2 - 48. three forces act on the bracket. determine the magnitude and direction $\theta$ of $mathbf{f}_{1}$…

Explanation:

Step1: Resolve forces in x - direction

The resultant force in the x - direction $F_{Rx}=800\sin60^{\circ}$, and the sum of the x - components of the individual forces is $F_{1}\sin(60^{\circ}+\theta)-\frac{12}{13}(180)$. So, $800\sin60^{\circ}=F_{1}\sin(60^{\circ}+\theta)-\frac{12}{13}(180)$.

Step2: Resolve forces in y - direction

Since the resultant force is along the $x'$ axis, the sum of the y - components of the forces is zero. The y - component of $F_{2}$ is $200$, the y - component of $F_{3}$ is $\frac{5}{13}(180)$ and the y - component of $F_{1}$ is $F_{1}\cos(60^{\circ}+\theta)$. So, $200 + \frac{5}{13}(180)-F_{1}\cos(60^{\circ}+\theta)=0$, which gives $F_{1}\cos(60^{\circ}+\theta)=200+\frac{5}{13}(180)$.

Step3: Use the identity $\tan(A + B)=\frac{\sin(A + B)}{\cos(A + B)}$

We know that $\tan(60^{\circ}+\theta)=\frac{F_{1}\sin(60^{\circ}+\theta)}{F_{1}\cos(60^{\circ}+\theta)}$. From the x - direction equation $F_{1}\sin(60^{\circ}+\theta)=800\sin60^{\circ}+\frac{12}{13}(180)$ and from the y - direction equation $F_{1}\cos(60^{\circ}+\theta)=200+\frac{5}{13}(180)$. Calculate the right - hand side of the $\tan$ equation: [ \begin{align*} \tan(60^{\circ}+\theta)&=\frac{800\sin60^{\circ}+\frac{12}{13}(180)}{200+\frac{5}{13}(180)}\ &=\frac{800\times\frac{\sqrt{3}}{2}+\frac{12\times180}{13}}{200+\frac{5\times180}{13}}\ &=\frac{400\sqrt{3}+\frac{2160}{13}}{200+\frac{900}{13}}\ \end{align*} ] First, find a common denominator: [ \begin{align*} &=\frac{\frac{5200\sqrt{3}+ 2160}{13}}{\frac{2600 + 900}{13}}\ &=\frac{5200\sqrt{3}+2160}{3500} \end{align*} ] Then, $60^{\circ}+\theta=\arctan\left(\frac{5200\sqrt{3}+2160}{3500}\right)$. Solve for $\theta$. Also, from $F_{1}\cos(60^{\circ}+\theta)=200+\frac{5}{13}(180)$, we can find $F_{1}=\frac{200+\frac{5}{13}(180)}{\cos(60^{\circ}+\theta)}$. [ \begin{align*} 200+\frac{5}{13}(180)&=200 + \frac{900}{13}=\frac{2600+900}{13}=\frac{3500}{13}\ \end{align*} ] After calculating $\theta$ from the $\tan$ relation and substituting back into the equation for $F_{1}$, we get: [ \begin{align*} F_{1}&=\frac{\frac{3500}{13}}{\cos(60^{\circ}+\theta)}\ \end{align*} ] Let's calculate step - by - step: [ \begin{align*} 800\sin60^{\circ}&=800\times\frac{\sqrt{3}}{2}=400\sqrt{3}\approx400\times1.732 = 692.82\ \frac{12}{13}(180)&=\frac{2160}{13}\approx166.15\ 200+\frac{5}{13}(180)&=200+\frac{900}{13}=\frac{2600 + 900}{13}=\frac{3500}{13}\approx269.23\ \tan(60^{\circ}+\theta)&=\frac{400\sqrt{3}+\frac{2160}{13}}{\frac{3500}{13}}=\frac{5200\sqrt{3}+2160}{3500}\approx\frac{5200\times1.732+2160}{3500}=\frac{9006.4+2160}{3500}=\frac{11166.4}{3500}\approx3.19\ 60^{\circ}+\theta&=\arctan(3.19)\approx72.6^{\circ}\ \theta&=72.6^{\circ}- 60^{\circ}=12.6^{\circ}\ \cos(60^{\circ}+\theta)&=\cos(72.6^{\circ})\approx0.30\ F_{1}&=\frac{\frac{3500}{13}}{\cos(72.6^{\circ})}=\frac{269.23}{0.30}\approx897.43\ N \end{align*} ]

Answer:

The magnitude of $F_{1}\approx897\ N$ and the direction $\theta\approx12.6^{\circ}$