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![**INTRODUCTION:**
This experiment will investigate the rotational analog of forces in equilibrium — i.e., torques in equilibrium. Here we will consider the net Torque acting on a rotating mass, \( \Sigma \tau \) around a specified pivot point. A single torque, \( \tau \), is directly related to an individual force acting on a rotating object:
\[ \tau = \mathbf{r} \times \mathbf{F}; \quad |\tau| = |\mathbf{r}||\mathbf{F}|\sin\theta = rF_{\perp} \quad (1) \]
where \( \mathbf{r} \) is the position vector indicating the distance from where the force acts to the pivot point. Only perpendicular components of force with respect to the \( \mathbf{r} \) vector will contribute to torque.
Generally speaking, net torque will give rise to an angular acceleration \( \alpha \).
\[ \Sigma \tau = I\alpha \quad (\text{general equation}) \]
Equilibrium problems consider the subset of the general equation where there is no rotation; i.e., \( \alpha = 0 \).
\[ \Sigma \tau = 0 \quad (\text{equilibrium equation}) \quad (2) \]](https://content.bartleby.com/qna-images/question/e9b53ebc-804c-49f3-b917-c4e1c4975b52/8cdce084-3cc3-4fcd-bb2b-a73317a623dd/h9d1dk_processed.png)
Transcribed Image Text:**INTRODUCTION:**
This experiment will investigate the rotational analog of forces in equilibrium — i.e., torques in equilibrium. Here we will consider the net Torque acting on a rotating mass, \( \Sigma \tau \) around a specified pivot point. A single torque, \( \tau \), is directly related to an individual force acting on a rotating object:
\[ \tau = \mathbf{r} \times \mathbf{F}; \quad |\tau| = |\mathbf{r}||\mathbf{F}|\sin\theta = rF_{\perp} \quad (1) \]
where \( \mathbf{r} \) is the position vector indicating the distance from where the force acts to the pivot point. Only perpendicular components of force with respect to the \( \mathbf{r} \) vector will contribute to torque.
Generally speaking, net torque will give rise to an angular acceleration \( \alpha \).
\[ \Sigma \tau = I\alpha \quad (\text{general equation}) \]
Equilibrium problems consider the subset of the general equation where there is no rotation; i.e., \( \alpha = 0 \).
\[ \Sigma \tau = 0 \quad (\text{equilibrium equation}) \quad (2) \]
Transcribed Image Text:**THEORY:**
1. **Force Body Diagram Analysis:**
Starting with equation (2) above and using the given values from the diagram, solve the force body diagram for the ratio \( m_1/m_2 \).
The diagram shows:
- A horizontal beam with a pivot at its center of mass.
- Two forces (\( m_1g \) and \( m_2g \)) acting downward at distances \( r_1 \) and \( r_2 \) from the pivot, respectively.
- A normal force \( N \) acting upward at the pivot point.
2. **Pivot Point Adjustment:**
Change the pivot location by moving it to the right of the center of mass by a distance \( r_3 \). Use equation (2) to solve the new system for the mass of the meter stick, resulting in equation (3).
The diagram shows:
- A horizontal beam with the pivot shifted right by \( r_3 \).
- The same forces (\( m_1g \), \( m_2g \), and \( m_{\text{com}}g \)) acting downward.
- The normal force \( N \) now acting at the new pivot location.
The diagrams illustrate the equilibrium conditions and moments acting on the beam for both pivot positions, demonstrating the principles of torque and balance.
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