To solve this seesaw problem, we need to calculate the mass of the second child to balance the system.

Assumptions and Approach:

1. Seesaw's principle: The system is balanced when the sum of torques about the fulcrum is zero.
2. Torque equation: $\tau = r \cdot F$ where $r$ is the distance from the fulcrum, and $F$ is the force (equal to the weight of the child: $F = m \cdot g$).
3. The first child, with mass $m_1 = 27.6 \, \text{kg}$, is sitting $1.12 \, \text{m}$ to the left of the fulcrum. Thus, their torque is: $\tau_1 = m_1 \cdot g \cdot r_1$
4. The second child, with unknown mass $m_2$, is sitting at the end of the seesaw, say at a distance $r_2$ to the right of the fulcrum. Their torque is: $\tau_2 = m_2 \cdot g \cdot r_2$
5. Since the system is balanced: $\tau_1 = \tau_2$ Simplify to: $m_1 \cdot r_1 = m_2 \cdot r_2$

Known values:

• $m_1 = 27.6 \, \text{kg}$
• $r_1 = 1.12 \, \text{m}$
• $r_2 = \text{distance of the second child to the fulcrum}$

Calculation:

Rearrange the torque equation to solve for $m_2$: $m_2 = \frac{m_1 \cdot r_1}{r_2}$

Please specify the value of $r_2$ (distance of the second child to the fulcrum), so we can compute the result.

Would you like to proceed with details or clarify $r_2$?

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Here are 5 related questions:

1. How does the mass distribution affect the balance of the seesaw?
2. What happens to the system if one child moves closer to the fulcrum?
3. How would the calculation change if the seesaw's mass is included?
4. What is the effect of gravitational acceleration on the balance?
5. Can the seesaw be balanced if the fulcrum is shifted?

Tip: Always choose the fulcrum as the point of reference when calculating torque—it simplifies the math!