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Tension of a rope with a hanging mass and acceleration

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the tension of a rope with a hanging mass of 0.5kg that falls with an acceleration of 0.44ms^2


    2. Relevant equations

    T = ma
    T = ma-mg



    3. The attempt at a solution

    T = mg = 0.5*9.81 = 4.905N, I would assume that this is correct for a static load on the rope ?

    My question is. How does the relationship between gravity and acceleration act on the tension of the rope? Is the correct equation
    T = mg-ma ?

    Any help would be appreciated
     
  2. jcsd
  3. Mar 24, 2013 #2
    Yes, t = mg - ma. But, for clarity, draw a free-body diagram of the block and sum your forces.
     
  4. Mar 25, 2013 #3
    Thanks,

    Tension acts in an upward direction
    Gravity acts in an downward direction
    Acceleration acts in a downward direction
    so i get T-ma-mg = 0
    transpose to get T = mg+ma
    Or am i looking at acceleration going in the wrong direction ?
     
  5. Mar 25, 2013 #4
    I apologize, my previous statement was incorrect. Since the mass is moving downward, the tension has to support the accelerations of both gravity and of the mass, and thus T = m(g+a).
    Your second statement is correct.
     
  6. Mar 25, 2013 #5
    From an inertial frame of reference ( say, a person standing on the ground) assuming the downward direction to be +ve, ma = mg - T.
     
  7. Mar 25, 2013 #6
    For a person standing on the ground ma would equal m, correct? no acceleration

    Transpose ma=mg-T for T
    T=mg-ma ?

    can someone explain why it is either g-a or g+a. This may be simple but im not getting it
     
  8. Mar 25, 2013 #7
    Sorry i mean to say that the person on the ground ma would equal mg
     
  9. Mar 25, 2013 #8
    No I think according to the question, the person on the ground sees it coming down with an acceleration of a=0.44m/s2. So ma=mg -T.
     
  10. Mar 25, 2013 #9
    So we are saying that
    ma=mg-T
    T= m(g-a)
    T= 0.5(9.81-0.44) = 4.685N in an upward direction

    Sunil Simha, where did i go wrong in post #3?
     
  11. Mar 25, 2013 #10
    I think you took the frame of reference of the falling block. Is that right?

    If yes, ma would be a pseudo force in upward direction and thus the equation would be T+ma=mg as the block is in rest in its own frame.

    If not, then maybe you took the frame of reference of the ground in which case the force equation would still be the same i.e. ma=mg-T.
     
  12. Mar 25, 2013 #11
    I maybe should have mentioned that the rope is on a flywheel

    would it be that acceleration (0.44) is working against gravity as it is slower?
     
  13. Mar 25, 2013 #12
    Check out my attachment... Is it the correct diagram?

    If yes then you can see the force equations.
    If no, I'm sorry for the inconvenience:redface:
     

    Attached Files:

  14. Mar 26, 2013 #13
    Not an inconvenience at all mate, thank you for all your patience and help.
    I am trying to understand why the equation is the way it is.

    Yes my diagram is similar but without the rope on the left hand side being pulled
    to re cap
    T= up
    mg= down
    ma=down
    thus T-mg-ma = 0 correct?
     
  15. Mar 26, 2013 #14
    The mass is allowed to fall at a rate of 0.44ms^-2
     
  16. Mar 26, 2013 #15
    You see, ma is actually the resultant of mg and T, right? Because (in the frame of reference of the ground) they are the only two forces acting on the block. So ma=mg-T.
     
  17. Mar 26, 2013 #16
    Ahh.... big help
    Not sure where you are in the world but you may have heard the penny drop.

    Thanks again Sunil Simha.
     
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