Chapter 2 - Force Vectors - Section 2.4 - Addition of a System of Coplanar Forces - Fundamental Problems - Page 39: 10

$F\approx236.09\;N$ $\theta\approx31.76^\circ$

First, we resolve each force into its $x$ and $y$ components by using the parallelogram law: Let $F_{1x}$ and $F_{1y}$ are the $x$ and $y$ components of $\vec {F_1}$ force, in which $F_1=600\;N$. Thus we have $F_{1x}=600\cos45^\circ\;N\approx 424.26\;N$ $F_{1y}=-600\sin45^\circ\;N\approx -424.26\;N$ Let $F_{2x}$ and $F_{2y}$ are the $x$ and $y$ components of $\vec {F_2}$ force, in which $F_2=325\;N$. Using proportional parts of similar triangles and considering the directions, we have $\frac{F_{2x}}{325\;N}=\frac{5}{13}$ or, $F_{2x}=325\;N\Big(\frac{5}{13}\Big)=125\;N$ Similarly, $\frac{F_{2y}}{325\;N}=\frac{12}{13}$ or, $F_{2y}=325\;N\Big(\frac{12}{13}\Big)=300\;N$ Let $F_{x}$ and $F_{y}$ are the $x$ and $y$ components of $\vec {F}$ force. Thus we have $F_{x}=F\cos\theta\;N$ $F_{y}=F\sin\theta\;N$ Second, using scalar notation and indicating the positive directions of components along the $x$ and $y$ axes with symbolic arrows, we sum these components algebraically: Summing the x components, we have $\xrightarrow{+} (F_R)_x=\sum F_x$ or, $(F_R)_x=(424.26+125+F\cos\theta)\;N$ or, $(F_R)_x=(549.26+F\cos\theta)\;N$ Summing the y components yields $+\uparrow (F_R)_y=\sum F_y$ or, $(F_R)_y=(-424.26+300+F\sin\theta)\;N$ or, $(F_R)_y=(-124.26+F\sin\theta)\;N$ The resultant force $\vec F_R$ acting on the bracket is to be $F_R=750\;N$ directed along the positive x axis. Therefore the $x$ and $y$ components of $\vec F_R$ are $(F_R)_x=750\;N$ and $(F_R)_y=0\;N$ respectively. Thus, $750=549.26+F\cos\theta$ or, $F\cos\theta=200.74$ .................$(1)$ and, $0=-124.26+F\sin\theta$ or, $F\sin\theta=124.26$ .................$(2)$ $\therefore$ $\tan\theta=\frac{124.26}{200.74}$ $\implies\;\theta=\tan^{-1}\Big(\frac{124.26}{200.74}\Big)\approx 31.76^\circ$ Substituting $\theta=31.76^\circ$ in eq. $1$, we get $F\approx236.09\;N$