To solve this problem, we can use the principle of torque balance. Since the seesaw is balanced, the torque on one side must equal the torque on the other side.

The torque ($\tau$) generated by a force is given by the equation:

$\tau = F \cdot d$

where:

• $F$ is the force (in this case, the weight of the child, which is mass $\times$ gravitational acceleration),
• $d$ is the distance from the pivot (the center of the seesaw),
• Gravitational acceleration $g = 9.8 \, \text{m/s}^2$.

Step 1: Calculate the torque for the first child

For the first child with mass $m_1 = 27.9 \, \text{kg}$ and sitting $d_1 = 1.46 \, \text{m}$ to the left of the center:

$\tau_1 = m_1 \cdot g \cdot d_1$

Step 2: Set up the torque equation for the second child

Let the mass of the second child be $m_2$, and the distance from the center to the end of the seesaw on the opposite side is denoted as $d_2$. Since the seesaw is balanced, the torque for the second child will also be:

$\tau_2 = m_2 \cdot g \cdot d_2$

Step 3: Apply the balance condition

The seesaw is balanced, so the torques must be equal:

$\tau_1 = \tau_2$

This gives the equation:

$m_1 \cdot g \cdot d_1 = m_2 \cdot g \cdot d_2$

The gravitational acceleration $g$ cancels out:

$m_1 \cdot d_1 = m_2 \cdot d_2$

Step 4: Solve for $m_2$

Now, solve for the mass $m_2$ of the second child:

$m_2 = \frac{m_1 \cdot d_1}{d_2}$

You mentioned that the second child sits at the end, but we don't have the value of $d_2$. Can you provide the distance from the center to the end of the seesaw, or should we assume it?