Q14. - Algebra 1

Question

A 135-pound person sits 5 feet from the centre of a seesaw. How far from the centre should a 108-pound person sit to balance the seesaw?

Step-by-Step Solution

Answer

To balance the seesaw, a person should sit at $\mathbf{6}\mathbf{.}\mathbf{25}\mathbf{feet}$ from the centre.

Step 1. Torque.

Torque is given by $\tau =rF$.

Step 2. Apply product rule.

Apply product rule for inverse variation.

$\begin{array}{c}{r}_1{F}_1=r_2{F}_2\\ \left(5\right)(135\times 32.17)=r_2(108\times 32.17)\end{array}$

Step 3. Simplify.

Simplify the above obtained equation.

$\begin{array}{c}\frac{21714.75}{3474.36}=\frac{3474.36r_2}{3474.36}\\ r_2=6.25\end{array}$

Therefore, to balance the seesaw, a person should sit at $6.25\text{ feet}$ from the centre.