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183_notes:torquediagram [2016/03/14 16:37] – caballero | 183_notes:torquediagram [2016/04/01 13:39] – [Lecture Video] pwirving | ||
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==== Lecture Video ==== | ==== Lecture Video ==== | ||
- | Forthcoming... | + | {{youtube> |
==== A balanced situation ==== | ==== A balanced situation ==== | ||
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Why did location 4 get picked in this case? | Why did location 4 get picked in this case? | ||
- | * **Problem with Location 1 (Left end of plank)** The left end of the plank is a reasonable choice. Because all the torques about that location have to be zero, we can use the torques due to the weight of the plank, the pivot force, and the weight of mass 2 to do the analysis. The problem is that we neither know mass 2 or the pivot force, so we are again left with two unknowns and one equation. We could solve the system of equations resulting from the force analysis above and use that here, but the math could be a little hairy. You will probably have to do that in the future at some point. | + | * **Problem with Location 1 (Left end of plank)** The left end of the plank is a reasonable choice. Because all the torques about that location have to be zero, we can use the torques due to the weight of the plank, the pivot force, and the weight of mass 1 to do the analysis. The problem is that we neither know mass 1 or the pivot force, so we are again left with two unknowns and one equation. We could solve the system of equations resulting from the force analysis above and use that here, but the math could be a little hairy. You will probably have to do that in the future at some point. |
- | * ** Problem with Location 2 (Right end of plank)** The right end of the plank is also reasonable. Again, all the torques have to be zero about that location. Now, these torques are due to the weight of the plank, the pivot force, and the weight of mass 1. Notice that when you pick a point of application of a force to be the pivot point (e.g., location where mass 2 is) that force no longer contributes because there' | + | * ** Problem with Location 2 (Right end of plank)** The right end of the plank is also reasonable. Again, all the torques have to be zero about that location. Now, these torques are due to the weight of the plank, the pivot force, and the weight of mass 2. Notice that when you pick a point of application of a force to be the pivot point (e.g., location where mass 1 is) that force no longer contributes because there' |
- | * ** Problem with Location 3 (Center of plank)** At the center of the plank, the torques are due to the weight of the two masses and the pivot force. So we have a similar situation to location 1, we don't know mass 2 or the pivot force - so we will have to solve a system of equations. | + | * ** Problem with Location 3 (Center of plank)** At the center of the plank, the torques are due to the weight of the two masses and the pivot force. So we have a similar situation to location 1, we don't know mass 1 or the pivot force - so we will have to solve a system of equations. |
== Let's use Location 4 (Pivot location) == | == Let's use Location 4 (Pivot location) == | ||
- | In this case the pivot force is not included in the analysis and the only unknown is mass 2. We can perform a torque analysis around this location noticing that the weight of the plank and mass 1 will contribute to an out-of-the-page torque (positive torque) while mass 2 will contribute to an into-the-page torque (negative torque). [[183_notes: | + | In this case the pivot force is not included in the analysis and the only unknown is mass 1. We can perform a torque analysis around this location noticing that the weight of the plank and mass 1 will contribute to an out-of-the-page torque (positive torque) while mass 2 will contribute to an into-the-page torque (negative torque). [[183_notes: |
$$\vec{\tau}_{net} = 0 \longrightarrow \tau_{net, | $$\vec{\tau}_{net} = 0 \longrightarrow \tau_{net, | ||
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Here, you obtain $m_1$ without any additional work. So, to summarize, every pivot location can be used -- it's just that some make the work a little easier than others. You would not have been wrong to choose any of the other locations. | Here, you obtain $m_1$ without any additional work. So, to summarize, every pivot location can be used -- it's just that some make the work a little easier than others. You would not have been wrong to choose any of the other locations. | ||
+ | ===== Examples ===== | ||
+ | |||
+ | * [[: |