A 5.3 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. how far to the left of the pivot must a 3.7 kg ca

Question

A 5.3 kg cat and a 2.5 kg bowl of tuna fish are at opposite ends of the 4.0-m-long seesaw. how far to the left of the pivot must a 3.7 kg cat stand to keep the seesaw balanced?

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2022-04-15T21:28:12+00:00 1 Answer 0

Answers ( 1 )

  1. Elliana97
    0
    2022-04-15T21:29:21+00:00

    Moment about the pivot must be equal for the seesaw to balance. Initially, the first cat and the bowl are at 2 m from the pivot.

    The moment due to cat = 5.3*2 = 10.6 kg.m
    The moment due to bowl = 2.5*2 = 5 kg.m
    The unbalanced moment = 10.6 – 5 = 5.6 kg.m

    Therefore, the 3.7 kg cat should stand at a distance x from the pivot in left to balance the 5.6 kg.m.
    That is,
    3.7*x = 5.6 => x = 5.6/3.7 = 1.5134 m to the left (on the side of the bowl)

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