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?

Question

Ruben Ross

Physics

2 September, 01:48

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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Answers (1)

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Jay Dickson · Answer 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)