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electrical_engineering_2:task_1.2.1_with_calc [2023/03/15 13:12] – [Bearbeiten - Panel] mexleadminelectrical_engineering_2:task_1.2.1_with_calc [2024/06/25 23:11] (aktuell) – [Bearbeiten - Panel] mexleadmin
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-<panel type="info" title="Task 1.2.1 Multiple Forces on a Charge I (exam task, ca 8% of a 60 minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Task 1.2.1 Multiple Forces on a Charge I (exam task, ca 8% of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
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 <button size="xs" type="link" collapse="Loesung_5_2_1_1_Tipps">{{icon>eye}} Tips for the Solution</button><collapse id="Loesung_5_2_1_1_Tipps" collapsed="true"> <button size="xs" type="link" collapse="Loesung_5_2_1_1_Tipps">{{icon>eye}} Tips for the Solution</button><collapse id="Loesung_5_2_1_1_Tipps" collapsed="true">
-  * How have the forces be prepared, in order to add them correctly? +  * How have the forces be prepared, to add them correctly? 
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 The forces have to be resolved into coordinates. Here, it is recommended to use an orthogonal coordinate system ($x$ and $y$). \\ The forces have to be resolved into coordinates. Here, it is recommended to use an orthogonal coordinate system ($x$ and $y$). \\
-The coordinate system  shall be in such a way, that the origin lays in $Q_0$, the x-axis is directed towards $Q_3$ and the y-axis is orthogonal to it. \\ +The coordinate system shall be in such a way, that the origin lies in $Q_0$, the x-axis is directed towards $Q_3$ and the y-axis is orthogonal to it. \\ 
-For the resolution of the coordinates it is necessary to get the angles $\alpha_{0n}$ of the forces with respect to the x-axis. \\ +For the resolution of the coordinates, it is necessary to get the angles $\alpha_{0n}$ of the forces with respect to the x-axis. \\ 
-In the chosen coordinate system this leads to: $\alpha_{0n} = atan(\frac{\Delta y}{\Delta x})$ \\ +In the chosen coordinate system this leads to: $\alpha_{0n} = \arctan(\frac{\Delta y}{\Delta x})$ \\ 
-$\alpha_{01} = \rm{atan}(\frac{3}{1})= 1.249 = 71.6°$ \\ +$\alpha_{01} = \arctan(\frac{3}{1})= 1.249 = 71.6°$ \\ 
-$\alpha_{02} = \rm{atan}(\frac{4}{3})= 0.927 = 53.1°$ \\ +$\alpha_{02} = \arctan(\frac{4}{3})= 0.927 = 53.1°$ \\ 
-$\alpha_{03} = \rm{atan}(\frac{0}{3})= 0= 0°$ \\+$\alpha_{03} = \arctan(\frac{0}{3})= 0= 0°$ \\
  
 Consequently, the resolved forces are: \\ \\ Consequently, the resolved forces are: \\ \\
  
 \begin{align*} \begin{align*}
-F_{x,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{with } F_{x,0n} = F_{0n} \cdot \rm{sin}(\alpha_{0n})  \\  
-F_{x,0} &= (-5~\rm{N}) \cdot \rm{sin}(71.6°) + (-6~\rm{N}) \cdot \rm{sin}(53.1°) + (+3~\rm{N}) \cdot \rm{sin}(0°)  \\  
-F_{x,0} &= -2.18 ~\rm{N}  \\ \\ 
  
-F_{y,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{with } F_{y,0n} = F_{0n} \cdot \rm{cos}(\alpha_{0n})  \\  +F_{x,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{with } F_{x,0n} = F_{0n} \cdot \cos(\alpha_{0n})  \\  
-F_{y,0} &= (-5~\rm{N}) \cdot \rm{cos}(71.6°) + (-6~\rm{N}) \cdot \rm{cos}(53.1°) + (+3~\rm{N}) \cdot cos(0°)  \\  +F_{x,0} &= (-5~\rm{N}) \cdot \cos(71.6°) + (-6~\rm{N}) \cdot \cos(53.1°) + (+3~\rm{N}) \cdot \cos(0°)  \\  
-F_{y,0} &= -9.54 ~\rm{N}  \\ \\+F_{x,0} &= -9.54 ~\rm{N}  \\ \\ 
 + 
 +F_{y,0} &= F_{y,01} + F_{y,02} + F_{y,03} && | \quad \text{with } F_{y,0n} = F_{0n} \cdot \sin(\alpha_{0n})  \\  
 +F_{y,0} &= (-5~\rm{N}) \cdot \sin(71.6°) + (-6~\rm{N}) \cdot \sin(53.1°) + (+3~\rm{N}) \cdot \sin(0°)  \\  
 +F_{y,0} &= -2.18 ~\rm{N}  \\ \\
  
 \end{align*} \end{align*}
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 \begin{align*} \begin{align*}
-F_0 &= \sqrt{ (-2.18 ~\rm{N})^2 +  (-9.54 ~\rm{N})^2 } = 9.79 ~\rm{N} \rightarrow 9.8 ~\rm{N} \\+F_0 &= \sqrt{ (-9.54 ~\rm{N})^2 + (-2.18 ~\rm{N})^2} = 9.79 ~\rm{N} \rightarrow 9.8 ~\rm{N} \\
 \end{align*} \end{align*}
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