Question

Two children are playing on a seesaw trying to balance. If the child on the left sits 1.5 meters away from the fulcrum and weighs 25 kg, and the child on the right weighs 35 kg, how far away from the pivot point will the child on the right have to sit

Answer 1

For the seesaw to be balanced we need the torques each child apply on the seesaw to be equal, that is:

`\tau_1=\tau_2`

Where he torque is given by:

`\tau=Fr`

where F is the force each child applies and r is the distance from the pivot. For child one we know that it applies its weight, tha ihis mass is 25 kg and the distance is 1.5 m; for child two we know that it applies its weight and the distance is hat we need to find, then we have:

`\begin{gathered} (25)(9.8)(1.5)=(9.8)(35)r_2 \\ r_2=((25)(1.5))/((35)) \\ r_2=1.07 \end{gathered}`

Therefore, the distance from child two from the pivot has to be 1.07 m