Answer to Question #339411 in Calculus for alli

Question

Two Children balance on the seesaw. The 40 lb child is 5 ft from the fulcrum. How far from the fulcrum is the 50 lb child ?

Answer 1

Suppose that $x$ is a distance from the fulcrum to the second child. The equation of equilibrium has the form: $40\cdot5=x\cdot50$. We receive: $x=\frac{200}{50}=4$. Thus, the second child is $4\,\textrm{ft}$ from the fulcrum.

Answer: the second child is $4\,\textrm{ft}$ from the fulcrum.