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Hallo,

hier die hoffentlich richtige Lösung:

\vec{F_1}=3\;kN  \left(\begin{array}{c}  \cos(0\ensuremath{^\circ}})\\  \sin(0\ensuremath{^\circ}})\\  \end{array}\right)  =3\;kN  \left(\begin{array}{c}  1\\  0\\  \end{array}\right)

\vec{F_2}=5\;kN  \left(\begin{array}{c}  \cos(45\ensuremath{^\circ}})\\  \sin(45\ensuremath{^\circ}})\\  \end{array}\right)

\vec{F_3}=4\;kN  \left(\begin{array}{c}  -\cos(30\ensuremath{^\circ}})\\  -\sin(30\ensuremath{^\circ}})\\  \end{array}\right)  =-4\;kN  \left(\begin{array}{c}  \cos(30\ensuremath{^\circ}})\\  \sin(30\ensuremath{^\circ}})\\  \end{array}\right)

\vec{F_4}=2\;kN  \left(\begin{array}{c}  \cos(60\ensuremath{^\circ}})\\  -\sin(60\ensuremath{^\circ}})\\  \end{array}\right)

F=F1+F2+F3+F4

\vec{F}=  \left(\begin{array}{c}  3\;kN+5\;kN\cos(45\ensuremath{^\circ}})-4\;kN\cos(30\ensuremath{^\circ}})+2\;kNcos(60\ensuremath{^\circ}})\\  0\;kN+5\;kN\sin(45\ensuremath{^\circ}})-4\;kN\sin(30\ensuremath{^\circ}})-2\;kN\sin(60\ensuremath{^\circ}})\\  \end{array}\right)

\sin(45\ensuremath{^\circ}})=\cos(45\ensuremath{^\circ}})=\frac{\sqrt{2}}{2}

\sin(30\ensuremath{^\circ}})=\frac{1}{2}\quad\cos(30\ensuremath{^\circ}})=\frac{\sqrt{3}}{2}\quad\sin(60\ensuremath{^\circ}})=\frac{\sqrt{3}}{2}\quad\cos(60\ensuremath{^\circ}})=\frac{1}{2}

F=(3kN+522kN23kN+1kN522kN2kN3kN)=(4,07143229079kN0,19651690164kN)

 

R=|F|=4,071432290792+(0,19651690164)2=4,07617219842kN

θ=arctan(0,196516901644,07143229079)+360\ensuremath=2\ensuremath,76336594484+360\ensuremath=357\ensuremath,236634055

Die rerultierende R hat die Größe 4,076kNmit einem Winkel θ=357\ensuremath,24

 

S. aus. H.

07.05.2014
06.05.2014
 #2
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+3
06.05.2014
05.05.2014

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