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Tension force in Atwood machine

  1. Dec 13, 2015 #1
    I've read a couple other threads about this issue. And one even addressed my exact question but the answer didn't help me understand this at all.

    In an Atwood machine, why isn't the total force on a mass = its gravity minus the gravity of the mass on the other side?

    I keep seeing tension in every example online as being equal for both but it seems to me like the tension of the string touching a mass should be equal to the weight of the other mass.

    I've tried to model the forces. I see each weight has gravity weight and an upward force caused by the weight of the other mass.

    Yet nothing I see does it this way or gets answers anything like that.
     
    Last edited: Dec 13, 2015
  2. jcsd
  3. Dec 13, 2015 #2
    If this were true, then both weights would not experience any acceleration because there would be no net force, correct?

    There isn't, in the ideal case, we only consider the object's weight and the tension due to the rope.

    Draw out the free body diagrams for both masses and with the knowledge that the masses move (1 mass moves down, the other moves up), try and rationalize why the above cannot be true.

    Also another point to note is that the rope is generally treated to be massless and inextensible in the ideal case, and so the tension within the rope should be constant throughout the rope.

    Am I addressing your misunderstanding? Or did I misread your question. Let me know. :D
     
  4. Dec 13, 2015 #3
    Hmm. I still don't get it.

    For example, if there's a 2kg mass and a 3kg mass on the other side and we let gravity = 10 m/s^2

    Then the way I see it, the forces acting on the 2kg mass are 20 N downwards from its gravity, 30 N upwards from the weight of the other mass. And the forces on the 3kg mass are 30 N downwards and 20 N upwards.

    So the resultant force on the 2kg mass is 10 N upwards and the resultant on the 3kg mass is 10 N downwards. Then the resultant acceleration is 10 N / (2kg+3kg) = 2 m/s^2

    Is the tension in the rope the force with which the rope is pulling up on a mass? Then that would give a tension of 30 N for the 2kg mass and 20 N for the 3kg mass

    As you can see... I'm missing some very crucial point here -.-
     
  5. Dec 13, 2015 #4
    You were correct up till this point. The reason why you seem to be getting different tensions for each mass is because you failed to account for the acceleration of the mass itself.
    Applying Newton's 2nd law to both masses, knowing the 2kg rises and 3kg falls:
    2kg mass: T - 2g = 2a ---- (1)
    3kg mass: 3g - T = 3a ---- (2)

    Notice from above that the tension is neither 20N nor 30N, but 24N. And the acceleration is as you calculated, 2m/s^2.

    Coming back to your misunderstanding, simply stating that "that would give a tension of 30 N for the 2kg mass and 20 N for the 3kg mass" is equivalent to saying:
    2kg mass: T=2g => T-2g = 0
    3kg mass: T=3g => T - 3g = 0
    and you see that your statements imply zero acceleration for BOTH masses. which is clearly untrue.
     
  6. Dec 13, 2015 #5
    in your scenario, the 2kg block would accelerate upwards at 5m/s2 and the 3kg block would accelerate downwards at 10/3 m/s2.
    Initially:
    Total Energy = 0 ( Kinetic Energy) + 2gh+3gh. For some arbitrary height h at which the masses are above the ground.
    Now, calculate total energy at some time later, for simplicity, lets say 1 second.
    Kinetic Energy of 2kg mass=0/5*2*52 = 25J
    Potential Energy of 2kg mass= 2g(h+2.5) . Apply the x=ut+0.5at2 formula here.
    Similarly,
    Kinetic Energy of 3kg block= 0.5*3*100/9= 50/3
    Potential Energy= 3g(h-5/3)
    Total Energy= 5gh -2.5 + 25+50/3.
    Energy is not conserved if your assumption, that the tension is equal to the weight of the other mass, is true. Ergo it is wrong. But do the calculations assuming the tension on both to be right and hey presto! You'll see that energy will be conserved.
     
  7. Dec 13, 2015 #6
    I see how to calculate it now and understand where all the m_1 g + m_2 g / (m_1 + m_2) stuff comes from now.

    I guess my remaining question really comes down to "what is the tension"?

    I thought it was just the force pulling up on a mass at a given point. But if the tension is the same on both sides, then I just have no clue what tension is supposed to be, conceptually.
     
  8. Dec 13, 2015 #7
    The tension within the rope is caused by the masses pulling down on the rope, and the tension force acting on the weight is simply the result of Newton's third law. i.e. A reaction force to the internal tension force caused by the weight.
    Note here that the tension in a rope is a result of two bodies pulling on the rope (Imagine pulling on a rope that is not tied to anything on the other side, the tension within the rope then is zero)

    The idea and concept of rope tension is not so straightforward to grasp, hopefully someone else here can provide a better explanation on what tension actually is.
     
  9. Dec 13, 2015 #8
    I think you did it well and believe I get it now.

    Thanks for your help.
     
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