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Question on fundamentals of tension problem

  1. Aug 5, 2011 #1
    The question I am looking at has a man of X kilograms of mass, with a rope tied around his waist. That rope is thrown over a tree limb. The man pulls on the other end of the rope to hoist himself up, with say, Y netwons of force (tension). The aim of the problem is to solve for his acceleration.

    The answer to this question says that the forces acting upon him are his weight mg, and 2 tension forces upwards.

    So net force is F = 2T - mg

    I am racking my brain to understand why the downward force of his hand doesn't contribute to the net force. 1) Is it because this force isn't acting upon the system (AKA the man). Rather, it is a force that the system exerts on another object, and thus, it shouldn't be counted?

    2) How do we know that the tension pulling up on the man's waist is pointed upwards? I know that the tension of a rope is the same all along a rope, but does that include direction?

    3) Also, I don't understand why weight doesn't contribute to the tension in the rope.

    I am studying physics for the MCAT and tension gets me every, single time. I need to master it once and for all..Been attempting every and any tension problem I come across lol..
     
    Last edited: Aug 5, 2011
  2. jcsd
  3. Aug 5, 2011 #2
    Ok, so I have cleared up that I was correct on my first hunch that the downward force that the man exerts on the rope doesn't factor into the net force equation. But I still don't understand why his weight isn't a contributing factor in the tension of the rope?
     
    Last edited: Aug 5, 2011
  4. Aug 5, 2011 #3

    tiny-tim

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    welcome to pf!

    hi bulbasaur88! welcome to pf! :smile:
    tension is always inward

    if it was outward, it would be called compression, and ropes (and chains) can't be in compression
    what do you mean by "contribute"? :confused:

    if you mean what i think you mean, then it does contribute

    anyway, what equation are you using?​

    (btw, never first-reply to your own thread … it knocks it off the no-replies list :redface:)
     
  5. Aug 5, 2011 #4
    Ahh thanks for the tip [: I was wondering why my question was getting passed by! *chuckle*

    By inward, do you mean that tension always points towards the midpoint of the rope?
    *confused* :/ My textbook only provides LESS than a page on tension!!! Very frustrating.
     
  6. Aug 5, 2011 #5

    tiny-tim

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    yes, the force which the rope exerts on another body is always inward

    (of course, the equal and opposite https://www.physicsforums.com/library.php?do=view_item&itemid=73" which the body exerts on the rope is always outward)
     
    Last edited by a moderator: Apr 26, 2017
  7. Aug 5, 2011 #6
    so the man trying to pull himself up is going to be the equation T-Yg=Ya since Tension is on his side and T>Yg (assuming the system is going in a clockwise direction)

    and the man of x kg is going to have the equation Xg-T=Xa since Xg>T (again assuming the system is going in a clockwise direction)

    T=Xg-Xa and T=Yg+Ya

    Xg-Xa=Yg+Ya

    Xg-Yg=a(Y+X)

    a=(Xg-Yg)/(Y+X)

    does this help you answer the question? this isnt necessarily the answer, since its just a set of variables and yea ;) its not an exact answer. So, plug in what you got. I won't have to get in trouble as a result.
     
  8. Aug 5, 2011 #7
    Thanks everyone :) it makes perfect sense now. TOok me awhile. but i got there... :D
     
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