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Rope, tension and newtons third law

  1. Oct 9, 2008 #1
    This is more of a conceptual question, meant to help us understand how tension forces work.

    1. The problem statement, all variables and given/known data

    Assume that segment R exerts a force of magnitude T on segment L. What is the magnitude FRL of the force exerted on segment R by segment L?
    Give your answer in terms of T and other constants such as g.

    2. Relevant equations

    Fnet = ma

    3. The attempt at a solution

    Fnet = FR on L + FL on R = 0 (I assume it's 0)

    I also know that Fnet = ma, but I dont know if I should use any mass, since it's a rope, and in first year undergraduate class, I understood that ropes are massless.

    The question says in terms of g and I know that FG=m*g (g = -9.81)

    I dont know how to indroduce the g in the equations.

    Attached Files:

    Last edited: Oct 9, 2008
  2. jcsd
  3. Oct 9, 2008 #2

    Doc Al

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    Staff: Mentor

    Consider Newton's 3rd law. (It's way easier than you think.)
  4. Oct 9, 2008 #3
    Ok, Newtons third law states that "every force occurs as one member of an action/reaction pair of forces". So the two objects have forces acting on them of the same magnitude but opposite directions.

    I think I understand how the tension forces work horizontally, but I have no idea how to add the g in there. Is there also a normal force that is equal and opposite to g? If so, how? It's in the air so I don't understand where that would come from :(
  5. Oct 9, 2008 #4


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    Welcome to PF!

    Hi DakE_FeatH! Welcome to PF! :smile:

    What makes you think g is involved? :wink:
  6. Oct 9, 2008 #5
    Re: Welcome to PF!

    They say in the problem to give my answer in terms of T and "constants such as g"

    So I cant just use Fnet = FR on L + FL on R = 0
  7. Oct 9, 2008 #6

    Doc Al

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    Re: Welcome to PF!

    I think you are overly constraining yourself. I'm sure they meant: "... in terms of T and constants such as g if you need them".

    If A exerts a force on B, then B exerts an equal and opposite force on A. That's Newton's 3rd law. Case closed. :wink:
  8. Oct 9, 2008 #7
    Ok I got it, it was |-T|, they just put in the g to mess me up. Thank you guys!
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