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--- electrical_engineering_2:task_1.2.1_with_calc [2022/03/10 12:29]
tfischer
+++ electrical_engineering_2:task_1.2.1_with_calc [2023/03/15 13:20] (aktuell)
mexleadmin
@@ Zeile 1: / Zeile 1: @@
-<panel type="info" title="Task 1.2.1 Multiple Forces on a Charge I (exam task, ca 8% of a 6 minut 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%>
 <WRAP right>
-{{elektrotechnik_1:kraefteadditiongeometriei.jpg?400}}
+{{drawio>electrical_engineering_2:kraefteadditiongeometriei.svg}}
 </WRAP>
 Given is the arrangement of electric charges in the picture on the right. \\
 The following force effects result: \\
-$F_{01}=-5 N$ \\
+$F_{01}=-5 ~\rm{N}$ \\
-$F_{02}=-6 N$ \\
+$F_{02}=-6 ~\rm{N}$ \\
-$F_{03}=+3 N$
+$F_{03}=+3 ~\rm{N}$
 Calculate the magnitude of the resulting force.
 <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">
-  * Wie müssen die Kräfte vorbereitet werden, dass sie tatsächlich addiert werden können?
+  * How have the forces be prepared, in order to add them correctly?
 </collapse>
 <button size="xs" type="link" collapse="Loesung_5_2_1_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_5_2_1_1_Lösungsweg" collapsed="true">
 \begin{align*}
-F_0 &= |\vec{F_0}|  \quad \quad \text{   mit   }  \vec{F_0} = \left( \begin{matrix}{F_{x,0}}\\ {F_{y,0}} \end{matrix} \right) = \left( \begin{matrix}  \sum\limits_{n} F_{x,0n} \\ \sum\limits_{n} F_{y,0n}    \end{matrix} \right) \\
+F_0 &= |\vec{F_0}|  \quad \quad \text{   with   }  \vec{F_0} = \left( \begin{matrix}{F_{x,0}}\\ {F_{y,0}} \end{matrix} \right) = \left( \begin{matrix}  \sum\limits_{n} F_{x,0n} \\ \sum\limits_{n} F_{y,0n}    \end{matrix} \right) \\
 F_0 &= \sqrt{ \left(\sum\limits_{n} F_{x,0n} \right)^2 +  \left(\sum\limits_{n} F_{y,0n} \right)^2 } \\
 \end{align*}
-Die vorhandenen Kräfte müssen in Koordinaten zerlegt werden. Hier empfehlen sich die orthogonalen Koordinaten ($x$ und $y$). \\
+The forces have to be resolved into coordinates. Here, it is recommended to use an orthogonal coordinate system ($x$ and $y$). \\
-Das Koordinatensystem sei so ausgelegt, dass der Ursprung in $Q_0$ liegt mit der x-Achse in Richtung Q_3 und die y-Achse entsprechend rechtwinklig dazu. \\
+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. \\
-Zur Koordinatenzerlegung sind die Winkel $alpha_{0n}$ der Kräfte zur x-Achse notwendig. \\
+For the resolution of the coordinates it is necessary to get the angles $\alpha_{0n}$ of the forces with respect to the x-axis. \\
-Diese ergeben sich im gewählten Koordinatensystem aus den Koordinaten der Ladungen: $\alpha_{0n} = atan(\frac{\Delta y}{\Delta x})$ \\
+In the chosen coordinate system this leads to: $\alpha_{0n} = atan(\frac{\Delta y}{\Delta x})$ \\
-$\alpha_{01} = atan(\frac{3}{1})= 1.249 = 71.6°$ \\
+$\alpha_{01} = \rm{atan}(\frac{3}{1})= 1.249 = 71.6°$ \\
-$\alpha_{02} = atan(\frac{4}{3})= 0.927 = 53.1°$ \\
+$\alpha_{02} = \rm{atan}(\frac{4}{3})= 0.927 = 53.1°$ \\
-$\alpha_{03} = atan(\frac{0}{3})= 0= 0°$ \\
+$\alpha_{03} = \rm{atan}(\frac{0}{3})= 0= 0°$ \\
-Dann ergeben sich die zerlegten Kräfte zu: \\ \\
+Consequently, the resolved forces are: \\ \\
 \begin{align*}
-F_{x,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{mit } F_{x,0n} = F_{0n} \cdot sin(\alpha_{0n})  \\
+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} &= (-5N) \cdot sin(71.6°) + (-6N) \cdot sin(53.1°) + (+3N) \cdot sin(0°)  \\
+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 N  \\ \\
+F_{x,0} &= -2.18 ~\rm{N}  \\ \\
-F_{y,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{mit } F_{y,0n} = F_{0n} \cdot cos(\alpha_{0n})  \\
+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_{y,0} &= (-5N) \cdot cos(71.6°) + (-6N) \cdot cos(53.1°) + (+3N) \cdot cos(0°)  \\
+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_{y,0} &= -9.54 N  \\ \\
+F_{y,0} &= -9.54 ~\rm{N}  \\ \\
 \end{align*}
@@ Zeile 49: / Zeile 49: @@
 <button size="xs" type="link" collapse="Loesung_5_2_1_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_5_2_1_1_Endergebnis" collapsed="true">
 \begin{align*}
-F_0 &= \sqrt{ (-2.18 N)^2 +  (-9.54 N)^2 } = 9.79 N \rightarrow 9.8 N \\
+F_0 &= \sqrt{ (-2.18 ~\rm{N})^2 +  (-9.54 ~\rm{N})^2 } = 9.79 ~\rm{N} \rightarrow 9.8 ~\rm{N} \\
 \end{align*}
  \\