Question Three concurrent and coplanar forces in equilibrium is shown in the figure. Which of the following most nearly gives the horizontal component of the 94.5 kN Force 50 KN 0 a 60 KN 94.5 KN

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Transcribed Image Text: Three concurrent and coplanar forces in equilibrium is shown in the figure. Which of the following most nearly gives the horizontal component of the 94.5 kN Force 50 KN 0 a 60 KN 94.5 KN

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Sam:-we apply Egulibirium Equition.\begin{array}{c}\sum F_{x}=0, \quad \sum F_{y}=0 \\94.5 \cos \alpha-60-50 \cos \theta=0 \\50 \sin \theta-94.5 \sin \alpha=0 \\\sin \theta=\frac{94.5}{50} \sin \alpha \\94.5 \sin \alpha 50\end{array}find \cos \theta\begin{array}{l} A C^{2}=A B^{2}+B C^{2} \\50^{2}=(94.5 \sin \alpha)^{2}+B C^{2} \\\sqrt{(50)^{2}-(94.5 \sin \alpha)^{2}}=B C \\\cos \theta 0=\frac{\sqrt{(50)^{2}-(94 \cdot 5 \sin \alpha)^{2}}}{50}\end{array}put this value of \cos \theta in \operatorname{eqn}(i)Thank you \begin{array}{l}\Rightarrow \quad 94.5 \cos \alpha-60-50 \times \frac{\sqrt{(50)^{2}}-(94.5 \sin \alpha)^{2}}{50}=0 \\\Rightarrow 94 \cdot 5 \cos 2-60=\sqrt{(50)^{2}-(94.5 \sin 2)^{2}} \\\Rightarrow(94.5 \cos \alpha-60)^{2}=(50)^{2}-(94.5 \sin \alpha)^{2} \\\Rightarrow(94 \cdot 5 \cos 2)^{2}+(60)^{2}-2 \times 94 \cdot 5 \cos 2 \times 60=(50)^{2}-(94.5 \sin \alpha)^{2} \\\Rightarrow(94.5 \cos \alpha)^{2}+(94.5 \sin \alpha)^{2}-2 \times 94.5 \cos 2 \times 60=(50)^{2}-(60)^{2} \\\Rightarrow(94.5)^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)-11,340 \cos \alpha=(50)^{2}-(60)^{2} \\\Rightarrow \quad-11340 \cos \alpha=(50)^{2}-(60)^{2}-(94.5)^{2} \\-11340 \cos \alpha=-10,030.25 \\\cos \alpha=0.8845 \\\alpha=\$ 5227.809 \\\end{array}\Rightarrow horizontal component of forle 94.5 isF_{H}=94.5 \cos \alpha=83.585 ...