Three concurrent and coplanar

forces in equilibrium is as shown

in the figure.

a. What is the value of the angle

θ?

b. What is the value of the angle

α?

c. What is the value of the

horizontal component of the

94.5 KN force?

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The complete solution of the question is provided in the images. If you find the answer helpful then please do UPVOTE u200b Have a nice day ahead. Happy ing These three concurrent and coplanar forces are in equilibrium.Hence their re solved x \& y components are also in equilibrium.\begin{array}{l} \therefore \quad \sum F_{x}=0 \\60+50 \cos \theta=94.5 \cos \alpha-\text { (1) } \\\therefore \quad \sum F_{y}=0 \\50 \sin \theta=94.5 \sin \alpha \text { (2) }\end{array}Now squaring Equation (1) 2 (2) and then adding we get\begin{aligned}& 94.5^{2} \cos ^{2} \alpha+94.5^{2} \sin ^{2} \alpha=(60+50 \cos \theta)^{2}+50^{2} \sin ^{2} \theta \\\Rightarrow & 94.5^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)=60^{2}+50^{2} \cos ^{2} \theta+6000 \cos \theta+50^{2} \sin ^{2} \theta \\\Rightarrow & 94.5^{2}=60^{2}+50^{2}+6000 \cos \theta \\\Rightarrow & \theta=61.85^{\circ}\end{aligned}putting \theta value in Equation (2)\begin{aligned}& 50 \sin 61.85^{\circ}=94.5 \sin \alpha \\\Rightarrow & \alpha=27.80^{\circ}\end{aligned}Horizontal component of the 94.5 \mathrm{KN} Force\begin{array}{l}=94.5 \cos \alpha \\=94.5 \cos 27.80^{\circ} \\=83.59 \mathrm{kN} \text { (inthe -ve x direction) }\end{array}u200bu200b     ...