3-31 할리데이 11판 솔루션 일반물리학

$$\begin{cases} d_1&=6.00\ut{m},\theta_{1}=-135\degree\\ \vec d_2&=4.00\ut{m}\i\\ d_3&=8.00\ut{m},\theta_{3}=60.0\degree\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} d_{1x}&=6\cos\(-135\degree\)\\ &=-3\sqrt2\ut{m}\\ &\approx -4.242640687119286\ut{m}\\ &\approx -4.24\ut{m} \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} d_{1y}&=6\sin\(-135\degree\)\\ &=-3\sqrt2\ut{m}\\ &\approx -4.242640687119286\ut{m}\\ &\approx -4.24\ut{m} \end{aligned}$$ $$\ab{c}$$ $$d_{2x}=4.00\ut{m}$$ $$\ab{d}$$ $$d_{2y}=0$$ $$\ab{e}$$ $$\begin{aligned} d_{3x}&=8\cos\(60\degree\)\\ &=4\ut{m}\\ &=4.00\ut{m}\\ \end{aligned}$$ $$\ab{f}$$ $$\begin{aligned} d_{3y}&=8\sin\(60\degree\)\\ &=4\sqrt3\ut{m}\\ &\approx 6.928203230275509\ut{m}\\ &\approx 6.93\ut{m}\\ \end{aligned}$$ $$\ab{g,h,i,j}$$ $$\begin{aligned} \Sigma \vec d&=\vec d_1+\vec d_2+\vec d_3\\ &=\(-3\sqrt2\i-3\sqrt2\j\)+\(4\i\)+\(4\i+4\sqrt{3}\j\)\\ &=\(8-3 \sqrt{2}\)\i+\(4 \sqrt{3}-3 \sqrt{2}\)\j\\ \end{aligned}$$ $$\ab{g}$$ $$\begin{aligned} d_x&=\(8-3 \sqrt{2}\)\ut{m}\\ &\approx 3.757359312880714\ut{m}\\ &\approx 3.76\ut{m}\\ \end{aligned}$$ $$\ab{h}$$ $$\begin{aligned} d_y&=\(4 \sqrt{3}-3 \sqrt{2}\)\ut{m}\\ &\approx 2.685562543156223\ut{m}\\ &\approx 2.69\ut{m}\\ \end{aligned}$$ $$\ab{i}$$ $$\begin{aligned} d&=\sqrt{\(8-3 \sqrt{2}\)^2+\(4 \sqrt{3}-3 \sqrt{2}\)^2}\\ &=2 \sqrt{37-6 \sqrt{2} \left(2+\sqrt{3}\right)}\ut{m}\\ &\approx 4.618440773604783\ut{m}\\ &\approx 4.62\ut{m}\\ \end{aligned}$$ $$\ab{j}$$ $$\begin{aligned} \theta_d&=\tan^{-1}\frac{4 \sqrt{3}-3 \sqrt{2}}{8-3 \sqrt{2}}\\ &\approx 0.6205550848427059\ut{rad}\\ &\approx 0.621\ut{rad}\\ \end{aligned}$$ $$\ab{k,l}$$ $$\begin{aligned} \Ans &=\vec d_{\text{Back}}= -\vec d\\ \end{aligned}$$ $$\ab{k}$$ $$\begin{aligned} d_B&=d\\ &=2 \sqrt{37-6 \sqrt{2} \left(2+\sqrt{3}\right)}\ut{m}\\ &\approx 4.618440773604783\ut{m}\\ &\approx 4.62\ut{m}\\ \end{aligned}$$ $$\ab{l}$$ $$\begin{aligned} \theta_{d_2}&=\theta_d+\pi\\ &=\pi+\tan^{-1}\frac{4 \sqrt{3}-3 \sqrt{2}}{8-3 \sqrt{2}}\\ &\approx -2.521037568747087\ut{rad}\\ &\approx -2.52\ut{rad}\\ \end{aligned}$$