183_notes:static_eq

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183_notes:static_eq [2021/05/08 18:59] – [Defining Static Equilibrium] stumptyl183_notes:static_eq [2021/11/15 17:19] – [Lecture Video] pwirving
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 ==== Lecture Video ===== ==== Lecture Video =====
  
-{{youtube>RzA8uywxjzc?large}}+{{youtube>itsnuo1DnRw?large}} 
 + 
  
 ===== Defining Static Equilibrium ===== ===== Defining Static Equilibrium =====
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 $$\sum F_x = 0 \qquad \sum F_y = 0$$ $$\sum F_x = 0 \qquad \sum F_y = 0$$
  
-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:torque|torques]] to the bar if we consider the center of the bar to be the rotation point. In that case, both torques point out of the page ([[183_notes:torque|Torque directions are defined by the right-hand rule]]) and are (roughly) the same size (as long as they are the same distance from the center). Remember that to calculate a torque, you need to choose a location about which you will consider rotation (more on this later).+The forces apply [[183_notes:torque|torques]] to the bar //__if we consider the center of the bar to be the rotation point.__// In that case, both torques point out of the page ([[183_notes:torque|Torque directions are defined by the right-hand rule]]) and are (roughly) the same size (as long as they are the same distance from the center). Remember that to calculate a torque, you need to choose a location about which you will consider rotation (more on this later).
  
 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:
  • 183_notes/static_eq.txt
  • Last modified: 2021/11/15 17:25
  • by pwirving