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?

A see-saw of length 2 m is pivoted at 50 cm form one end in a public p

1.	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?
Answer» Solution : `MA=("Effort arm")/("LOAD arm")` `=(50)/(150)=(1)/(3)` `MA=(L)/(E)` `E=(L)/(MA)=150xx(3)/(1)` `:.` Effort =450 N