Force 1 \[ \begin{array}{l} A_{x}=20 \mathrm{~N} \cdot \cos \left(10^{\circ}\right)=19.7 \\ A_{y}=20 \mathrm{~N} \cdot \sin \left(10^{\circ}\right)=3.47 \end{array} \] force B \[ \begin{array}{l} B_{x}=55 \mathrm{~N}=\cos \left(70^{\circ}\right)=18.81 \\ B_{y}=55 \mathrm{~N}=\sin \left(70^{\circ}\right)=51.7 \end{array} \] Force (: \[ \begin{array}{l} C=-45 N \cdot \cos \left(33^{\circ}\right)=-37.7 \\ C_{4}=45 N \cdot \sin \left(33^{\circ}\right)=24.5 \end{array} \] Force \( D \) : \[ \begin{array}{l} D_{x}=-30 \mathrm{~N} \cdot \cos \left(80^{\circ}\right)=-5.21 \\ D_{y}=-30 \mathrm{~N} \cdot \sin \left(40^{\circ}\right)=-29.54 \end{array} \] \[ \begin{array}{l} f_{x}=19.7+18.81-37.7-5.21=-4.4 \\ f_{y}=3.47+51.7+24.5-29.54=50.13 \\ f_{r}=\sqrt{F_{x}^{2}+F_{y}^{2}} \end{array} \] Hhys: \[ \begin{aligned} f_{r} & =\sqrt{(-4.4)^{2}}+(50.13)^{2} \\ & =\sqrt{(-19.36)}+(2513.01) \end{aligned} \] Thus: \[ \begin{array}{l} \tan \theta=\left(\frac{50.13}{-4.4}\right) \\ \tan \theta=-11.39 \\ \theta=\operatorname{Arctan}-11.39 \\ \theta= \end{array} \]
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- \( A_x = 20 \, \text{N} \cdot \cos(10^\circ) = 19.7 \, \text{N} \) - \( A_y = 20 \, \text{N} \cdot \sin(10^\circ) = 3.47 \, \text{N} \) Show more…
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