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March 25, 2017

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If a 100-g mass was placed at the 25-cm mark, and a 20-g mass at the 10-cm mark, where should a 500-g mass be placed to balance the system?

  • Physics - ,

    100*(25- distance from fulcrum) + 20*(40- distance from fulcrum)= 500(x- distance from fulcrum)

    3300=500x Solve for x. then add to 50 because it's in the opposite side of the fulcrum.

  • Physics - ,

    100*(25- distance from fulcrum) + 20*(40- distance from fulcrum)= 500(x- distance from fulcrum)

    3300=500x Solve for x. then add to 50 because it's in the opposite side of the fulcrum.

    At the 50.66 cm mark the 500g mass should be placed

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