Atwood machine pully problem with a twist

AI Thread Summary
The discussion revolves around calculating the tension in the cord supporting a pulley system with two masses, M1 and M2. The user initially seeks guidance on whether the tension in the support cord (Tc) is double that of the tension in the rope (Tr) holding the masses. After clarifying that the pulley is ideal (massless and frictionless), they derive the relationship Tc = 2Tr, leading to the calculation of the system's acceleration and tensions. The user confirms their calculations, finding the acceleration to be 4.45 m/s², Tr to be 17.1 N, and Tc to be 34.2 N. The thread concludes with the user expressing satisfaction with their solution.
DukeJP2010
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Homework Statement


A simple pulley has two masses on either end, one M1 and one M2. It is easy to find the acceleration of the two masses, and the tension on the rope suspending these masses. The pulley is suspended by a cord, however, and the question asks to find the tension in this cord that is holding up the pulley system.

Homework Equations


F=ma
Fnet=F1+F2+...

The Attempt at a Solution


I want to solve this one myself, but I need a pointer in the right direction.
 
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Do a freebody diagram of the pulley. From what you've described the acceleration of the pulley is 0.
 
so, would the tension of the cord holding up the pulley be twice the tension of the rope holding the masses?
 
things to consider:
what is the mass of the pulley? what is the mass of the system (minus the cord), if it is an non-ideal pulley shall need to take into account moment of inertia.

in the simplest case of massless pulley, massless, non-stretchable string... what do you think is the answer? remember, tension must balance the force that trying to pull the pulley system down.
 
DukeJP2010 said:
so, would the tension of the cord holding up the pulley be twice the tension of the rope holding the masses?

If it is an ideal pulley yes. As mjsd mentioned if the pulley has mass, then we need to consider the moment of inertia of the pulley...

I'm guessing "simple" pulley means ideal pulley. Plus they haven't given any information about the pulley, so we can assume it's ideal.
 
Sorry for not stating that it is a massless, frictionless pulley and cord. The only things with mass are the two blocks. I omitted their numbers so i can get a point in the right direction.

There is a cord holding the pulley up (Tc) and then the tension of the rope (Tr) doing down, but twice (one for each mass) so this would yield:

0=Tc-2Tr

If that is correct I think i can solve the problem since i have already found the tension in the rope (Tr) using atwood machine equations.
 
yup. looks right to me.
 
After some quick work, can someone check this for me?

M1 = 1.2kg
M2 = 3.2kg

doing some math (that would be painful to type) yields:

a = 4.45 m/s2 (a = acceleration of the blocks)
Tr = 17.1 N
Tc = 34.2

Thank you all for your help.
 
looks right to me.
 
  • #10
Once again, that you for your help.

Topic closed.
 
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