The two children shown in the figure below are balanced on a seesaw of negligible mass. The first child has a mass of 25.3 kg and sits 1.74 m from the pivot. The second child has a mass of 33.7 kg and sits 1.31 m from the pivot. Use the second condition for equilibrium (net 𝜏 = 0) to calculate *F*p, the supporting force (in N) exerted by the pivot. (Enter the magnitude.)

To solve this problem, we can apply the principle of rotational equilibrium, which states that the net torque (𝜏) about any pivot point is zero for an object in rotational equilibrium. This condition ensures that the seesaw remains balanced.

1. **Identify the Given Information:**
   - Mass of the first child, \( m_1 = 25.3 \, \text{kg} \)
   - Distance of the first child from the pivot, \( d_1 = 1.74 \, \text{m} \)
   - Mass of the second child, \( m_2 = 33.7 \, \text{kg} \)
   - Distance of the second child from the pivot, \( d_2 = 1.31 \, \text{m} \)

2. **Calculate the Weight (Force due to Gravity) for Each Child:**
   The force due to gravity, \( F = mg \), where \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

   - Weight of the first child, \( F_1 = m_1 \cdot g = 25.3 \times 9.8 = 247.94 \, \text{N} \)
   - Weight of the second child, \( F_2 = m_2 \cdot g = 33.7 \times 9.8 = 330.26 \, \text{N} \)

3. **Calculate the Torque Created by Each Child Around the Pivot:**
   Torque, \( \tau = F \cdot d \).

   - Torque due to the first child, \( \tau_1 = F_1 \cdot d_1 = 247.94 \times 1.74 = 431.43 \, \text{Nm} \)
   - Torque due to the second child, \( \tau_2 = F_2 \cdot d_2 = 330.26 \times 1.31 = 432.64 \, \text{Nm} \)

4. **Set Up the Equation for Rotational Equilibrium:**
   Since the seesaw is balanced, the sum of torques around the pivot is zero:

   \[
   \tau_1 = \tau_2
   \]

   Observing the values calculated, we see the torques are nearly equal, suggesting the seesaw is in balance.

5. **Determine the Supporting Force by the Pivot, \( F_p \):**
   For vertical equilibrium, the total upward force exerted by the pivot must balance the combined weight of the two children.

   \[
   F_p = F_1 + F_2 = 247.94 + 330.26 = 578.2 \, \text{N}
   \]

Thus, the supporting force exerted by the pivot, \( F_p \), is approximately **578.2 N**.

Would you like further details on any part of this solution?

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**Additional Questions for Exploration:**

1. How would the distance of each child affect the balance of the seesaw?
2. If the mass of the seesaw were not negligible, how would that impact the calculation?
3. What would happen if the first child moved closer to the pivot?
4. How would you calculate the torque if the seesaw were tilted at an angle?
5. How does this problem illustrate the concept of leverage in physics?

**Tip:** In torque problems, always choose the pivot point where unknown forces (like the support force in this case) act, as this simplifies the calculations by eliminating that force from the torque equation.

The two children shown in the figure below are balanced on a seesaw of negligible mass. The first child has a mass of 25.3 kg and sits 1.74 m from the pivot. The second child has a mass of 33.7 kg and sits 1.31 m from the pivot. Use the second condition for equilibrium (net 𝜏 = 0) to calculate Fp, the supporting force (in N) exerted by the pivot. (Enter the magnitude.)

Learn to calculate the supporting force exerted by the pivot on a balanced seesaw with two children. Using the principles of rotational equilibrium and net torque, this problem guides you through finding the force required for equilibrium. Suitable for Grades 10-12 physics students.

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