3-22 할리데이 10판 솔루션 일반물리학

$$\begin{aligned} \vec E:& (6.00 \ut{m}, + 0.900\ut{rad})\\ \vec F:& (5.00 \ut{m}, -75.0^\circ)\\ \vec G:& (4.00 \ut{m},+ 1.20\ut{rad})\\ \vec H:& (6.00 \ut{m}, -210^\circ)\\ \end{aligned}$$

$$\begin{aligned} \vec E &= (6\cos0.9)\hat i + (6\sin0.9)\hat j\\ \vec F &= (5\cos-75^\circ)\hat i + (5\sin-75^\circ)\hat j\\ &=\(\frac{5}{2} \sqrt{2-\sqrt{3}}\)\hat i + \(-\frac{5}{2} \sqrt{2+\sqrt{3}}\)\hat j\\ \vec G &= (4\cos1.2)\hat i + (4\sin1.2)\hat j\\ \vec H &= (6\cos-210^\circ)\hat i + (6\sin-210^\circ)\hat j\\ &=\(-3\sqrt3\)\hat i + 3\hat j \end{aligned}$$
(a) $\vec r=?$ $$\begin{aligned} \Ans =& \vec E + \vec F + \vec G + \vec H\\ =& \(6\cos0.9+\frac{5}{2} \sqrt{2-\sqrt{3}}+4\cos1.2-3\sqrt3\)\hat i\\ &+\(6\sin0.9-\frac{5}{2} \sqrt{2+\sqrt{3}}+4\sin1.2+3\)\hat j\\ \approx& 1.28\hat i + 6.60\hat j\\ \end{aligned}$$
(b) $r=?$ $$\begin{aligned} \Ans =& \vec E + \vec F + \vec G + \vec H\\ =& \sqrt{\(6\cos0.9+\frac{5}{2} \sqrt{2-\sqrt{3}}+4\cos1.2-3\sqrt3\)^2+\(6\sin0.9-\frac{5}{2} \sqrt{2+\sqrt{3}}+4\sin1.2+3\)^2}\\ \approx& 6.72\ut{m}\\ \end{aligned}$$
(c) $\theta_r=? \ \text{in} \ut{^\circ}$ $$\begin{aligned} \theta_r=&\tan^{-1}\(\frac{6\sin0.9-\frac{5}{2} \sqrt{2+\sqrt{3}}+4\sin1.2+3}{6\cos0.9+\frac{5}{2} \sqrt{2-\sqrt{3}}+4\cos1.2-3\sqrt3}\)\\ \approx&79.1^\circ \end{aligned}$$
(d) $\theta_r=? \ \text{in} \ut{rad}$ $$\begin{aligned} \theta_r=&\tan^{-1}\(\frac{6\sin0.9-\frac{5}{2} \sqrt{2+\sqrt{3}}+4\sin1.2+3}{6\cos0.9+\frac{5}{2} \sqrt{2-\sqrt{3}}+4\cos1.2-3\sqrt3}\)\\ \approx&1.38\ut{rad} \end{aligned}$$