10판/3. 벡터

3-20 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 7. 20:42

{r=4.8j^a:(7.8,50)a+b=r\begin{cases} \vec r = 4.8\hat j\\ \vec a:(7.8,50^\circ)\\ \vec a+\vec b=\vec r\\ \end{cases} a=(7.8cos50)i^+(7.8sin50)j^ a= (7.8\cos50^\circ)\hat i + (7.8\sin50^\circ)\hat j b=ra=(7.8cos50)i^+(4.87.8sin50)j^ \begin{aligned} \vec b &= \vec r - \vec a\\ &= (-7.8\cos50^\circ)\hat i + (4.8-7.8\sin50^\circ)\hat j\\ \end{aligned}
(a)b=?b=? b=(7.8cos50)2+(4.87.8sin50)2=35233208cos(40)5.2[m] \begin{aligned} b&=\sqrt{(-7.8\cos50^\circ)^2+(4.8-7.8\sin50^\circ)^2}\\ &=\frac{3}{5} \sqrt{233-208 \cos(40 {}^{\circ})}\\ &\approx 5.2\ut{m}\\ \end{aligned}
(b)θb=?\theta_b=? θb=tan1(4.87.8sin507.8cos50)=tan1(cot40813csc40)π2.9[rad] \begin{aligned} \theta_b&=\tan^{-1}\(\frac{4.8-7.8\sin50^\circ}{-7.8\cos50^\circ}\)\\ &=\tan^{-1}\left(\cot40^\circ-\frac{8}{13} \csc40^\circ\right)-\pi \\ &\approx -2.9\ut{rad} \end{aligned}