**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 of \( m_p = 11.6 \, \text{kg} \). - **(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 seesaw. - **(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. **Exercise 8.3** Suppose a 29.7 kg child sits 2.00 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. \( m_{\text{child 2}} = \_\_\_\_\_\_\_ \, \text{kg} \) (b) Find the normal force acting at the pivot point. \( F_n = \_\_\_\_\_\_\_ \, \text{N} \)
**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 of \( m_p = 11.6 \, \text{kg} \). - **(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 seesaw. - **(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. **Exercise 8.3** Suppose a 29.7 kg child sits 2.00 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced. (a) Find the mass of the second child. \( m_{\text{child 2}} = \_\_\_\_\_\_\_ \, \text{kg} \) (b) Find the normal force acting at the pivot point. \( F_n = \_\_\_\_\_\_\_ \, \text{N} \)
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:**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 of \( m_p = 11.6 \, \text{kg} \).
- **(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 seesaw.
- **(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.
Transcribed Image Text:**Exercise 8.3**
Suppose a 29.7 kg child sits 2.00 m to the left of center on the same seesaw. A second child sits at the end on the opposite side, and the system is balanced.
(a) Find the mass of the second child.
\( m_{\text{child 2}} = \_\_\_\_\_\_\_ \, \text{kg} \)
(b) Find the normal force acting at the pivot point.
\( F_n = \_\_\_\_\_\_\_ \, \text{N} \)
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