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Tension in rope when the system is accelerating

  1. Jun 27, 2014 #1
    hey, I'm currently going through a mechanics course on edx.org . And part of last's week homework was this problem. well okay I don't know how to add images so I'm gonna try my best to describe it.

    There is a platform has a mass of 200kg. From that platform a rope hangs( with tension TA with a mass of 70kg at the end of it ( say a human ). Then from that mass ( human ) another rope holds another mass 90kg. it says it in the problem that g = 10 m/s^2 and the acceleration of the whole thing is 2 m/s^2 .Also the ropes are ideal ropes so have no mass.

    What is the tension Ta ( of the first rope ) ?


    here's the question copied " Man A (70kg) and Man B (90kg) are hanging motionless from a platform (200kg) at rest. What is the tension, TA, in the top rope if the platform accelerates upward at a constant rate of 2 m/s2? Assume the ropes are massless and use g = 10 m/s2. "


    hope I made it clear . So after I submitted my answer ( forgot what it was now ) It showed me the correct answer which is 1920 N. I could just accept it and move on. But I'll rather learn the physics behind it. And it's a problem I'm struggling with so I'll help me develop my understanding of physics.
    Okay so I have no idea on how to approach it. I tried so many ways but I never get the answer they get. So if you could , please explain the solution for me :)
     

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  3. Jun 27, 2014 #2

    A.T.

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    Assume g=12m/s^2 and just compute the weight.
     
  4. Jun 27, 2014 #3
    thanks , that works .Could you explain why ? isn't acceleration due to gravity downwards , and the system acceleration upwards . So why do they add up together?
     
  5. Jun 27, 2014 #4
    Add the masses of the two men together and find the net force acting on them using Newton's 2nd law. That net force is a resultant of the weight of the men downwards and the rope's tension upwards.
     
  6. Jun 28, 2014 #5

    A.T.

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    If the reference frame accelerates upwards, the objects tend to accelerate downwards. This is just like artificial gravity from acceleration added to the normal gravity. In General Relativity they are in fact the same thing.

    The formal ways to solve it are these:

    - Intertial frame:Find the tensions that together with gravity produce the required net forces for the acceleration of 2m/s^2 of each mass.

    - Non-Inertial frame: In the Accelerating rest frame of the masses the net forces must be zero, but there is an inertial force -ma acting on every mass m.

    - Intertial frame (quasi static): In this trivial case, its the same as the non-Inertial frame. But if you encounter a problem where interacting bodies accelerate differently, then you cannot construct a common rest frame for them. In this case you replace each individual acceleration a with a force -ma for every object, and assume that all net forces must be zero. This allows you to figure out the interaction forces at one time instant, but not the movement over time (hence "quasi static").
     
  7. Jun 28, 2014 #6

    Doc Al

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    Why don't you just apply Newton's 2nd law?

    Draw a free body diagram of each man or (even easier) both men together. What forces act? (Hint: The tension in the rope above man A is one of those forces.) Apply ∑F = ma.
     
  8. Jun 28, 2014 #7

    BvU

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    A.T. makes things difficult for newcomers. He isn't wrong, just that it's unnecessarily difficult to follow and very easy to make mistakes. So stick to the simplest possible frame of reference and g = 9.8 (or 10 in this case).

    To corroborate my argument: even the makers of the exercise fall into such a trap: They clearly state the two guys are hanging motionless. It is very difficult to explain how a motionless person can be motionless if the platform he hangs from accelerates with 2 m/s2! Motionless is pretty meaningless in this way ! (Intentionally provocative to trigger the people from the world's best universities -- even "best" is relative :smile:).
     
  9. Jun 28, 2014 #8

    A.T.

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    Motionless obviously just means that they hang passively, without climbing up or down the rope. Since the platform is initially at rest they are motionless. Then the platform accelerates upwards.
     
  10. Jun 28, 2014 #9

    BruceW

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    yeah... using g=12m/s^2 is easier and simpler. But to someone not familiar with the ideas of general relativity, it might be best to just use g=10m/s^2 and solve such that there is an upward acceleration of 2m/s^2.
     
  11. Jun 28, 2014 #10

    BvU

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    Let's err on the safe side and not expect KK to be familiar with GR ideas :smile:
     
  12. Jun 29, 2014 #11
    thanks guys for your help, I understand it now :)
     
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