip" it and pretend its not there and apply this tension of force A onto both of the masses.

But I have no idea how this tension applies.

Do I have to divide it in half because I have two masses?

And if I did do that, it would change the acceleration of the two masses, but that doesn’t make sense to me- because if we just pretended the Atwoods machine was hanging from a ceiling and ignored the tension A, then the acceleration would be different from what I would get if I didn’t ignore it. And I know this is wrong.

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