11판/12. 평형과 탄성

12-32 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 10. 9. 18:03
{F1=8.40i^5.70j^[N]F2=16.0i^+4.10j^[N] \begin{cases} \vec F_1&=8.40\i-5.70\j\ut{N}\\ \vec F_2&=16.0\i+4.10\j\ut{N}\\ \end{cases} (a,b)\ab{a,b} F3=(F1+F2)=24.4i^+1.6j^[N] \begin{aligned} \vec F_3&=-\br{\vec F_1+\vec F_2}\\ &=-24.4\i+1.6\j\ut{N}\\ \end{aligned} {F3x=24.4[N]F3y=1.6[N] \begin{cases} F_{3x}&=-24.4\ut{N}\\ F_{3y}&=1.6\ut{N}\\ \end{cases} (c)\ab{c} θ=tan11.624.43.0761126286663285[rad]3.08[rad] \begin{aligned} \theta&=\tan^{-1}{1.6\over-24.4}\\ &\approx 3.0761126286663285\ut{rad}\\ &\approx 3.08\ut{rad}\\ \end{aligned}