## 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

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## Answers ( 1 )

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)