Question

A see saw at the park has a length of 3.0 m. Sally has a mass of 30 kg and wants to see saw with Sarah who has a mass of 40 kg. If Sally sits on the very right end of the see saw, how far from the center should Sarah sit?

Answer 1

`1.125\; {\rm m}`.

Step-by-step explanation:

The torque that Sarah and Sally exert on this seesaw need to be equal in magnitude.

The weight of Sarah is
`m(\text{Sarah})\, g`, where
`m(\text{Sarah}) = 40\; {\rm kg}`.

The weight of Sally is
`m(\text{Sally})\, g`, where
`m(\text{Sally}) = 30\; {\rm kg}`.

Assuming that the seesaw is level. The force that Sarah exerts on the seesaw will be perpendicular to the seesaw. The resultant torque will be of magnitude
`\tau(\text{Sarah}) = m(\text{Sally})\, g\, l(\text{Sarah})`, where
`l(\text{Sarah})` is the distance between Sarah and the center of the seesaw (the fulcrum).

Similarly, the torque from Sally will have a magnitude of
`\tau(\text{Sally}) = m(\text{Sally})\, g\, l(\text{Sally})`, where
`l(\text{Sally}) = (3.0 / 2)\; {\rm m} = 1.5\; {\rm m}` is the distance between Sally and the center of the seesaw.

The magnitude of the two torques should be equal. Thus:

`m(\text{Sarah})\, g\, l(\text{Sarah}) = m(\text{Sally})\, g\, l(\text{Sally})`.

Rewrite this equation and solve for
`l(\text{Sarah})`:

`\begin{aligned}l(\text{Sarah}) &= \frac{m(\text{Sally})\, g\, l(\text{Sally})}{m(\text{Sarah})\, g} \\ &= \frac{m(\text{Sally})}{m(\text{Sarah})}\, l(\text{Sally})\\ &= \frac{30\; {\rm kg}}{40\; {\rm kg}}\, (1.5\; {\rm m}) \\ &= 1.125\; {\rm m}\end{aligned}`.