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183_notes:torquediagram [2016/03/14 06:28] – [Torque Diagrams] pwirving | 183_notes:torquediagram [2021/06/04 03:54] – [A Balanced Situation] stumptyl | ||
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===== Torque Diagrams ===== | ===== Torque Diagrams ===== | ||
- | To investigate situations in static equilibrium more throughly, you can make use of an extended free-body diagram that shows the "point of application" | + | To investigate situations in static equilibrium more thoroughly, you can make use of an extended free-body diagram that shows the "point of application" |
- | In these notes, you will read through an example where we will apply the ideas that the net force and net torque are zero and make use of the torque diagram to construct a mathematical representation of the problem. | + | **In these notes, you will read through an example where we will apply the ideas that the net force and net torque are zero and make use of the torque diagram to construct a mathematical representation of the problem.** |
==== Lecture Video ==== | ==== Lecture Video ==== | ||
- | Forthcoming... | + | {{youtube> |
- | ==== A balanced situation | + | ===== A Balanced Situation ===== |
- | Consider the situation in the figure below. A uniform plank of mass $m_p$ and length $3d$ is placed on a pivot as shown below. A mass $m_2$ is placed on the far left edge of the plank. To balance the whole set up, you want to place a mass $m_1$ on the far right edge. You need to figure out how large $m_1$ should be compared to the other masses in the set up to determine what mass to place on the far left edge (As shown). | + | Consider the situation in the figure below. A uniform plank of mass $m_p$ and length $3d$ is placed on a pivot as shown below. A mass $m_2$ is placed on the far left edge of the plank. To balance the whole setup, you want to place a mass $m_1$ on the far right edge. You need to figure out how large $m_1$ should be compared to the other masses in the setup to determine what mass to place on the far left edge (As shown). |
- | {{ 183_notes:balance.png?400 }} | + | {{ 183_notes:week12_torquediagrams1.png?400 }} |
- | Because you are designing for no motion, you want the entire system to be in static equilibrium. We can do a typical force analysis to see if the yields anything helpful, but first let's draw the extended free-body diagram -- the torque diagram. Below, you will see a diagram that labels the four forces acting on the system: the gravitational force on each of the three objects, which points downward, and the force due to the pivot, which points upward. In addition, a torque diagram labels all the lever arm distances (given a specific choice of pivot point, red dot in this case). | + | //Because you are designing for no motion, you want the entire system to be in static equilibrium.// We can do a typical force analysis to see if this yields anything helpful, but first, let's draw the extended free-body diagram -- the torque diagram. Below, you will see a diagram that labels the four forces acting on the system: the gravitational force on each of the three objects, which points downward, and the force due to the pivot, which points upward. In addition, a torque diagram labels all the lever arm distances (given a specific choice of the pivot point, red dot in this case). |
- | Notice the forces are positioned at the location where they act. For each of the three objects, we treat the gravitational force as acting at the center of masses (in this case at their centers because they are uniform density). For objects of non-uniform density, [[https:// | + | Notice the forces are positioned at the location where they act. For each of the three objects, we treat the gravitational force as acting at the center of masses (in this case at their centers because they are of uniform density). For objects of non-uniform density, [[https:// |
- | {{ 183_notes:torquediagram.png?400 }} | + | {{ 183_notes:week12_torquediagrams2.png?400 }} |
- | === An analysis | + | ==== An Analysis |
- | In this case, you know you want the system to be in static equilibrium, | + | In this case, you know you want the system to be in static equilibrium, |
$$\vec{F}_{net} = 0 \longrightarrow F_{net,x} = 0\; | $$\vec{F}_{net} = 0 \longrightarrow F_{net,x} = 0\; | ||
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$$F_p = m_p g + m_2 g + m_1 g$$ | $$F_p = m_p g + m_2 g + m_1 g$$ | ||
- | This makes sense! The pivot force must be large enough to support the weight of all the objects. Once you find $m_1$, you can determine the size of the picot force. | + | This makes sense! The pivot force must be large enough to support the weight of all the objects. Once you find $m_1$, you can determine the size of the pivot force. |
=== An analysis of the torques === | === An analysis of the torques === | ||
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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. | + | * **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 |
- | * ** 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., |
- | * ** 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 no 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 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 |
$$\vec{\tau}_{net} = 0 \longrightarrow \tau_{net, | $$\vec{\tau}_{net} = 0 \longrightarrow \tau_{net, | ||
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$$m_1 = 2m_2 + \frac{m_p}{2}$$ | $$m_1 = 2m_2 + \frac{m_p}{2}$$ | ||
- | Here, you obtain $m_1$ without any additional work. So, to summarize, every pivot location can be used in -- 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 ===== | ||
+ | |||
+ | * [[: |