Atwood machine headache woe

  • Thread starter Shentar
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  • #1
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I was given this problem below in my physics homework. :cry: :yuck:

http://www.turbozcar.com/images/knight.Figure.08.40.jpg [Broken]

It wants to know the acceleration of m1. I couldn't figure it out, so the next day a friend in class gave me the answer to be:
(2*m_2*g)/(4*m_1+m_2)

I took that answer back to my house and was going to work it out to understand it all. I got my girlfriend to help me out, but neither of us can get the answer to come out correctly. We both get:
(m_2*g)/(2*m_1+m_2)

I don't know where we are going wrong.
What I know:
the forces acting on m_1 is the tension in the rope.
The forces acting on m_2 is gravity (m_2*g) and then two tension forces acting upward.

If I'm missing something, please let me know. I have a test tomorrow on tthis.
 
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Answers and Replies

  • #2
arildno
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Welcome to PF!
Your problem is that you have assumed that object 2 has the same acceleration as object 1.
This is however, physically impossible:

Suppose the rope segment attached to object 1 and "ending" at the fixed pulley shortens an amount L.

Since the rope remains of fixed length, the only viable situation is that the rope segment slung around the accelerating pulley has lengthened by L/2 on both sides of the pulley.
But this means that object 2's acceleration must be half of object 1's
 
  • #3
Gza
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I wouldn't feel too disappointed about not figuring that out, Shentar. I haven't found a physics text yet that's dealt with these types of pulley problems in a straightforward fashion. That's why I would recommend the REA's mechanics problem solver, which actually goes through the trouble of teaching you the methods of solving physics problems without the b.s. texts expecting you to just derive these methods on your own, while providing nothing but weak examples for you to learn from.
 
  • #4
Doc Al
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acceleration constraints

Just for the record, the problem you are working is not an Atwood's machine. An Atwood's machine looks like this: http://hyperphysics.phy-astr.gsu.edu/hbase/atwd.html

Nonetheless, one "secret" to solving problems involving multiple objects constrained by ropes and pulleys is to figure out the so-called acceleration constraints. This is what arildno explained. In this case, the constraint is that the magnitude of m1's acceleration equals twice the magnitude of m2's acceleration. Of course, when m2 moves down, m1 moves to the right.

(For the normal Atwood's machine, the acceleration constraint is: the magnitudes of the accelerations of the two masses must be equal. They, of course, move in opposite directions.)
 

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