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183_notes:static_eq [2018/05/30 18:23] – hallstein | 183_notes:static_eq [2021/11/15 17:25] (current) – [Lecture Video] pwirving | ||
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===== Static Equilibrium ===== | ===== Static Equilibrium ===== | ||
- | While you are beginning to learn about [[183_notes: | + | While you are beginning to learn about [[183_notes: |
- | ==== Lecture Video ===== | ||
- | {{youtube> | ||
- | ==== Defining Static Equilibrium ==== | ||
- | We define a system to be in static equilibrium if it is not moving. That seems obvious, but there' | + | |
+ | ===== Defining Static Equilibrium ===== | ||
+ | |||
+ | **We define a system to be in static equilibrium if it is not moving**. That seems obvious, but there are two specific conditions on the motion that has to be satisfied: | ||
- The system cannot be translating (moving up/down, left/right, etc.). | - The system cannot be translating (moving up/down, left/right, etc.). | ||
Line 24: | Line 24: | ||
∑Fx=0∑Fy=0 | ∑Fx=0∑Fy=0 | ||
- | If the sum of all the forces is zero then static equilibrium is possible but **not** guaranteed. | + | //If the sum of all the forces is zero then static equilibrium is possible but **not** guaranteed.// |
- | [{{ 183_notes:statics_bar.png?250|A bar with two identically sized forces acting on it.}}] | + | [{{ 183_notes:week12_staticpivotforces.png?350|A bar with two identically sized forces acting on it.}}] |
- | === Why torque matters | + | ==== Why Torque Matters ==== |
- | Consider the simple system of the bar to the right. Two equal sized forces are acting on the bar in opposite directions. In this case, the sum of the forces in each coordinate direction (namely, the vertical direction) is zero. Hence, this situation satisfies the first condition for static equilibrium. However, you can probably easily see that with these forces applied, the bar will rotate (counter-clockwise). So, in this case, we violate condition 2 above and the bar is not in static equilibrium (because it is going to rotate). | + | Consider the simple system of the bar to the right. Two equal-sized forces are acting on the bar in opposite directions. In this case, the sum of the forces in each coordinate direction (namely, the vertical direction) is zero. Hence, this situation satisfies the first condition for static equilibrium. However, you can probably easily see that with these forces applied, the bar will rotate (counter-clockwise). So, in this case, we violate condition 2 above and the bar is not in static equilibrium (because it is going to rotate). |
- | The forces apply [[183_notes: | + | The forces apply [[183_notes: |
So, to have a static equilibrium situation it must be that we have both: | So, to have a static equilibrium situation it must be that we have both: |