Question

Two children seat themselves on a seesaw. The one on the left has a weight of 400 N
while the one on the right weights 300 N. The fulcrum is at the midpoint of the seesaw. If the child on the left is not at the end but is 1.50 m from the fulcrum andthe seesaw is balanced, what is the torque provided by the weight of the child on the right?

2.0 m is the correct answer.

Answer 1

Consider the formula for the torque:

\(\tau=F \times d\) ...(1)

Here, 

\(F\) - Force (Weight) in newton 

\(d\) - Distance in meter

Let \(\tau_{l}\) be the torque on the left side, \(F_{l}\) be the force on the left side and \(d_{l}\) be the distance between the fulcrum and the left side child. Therefore (1) becomes \(\tau_{l}=F_{l} \times d_{i}\) ...(2)

Substitute, \(400 \mathrm{~N}\) for \(F_{l}\) and \(1.5 \mathrm{~m}\) for \(d_{l}\) in equation (2) \(\tau_{i}=400 \times 1.5\) \(=600 \mathrm{Nm}\) ...(3)

Let \(\tau_{r}\) be the torque on the right side, \(F_{r}\) be the force on the right side and \(d,\) be the distance between the fulcrum and the right side child. Therefore (1) becomes \(\tau_{r}=F_{r} \times d_{r}\) ...(4)

Substitute, \(300 \mathrm{~N}\) for \(F_{r}\) in equation (4) \(\tau_{r}=300 \times d_{r}\)

If the seesaw is balanced, then its angular acceleration will be zero. Therefore the torque on the right child is equal to the torque on the left child.

\(\tau_{r}=\tau_{i}\) ...(5)

Substitute, \(300 \times d,\) for \(\tau_{r}\) and \(600 \mathrm{Nm}\) for \(\tau_{l}\) in equation (5) \(300 \times d_{r}=600\)

$$ \begin{aligned} d_{r} &=\frac{600}{300} \\ &=2 \mathrm{~m} \end{aligned} $$

Thus, the distance between the fulcrum and the right side child is \({d_{r}=2 m}\) Torque on the right side is equal to the torque on the left side. So, the torque on the right side is \({\tau_{r}=600 \mathrm{Nm}}\)