Question

A boy who weighs 200 N is seated 2 m from a fulcrum of a sea saw with another boy who weighs 100 N and is seated 1 m from the fulcrum. How much must a girl sited at 4 m away from a fulcrum on the other side weigh to keep the seasaw at equilibrium.
200 * 2 + 100 * 1 = w * 4
500 = 4 w
w = 125 Newtons

Related Questions

1. Two children balance a see-saw in horizontal equilibrium. One weighs 80 lb. and the other weighs... Susie and Sally are seated on opposing sides of a seesaw. Sally weighs 500 N and is seated a distanc... A father and son sitting on opposite sides of a seesaw balance each other. What is the father's wei...