Physics balancing problem?

It says "A 2kg mass is 5m away from the fulcrum of a see-saw. Where must a 3kg mass be placed to balance the see-saw?

please help


can you guys explain how you did it please?

5 Answers

  • 9 years ago
    Best Answer

    lets balance the see-saw. So it'll rotate neither left nor right, right?

    That's, in examination jargon, moments about the fulcrum add to zero


    d=10/3 ;


    But what weighs 2 Kg and is found 5m from a seesaw fulcrum? A large seesaw too , if u ask me ;-)

  • 9 years ago

    To balance, you have to have the fulcrum point be the "equal sign" in your equation, so that you have the (mass1 x distance1 from fulcrum) = (second mass2 x distance2 from fulcrum)

    2kg * 5m = 3 kg * X

    solve for X:

    X = 2kg* 5m / 3kg

    X = 3.33m

    So this means if the 2kg is 5m from the pivot point, then the 3kg on the other seat of the see saw will be 3.33m from the pivot

  • 9 years ago

    there must be a balance of torque. torque = force * distance. so on one side there is a 2kg mass 5m away. 2kg * 5m is 10 and that is the torque on that side. now on the other side u need the same torque. therefore u need 10= 3kg* d where d is the distance. you solve it and divide 10/3 to get 3 and a third. not 3.5 lol

  • TBT
    Lv 7
    9 years ago

    weight of 2kg = 2g N

    weight of 3kg = 3g N

    moment of 2g N about fulcrum = moment of 3g N about fulcrum

    2g * 5 = 3g * x

    x = 10/3 = 3.333 m ---answer

    Source(s): my brain (Prof TBT)
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  • 9 years ago

    3.5 m

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