Question

A uniform seesaw is 3.2 m long, with its pivot at the center. A girl weighing 580 N sits at one end. A boy balances the seesaw by sitting 0. 4m from his end. What is his weight?

Answer 1

Step-by-step explanation:

for a seesaw (or a lever) to be in balance, the product of force and distance to the center must be the same on both sides.

our seesaw is split into 2 equal sides by the pivot.

each side is therefore 3.2/2 = 1.6 m long.

the girl with 580 N sits at the end of her side. let's call this side 1.

the corresponding force f1 = 580 N.

the corresponding distance d1 from the center = 1.6 m.

the boy with x N achieves balance by sitting on side 2 with 0.4 m from the end. which means

f2 = x N

d2 = 1.6 - 0.4 = 1.2 m

as they are in balance, the following must be true :

f1 × d1 = f2 × d2

580 × 1.6 = x × 1.2

x = 580 × 1.6 / 1.2 = 773.3333333... N

he weighs 773.333333... N or 773 ⅓ N.