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Man hanging on a rope

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the force that the man (weight 70 kg) should exert on the rope to stay in balance.
    (answer = 137,2 N)

    [​IMG]


    2. Relevant equations

    Like I see, there are two ropes who are in equilibrium. In each rope the tension is the same but different from each other.

    I suppose the sum of the tensions of both ropes are equally to the weight of the person



    3. The attempt at a solution
    The netforce is zero cause there's no movement or acceleration

    Rope 1 = the rope who holds the chair.
    Rope 2 = the rope the man holds.

    in positive y-direction : T(rope1)-mg/2=0 (/2 because the weight from the two ropes are from the same person)

    so T(rope1)=mg/2

    the other one T(rope2)=mg/2

    like you see my equations doesn't match the answer. I know that the rope 2 depends on rope 1 cause they are attached to each other. But i can't write that mathematical equation.

    Can you help me?
     
  2. jcsd
  3. Aug 25, 2011 #2

    Doc Al

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    How does the tension in rope 2 relate to the tension in rope 1? (Look at the middle pulley.)
     
  4. Aug 25, 2011 #3
    If he pulls on rope 2 , then will the pulley follow and the chair go up. So it's related to each other or else the two ropes where indepented and the weight (and tension) where the same for each rope. ?
     
  5. Aug 25, 2011 #4

    Doc Al

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    I'm looking for the mathematical relationship between the two tensions. Analyze the force on the middle pulley. (The net force on the pulley must be zero.) If the tension in rope 2 is "T", what's the tension in rope 1?
     
  6. Aug 25, 2011 #5
    i must be -"T". So the equation for the rope 2 => T=mg/2
    and for rope 1 => -T=-mg/2. (The opposite ). But a negative force doesn't exist :s

    But i don't understand why the answer is 137,2 N. Cause my answer would be for rope 2 = 343,35
     
  7. Aug 25, 2011 #6

    Doc Al

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    No. Analyze the forces on that middle pulley. How many rope segments pull up? What force do they exert? How many pull down? What's the downward force they exert?
     
  8. Aug 25, 2011 #7

    Doc Al

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    Maybe it will be easier if you call the tension in rope 1 T1 and the tension in rope 2, T2. Then write a force equation for the middle pulley.
     
  9. Aug 25, 2011 #8
    Allright. the middle pulley is the "center" point of all the tensions from the ropes.
    everythings comes to this point (correct ?).

    T1 is the tension in rope 1, T2 is the tension in rope 2
    There's no force in the x-direction.

    y-direction : The net force of the pulley must be zero cause there is no movement ([itex]\sum[/itex] Fy =0

    T1 is coming to this pulley but also goes to rope 2 so we have two components of T1.

    T2 is also coming to this pulley but also goes to rope 1 so we have positive T2 and a negative T2.

    so : T1 + T1 + T2 - T2 - mg-mg = 0.

    Am i close to it ?
     
  10. Aug 25, 2011 #9

    Doc Al

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    Analyzing the forces on the middle pulley is just a step in the solution, not the final answer.

    There are two rope segments pulling upward: What force do they exert on the pulley?

    There is one rope segment pulling down: What force does it exert on the pulley.

    Note that we assume that the pulleys have no mass.
     
  11. Aug 25, 2011 #10
    Hmm

    The Two forces that are pulling the pulley upwards are T2. The first T2 goes to the ceiling and haves a normalforce called T2' who's going back to the rope 2 (and pulls the pulley upwards).

    The force that pulls the pulley downwards is the normalforce of T1 called T1' .

    We have also the mg from the first pulley who's pulling downwards.

    so T2 + T2'-T1' -mg would be zero ?
     
  12. Aug 25, 2011 #11

    Doc Al

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    You're making it a bit more complicated than necessary.

    The ropes pulling upward have a tension of T2, thus the force they exert is 2*T2 upward.

    The rope pulling down has a tension of T1, thus the force it exerts on the pulley is T1 downward.

    Note that since the pulley is massless, there's no 'mg' term.

    The net force must be zero, thus:
    2*T2 - T1 = 0
    Or:
    2*T2 = T1

    Now you have the relationship between T1 and T2. You'll need this in a bit.

    Next step: Analyze the forces on the man+chair+bottom pulley (treat them as one 'object').
     
  13. Aug 25, 2011 #12
    There are two forces that are pulling the pulley up, both are T1 and one goes down. That is the weight of the object (man+chair).

    therefore the equation is

    2*T1-mg
    =>T1=mg/2

    And if I substitute this in 2*T2=T1
    We get for T2= mg/4

    ow no. just saw that the bottom pulley haves two tensions going up with the same size of Tension 2 ( same object like the object on rope 2. Hmm, i'm confused :)
     
    Last edited: Aug 25, 2011
  14. Aug 25, 2011 #13

    Doc Al

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    Don't forget about the rope that the man is holding--that also exerts an upward force.
     
  15. Aug 25, 2011 #14

    Thanks, i now see that you must do mg/5 for the answer.
    Thank you very much!.

    How do i must learn to solve some exercises like this?
    caus i don't understand the essential bout this exercise
     
  16. Aug 25, 2011 #15

    Doc Al

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    You are welcome.

    The only 'secret' is to try and solve as many problems as possible. Don't be afraid to just 'jump in' and try something. Whenever you solve a problem, try to analyze how you did it and see if you can generalize.

    In this case, one 'trick' is to realize that the net force on a massless pulley is always zero (even if it accelerates)--that will give you at least one equation to use.
     
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