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Thread: Force on rope

  1. #1
    Senior Member
    Nov 2013

    Force on rope

    hi all,

    Taru and Reiko simultaneously grab a 1.61 kg piece of rope and begin tugging on it in opposite directions. If Taru pulls with a force of 18.0 N and the rope accelerates away from her at 1.24 m/s2, with what force is Reiko pulling? Round your answer to 2 decimal places.

    My thoughts,
    force=mass*acc = 1.61 * 1.24 = 1.9964N
    But the force in the other direction has to be counteracted so
    Rieko must be pulling with a force of 18+1.9964 = 19.9964N => 20.00N

    Is this correct?
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  2. #2
    MHF Contributor
    Nov 2013

    Re: Force on rope


    maybe a better way to look at it is that the difference between the two forces causes the acceleration.

    (Fr - Ft)*1.61 = 1.24 where Fr is the force Reiku is applying, and Ft is the force Taru is applying.

    To be really careful you'd go ahead and associate all these forces and accelerations with vectors and just use the vector form of F=ma.
    Last edited by romsek; Dec 9th 2013 at 06:11 PM.
    Thanks from Melody2 and topsquark
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