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Tug of war physics

  1. Aug 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Two boys play tug of war.If according to newton's third law, for every force a there is a reaction b with the same magnitude but opposite direction. What determines who wins?
    Our professor gave a hint that it has something to do with the ground.
    2. Relevant equations
    f=ma and newtons 3rd law
    3. The attempt at a solution
    I have identified these forces:
    A= force exerted by boy1 to the rope
    A'=reaction to A ,force exerted by the rope to boy1
    B=force exerted by boy1 to the rope
    B'=reaction to B ,force exerted by the rope to boy2
    Fa=Friction exerted at boy1
    Fb=Friction exerted at boy2
    Our professor said that no one wins if the ground is frictionless but i cant see why.
    thanks!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 18, 2011 #2
    Actually the heavier person will win for frictionless ground.

    Here's how to think about it: What is external force on the system? Where is the center of mass? Does it move? Where will the two players meet if they keep pulling?
     
    Last edited: Aug 18, 2011
  4. Aug 18, 2011 #3

    PeterO

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    You started out listing some great 3rd law couples:
    Boy 1 pulls the rope, the rope pulls Boy 1
    Boy 2 pulls the rope, the rope pulls Boy 2

    You then degenerated into some general descriptions
    Fa=Friction exerted at boy1
    Fb=Friction exerted at boy2

    You must express those forces, that you just called Friction, as 3rd law Couples.

    Do that and I will get back to you if that doesn't already answer your problem.
     
  5. Aug 19, 2011 #4
    boy2 pushes the ground,ground pushes boy2 ??
    boy1 pushes the ground, ground pushes boy1 ??

    The forces acting on boy1 are A' and the force from the ground.
    The forces acting on boy2 are B' and the force from the ground.
    The forces acting on the rope are A and B
     
  6. Aug 19, 2011 #5

    PeterO

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    Great.
    That is a total of 8 forces, drouped according to Newtons 3rd Law couple.

    Now group the forces according to which body they act on: boy 1, boy 2, the rope and the ground

    eg
    Rope: Boy 1 pulls rope, Boy 2 pulls rope

    Now You -

    Boy 1:
    Boy 2:
    Ground:

    Then look at the sizes of the pair of forces grouped this way.

    NB: Newtons Third law says that the pair of forces : Boy 1 pulls Rope, Rope pulls Boy 1 are equal in size, but Newtons law does not compare the force of the Rope pulling Boy 1 to the force of Boy 2 pushing the Ground
     
  7. Aug 19, 2011 #6
    Boy1: Ground pushes boy1(Fa), rope pulls boy1(A')
    Boy2: Ground pushes boy2(Fb), rope pulls boy2(B')
    Rope:Boy1 and boy2 pull on rope(A and B)
    A=-B if the rope is in equilibrium or if the rope is massless
    I don't know what's next!
    Should i treat the rope massless?
     
  8. Aug 19, 2011 #7
    The system moves to the side of boy 1 if fa>A'=A>B=B'>Fb???
    Fa>fb
     
  9. Aug 19, 2011 #8

    PeterO

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    The tension in the rope ensures that all forces involving the rope have the same size.

    Now to the boys: The size with which the rope pulls them is equal, so it is the one who is pushed more strongly by the ground that wins.

    And of course Newtons Third Law tells us that the ground pushes most strongly on the boy that pushes most strongly on the ground.
    So he who pushes the ground the hardest wins.
     
    Last edited: Aug 19, 2011
  10. Aug 19, 2011 #9
    thanks a lot!!
     
    Last edited: Aug 19, 2011
  11. Aug 19, 2011 #10
    "The tension in the rope ensures that all forces involving the rope have the same size."

    Wouldn't that happen only if the rope is in equilibrium or if the rope is massless?
    Please reply!! Last question =)
     
  12. Aug 19, 2011 #11

    PeterO

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    The boys would simply not be able to accelerate at the rate required to vary the tension in the rope - especially since the rope has a much lower mass than each of the boys. In fact in a tug of war, motion is generally very slow

    If the rope was as heavy as the boys [50+ kg] and one of them took off faster than an olympic sprinter you might notice something.

    We can demonstrate by changing the situation to a mass-less rope with a 2kg mass in the middle [so two ropes actually] and the boys having mass 49 kg. [this is to mimic a rope of mass 2kg]

    suppose Boy 1 wins, by pushing on the ground with a force 20N more than Boy 2. The net force on the system is thus 20N so the 100 kg system will accelerate at 0.2 m/s^2.

    Individually, the 2 kg mass has an unbalanced force of 0.4N acting [it is accelerating at 0.2 m/s^2 also remember]

    That would mean the tension in the rope from Boy 1 to the mass is 0.4N more than the tension in the rope from the mass to boy 2.

    Now to accelerate at 0.2 m/s is an extremely high rate of acceleration - not the sort of acceleration one associates with a tug of war, so the effects are probably much less than this, indicating that the tension will be pretty well constant even with a real rope.
     
  13. Aug 20, 2011 #12
    Thanks!! Now i get it!=)
     
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