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Nächste Überarbeitung 		Vorhergehende Überarbeitung
electrical_engineering_2:task_1.2.1_with_calc [2022/06/27 21:50]
tfischer 		electrical_engineering_2:task_1.2.1_with_calc [2023/03/15 13:20] (aktuell)
mexleadmin
Zeile 2: Zeile 2:
  
 <WRAP right> <WRAP right>
-{{elektrotechnik_1:kraefteadditiongeometriei.jpg?400}}+{{drawio>electrical_engineering_2:kraefteadditiongeometriei.svg}}
 </WRAP> </WRAP>
  
 Given is the arrangement of electric charges in the picture on the right. \\ Given is the arrangement of electric charges in the picture on the right. \\
 The following force effects result: \\ 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. Calculate the magnitude of the resulting force.
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 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: $\alpha_{0n} = atan(\frac{\Delta y}{\Delta x})$ \\
-$\alpha_{01} = atan(\frac{3}{1})= 1.249 = 71.6°$ \\ +$\alpha_{01} = \rm{atan}(\frac{3}{1})= 1.249 = 71.6°$ \\ 
-$\alpha_{02} = atan(\frac{4}{3})= 0.927 = 53.1°$ \\ +$\alpha_{02} = \rm{atan}(\frac{4}{3})= 0.927 = 53.1°$ \\ 
-$\alpha_{03} = atan(\frac{0}{3})= 0= 0°$ \\+$\alpha_{03} = \rm{atan}(\frac{0}{3})= 0= 0°$ \\
  
 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{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*} \end{align*}
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 <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"> <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*} \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*} \end{align*}
  \\  \\