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A see-saw of length 2 m is pivoted at 50...

A see-saw of length 2 m is pivoted at 50 cm form one end in a public park. What is the effort applied at the near end to the fulcrum to lift a load of 150 N at the far end to fulcrum?

Text Solution

Verified by Experts

The correct Answer is:
450 N
`MA=("Effort arm")/("Load arm")`
`=(50)/(150)=(1)/(3)`
`MA=(L)/(E)`
`E=(L)/(MA)=150xx(3)/(1)`
`:.` Effort =450 N
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