10판/3. 벡터

3-29 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 9. 18:58

(a),(b) {w:(2.00[cm],60)v:(2.00[cm],90)i:(2.00[cm],120)h:(2.00[cm],90)\begin{cases} \vec w:(2.00\ut{cm},60^\circ)\\ \vec v:(2.00\ut{cm},90^\circ)\\ \vec i:(2.00\ut{cm},120^\circ)\\ \vec h:(2.00\ut{cm},90^\circ)\\ \end{cases} w=(2cos60)i^+(2sin60)j^=i^+3j^ \begin{aligned} \vec w =& (2\cos60^\circ)\hat i + (2\sin60^\circ)\hat j\\ =&\hat i + \sqrt3\hat j \end{aligned} v=h=(2cos90)i^+(2sin90)j^=2j^ \begin{aligned} \vec v =& \vec h\\ =& (2\cos90^\circ)\hat i + (2\sin90^\circ)\hat j\\ =&2\hat j\\ \end{aligned} i=(2cos120)i^+(2sin120)j^=i^+3j^ \begin{aligned} \vec i =& (2\cos120^\circ)\hat i + (2\sin120^\circ)\hat j\\ =&-\hat i + \sqrt3\hat j \end{aligned} A=w+v+i+h=(4+23)j^ \begin{aligned} \vec A =& \vec w + \vec v +\vec i + \vec h\\ =&\(4+2\sqrt3\)\hat j \end{aligned}
(a)A=?A=? A=4+23 A=4+2\sqrt3
(b)θA=?\theta_A=? θA=π2 \theta_A = \frac{\pi}{2}
(c),(d) {w:(2.00[cm],60)v:(2.00[cm],90)j:(2.00[cm],60)p:(2.00[cm],30)o:(2.00[cm],0)\begin{cases} \vec w:(2.00\ut{cm},60^\circ)\\ \vec v:(2.00\ut{cm},90^\circ)\\ \vec j:(2.00\ut{cm},60^\circ)\\ \vec p:(2.00\ut{cm},30^\circ)\\ \vec o:(2.00\ut{cm},0^\circ)\\ \end{cases} j=w=i^+3j^ \begin{aligned} \vec j =& \vec w\\ =&\hat i + \sqrt3\hat j \end{aligned} p=(2cos30)i^+(2sin30)j^=3i^+j^ \begin{aligned} \vec p =& (2\cos30^\circ)\hat i + (2\sin30^\circ)\hat j\\ =&\sqrt3\hat i + \hat j \end{aligned} o=(2cos0)i^+(2sin0)j^=2i^ \begin{aligned} \vec o =& (2\cos0^\circ)\hat i + (2\sin0^\circ)\hat j\\ =&2\hat i \end{aligned} B=w+v+j+p+o=(4+3)i^+(3+23)j^ \begin{aligned} \vec B =& \vec w+\vec v+\vec j+\vec p+\vec o\\ =&\(4+\sqrt3\)\hat i + \(3+2\sqrt3\)\hat j \end{aligned}
(a)B=?B=? B=(4+3)2+(3+23)2=25(2+3)8.64[cm] \begin{aligned} B=&\sqrt{\(4+\sqrt3\)^2+\(3+2\sqrt3\)^2}\\ =&2 \sqrt{5 \left(2+\sqrt{3}\right)}\\ \approx&8.64\ut{cm} \end{aligned}
(b)θB=?\theta_B=? θB=tan1(3+234+3)=tan1(6+5313)=0.845[rad] \begin{aligned} \theta_B=&\tan^{-1}\(\frac{3+2\sqrt3}{4+\sqrt3}\)\\ =&\tan^{-1}\left(\frac{6+5\sqrt{3}}{13} \right)\\ =&0.845\ut{rad} \end{aligned}