### Seesaw Equilibrium Analysis **(a) Where should the man sit to balance the seesaw?** To balance the seesaw, apply the second condition of equilibrium, setting the sum of the torques equal to zero: \[ \tau_{\text{pivot}} + \tau_{\text{gravity}} + \tau_{\text{man}} + \tau_{\text{woman}} = 0 \] Since the first two torques are zero, let \( x \) represent the man's distance from the pivot. The woman is at a distance \( L/2 \) from the pivot. Equation to solve for \( x \): \[ 0 + 0 - Mgx + mg(L/2) = 0 \] Solving for \( x \): \[ x = \frac{m(L/2)}{M} = \frac{(51.4 \, \text{kg})(2.22 \, \text{m})}{72.7 \, \text{kg}} = 1.57 \, \text{m} \] **(b) Find the normal force \( n \) exerted by the pivot on the seesaw.** Apply the first condition of equilibrium: \[ -Mg - mg - m_{\text{pl}}g + n = 0 \] Solving for \( n \): \[ n = (M + m + m_{\text{pl}})g = (72.7 \, \text{kg} + 51.4 \, \text{kg} + 11.6 \, \text{kg})(9.80 \, \text{m/s}^2) \] \[ n = 1330 \, \text{N} \] **(c) Repeat part (a), choosing a new axis through the left end of the plank.** Compute the torques and set their sum equal to zero. Now pivot and gravity forces on the plank have nonzero torques: \[ \tau_{\text{pivot}} + \tau_{\text{gravity}} + \tau_{\text{man}} + \tau_{\text{woman}} = 0 \] Equation: \[ -Mg(L/2 + x) + mg(0) - m_{\text{pl}}g(L/2) + n(L/2) = 0 \] Substitute **Example 8.3 Balancing Act** **Goal:** Apply the conditions of equilibrium and illustrate the use of different axes for calculating the net torque on an object. **Problem:** A woman of mass \( m = 51.4 \, \text{kg} \) sits on the left end of a seesaw—a plank of length \( L = 4.43 \, \text{m} \), pivoted in the middle as in Figure 8.6. **(a)** First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass \( M = 72.7 \, \text{kg} \) sit if the system (seesaw plus man and woman) is to be balanced? **(b)** Find the normal force exerted by the pivot point if the plank has a mass \( m_{pl} = 11.6 \, \text{kg} \). **(c)** Repeat part (b), but this time compute the torques about an axis through the left end of the plank. **Figure 8.6:** - **(a)** Two people on a see-saw. - **(b)** Free body diagram for the plank. **Strategy:** - In part (a), apply the second condition of equilibrium, \( \sum \tau = 0 \), computing torques around the pivot point. The mass of the plank forming the seesaw is distributed evenly on either side of the pivot point, so the torque exerted by gravity on the plank, \( \tau_{\text{gravity}} \), can be computed as if all the plank's mass is concentrated at the pivot point. Then \( \tau_{\text{gravity}} \) is zero, as is the torque exerted by the pivot, because their moment arms are zero. - In part (b), the first condition of equilibrium, \( \sum \vec{F} = 0 \), must be applied. - Part (c) is a repeat of part (a) showing that choice of a different axis yields the same answer.

### Seesaw Equilibrium Analysis (a) Where should the man sit to balance the seesaw? To balance the seesaw, apply the second condition of equilibrium, setting the sum of the torques equal to zero: \[ \tau_{\text{pivot}} + \tau_{\text{gravity}} + \tau_{\text{man}} + \tau_{\text{woman}} = 0 \] Since the first two torques are zero, let \( x \) represent the man's distance from the pivot. The woman is at a distance \( L/2 \) from the pivot. Equation to solve for \( x \): \[ 0 + 0 - Mgx + mg(L/2) = 0 \] Solving for \( x \): \[ x = \frac{m(L/2)}{M} = \frac{(51.4 \, \text{kg})(2.22 \, \text{m})}{72.7 \, \text{kg}} = 1.57 \, \text{m} \] (b) Find the normal force \( n \) exerted by the pivot on the seesaw. Apply the first condition of equilibrium: \[ -Mg - mg - m_{\text{pl}}g + n = 0 \] Solving for \( n \): \[ n = (M + m + m_{\text{pl}})g = (72.7 \, \text{kg} + 51.4 \, \text{kg} + 11.6 \, \text{kg})(9.80 \, \text{m/s}^2) \] \[ n = 1330 \, \text{N} \] (c) Repeat part (a), choosing a new axis through the left end of the plank. Compute the torques and set their sum equal to zero. Now pivot and gravity forces on the plank have nonzero torques: \[ \tau_{\text{pivot}} + \tau_{\text{gravity}} + \tau_{\text{man}} + \tau_{\text{woman}} = 0 \] Equation: \[ -Mg(L/2 + x) + mg(0) - m_{\text{pl}}g(L/2) + n(L/2) = 0 \] Substitute Example 8.3 Balancing Act Goal: Apply the conditions of equilibrium and illustrate the use of different axes for calculating the net torque on an object. Problem: A woman of mass \( m = 51.4 \, \text{kg} \) sits on the left end of a seesaw—a plank of length \( L = 4.43 \, \text{m} \), pivoted in the middle as in Figure 8.6. (a) First compute the torques on the seesaw about an axis that passes through the pivot point. Where should a man of mass \( M = 72.7 \, \text{kg} \) sit if the system (seesaw plus man and woman) is to be balanced? (b) Find the normal force exerted by the pivot point if the plank has a mass \( m_{pl} = 11.6 \, \text{kg} \). (c) Repeat part (b), but this time compute the torques about an axis through the left end of the plank. Figure 8.6: - (a) Two people on a see-saw. - (b) Free body diagram for the plank. Strategy: - In part (a), apply the second condition of equilibrium, \( \sum \tau = 0 \), computing torques around the pivot point. The mass of the plank forming the seesaw is distributed evenly on either side of the pivot point, so the torque exerted by gravity on the plank, \( \tau_{\text{gravity}} \), can be computed as if all the plank's mass is concentrated at the pivot point. Then \( \tau_{\text{gravity}} \) is zero, as is the torque exerted by the pivot, because their moment arms are zero. - In part (b), the first condition of equilibrium, \( \sum \vec{F} = 0 \), must be applied. - Part (c) is a repeat of part (a) showing that choice of a different axis yields the same answer.