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+52√2kN−2√3kN+1kN52√2kN−2kN−√3kN)=(4,07143229079kN−0,19651690164kN)
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.