A 5.0 kg cat and a 2.0 kg bowl of tuna fish are at opposite ends of a 4.0 m long seesaw. How far to the left of the pivot point must a 4.0 kg cat stand to keep the seesaw balanced? 5.0 kg 4.0kg 2.0 kg d A 2.0 m 2.0 m Pivot

Principles of Physics: A Calculus-Based Text
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Chapter10: Rotational Motion
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Problem 36P: A crane of mass m1 = 3 000 kg supports a load of mass m2 = 10 000 kg as shown in Figure P10.36. The...
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5.
A 5.0 kg cat and a 2.0 kg bowl of tuna fish are at opposite ends of a 4.0 m long seesaw.
How far to the left of the pivot point must a 4.0 kg cat stand to keep the seesaw balanced?
5.0 kg
4.0kg
2.0 kg
d A
2.0 m
2.0 m
Pivot
Anint
In order to tep the susaw in balance, the moment abouł the pivot point must be causl
4 kq cat is zt dishnece "dn from the pivot point
Mornent = masscm) x
distance (d)
Transcribed Image Text:5. A 5.0 kg cat and a 2.0 kg bowl of tuna fish are at opposite ends of a 4.0 m long seesaw. How far to the left of the pivot point must a 4.0 kg cat stand to keep the seesaw balanced? 5.0 kg 4.0kg 2.0 kg d A 2.0 m 2.0 m Pivot Anint In order to tep the susaw in balance, the moment abouł the pivot point must be causl 4 kq cat is zt dishnece "dn from the pivot point Mornent = masscm) x distance (d)
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