Question

problem 2
two children are playing on the seesaw. one child has a mass of 35 kg and is
seated 2.3 m from the pivot. the other child is seated at 3 m from the pivot.
if the seesaw is in equilibrium. find the mass of the second child, using the
fact that the sum of the torques is 0.

Explanation:

Step1: Recall the torque formula

Torque \(\tau = rF\), and \(F = mg\) (where \(g = 9.8\ m/s^{2}\)). For equilibrium \(\tau_1=\tau_2\), so \(r_1m_1g=r_2m_2g\). The \(g\) cancels out, giving \(r_1m_1 = r_2m_2\).

Step2: Solve for \(m_2\)

We know \(r_1 = 2.3\ m\), \(m_1=35\ kg\), \(r_2 = 3\ m\). Substitute into \(m_2=\frac{r_1m_1}{r_2}\).

$$m_2=\frac{2.3\times35}{3}=\frac{80.5}{3}\approx26.83\ kg$$

Answer:

The mass of the second child is approximately \(26.83\ kg\).