7-15 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} F_1&=4.00\ut{N}\\ F_2&=4.00\ut{N},\theta_2&=50.0\degree\\ F_3&=10.0\ut{N},\theta_3&=35.0\degree\\ S&=2.50\ut{m} \end{cases} $$ $$ \begin{cases} \vec F_1&=-4\i\ut{N}\\ \vec F_2&=-4\sin50\degree\i-4\cos50\degree\j\ut{N}\\ \vec F_3&=10\cos35\degree\i+10\sin35\degree\j\ut{N}\\ \end{cases} $$ $$ \begin{aligned} \theta_{F} &= \vec F_1+\vec F_2+\vec F_3\\ =&(-4-4\sin50\degree+10\cos35\degree)\i\\ &+(-4\cos50\degree+10\sin35\degree)\j\ut{N}\\ \end{aligned} $$ $$ \begin{aligned} \theta_{\Sigma F} &=\theta_S\\ &=\tan^{-1}\frac{-4\cos50\degree+10\sin35\degree}{-4-4\sin50\degree+10\cos35\degree}\\ &=\tan ^{-1}\frac{2\sin 40\degree-5\sin35\degree}{2+2\cos40\degree-5\cos35\degree} \end{aligned} $$ $$ \begin{aligned} \vec S=&2.5\cos\(\tan ^{-1}\frac{2\sin 40\degree-5\sin35\degree}{2+2\cos40\degree-5\cos35\degree}\)\i\\ &+2.5\sin\(\tan ^{-1}\frac{2\sin 40\degree-5\sin35\degree}{2+2\cos40\degree-5\cos35\degree}\)\j\\ \end{aligned} $$ $$ \begin{aligned} W_{F1}&=\vec F_1 \cdot \vec S\\ &\approx 3.3557687537820193\ut{J}\\ &\approx 3.36\ut{J}\\ \end{aligned} $$ $$ \begin{aligned} W_{F2}&=\vec F_2 \cdot \vec S\\ &\approx -8.625809721004044\ut{J}\\ &\approx -8.63\ut{J}\\ \end{aligned} $$ $$ \begin{aligned} W_{F3}&=\vec F_3 \cdot \vec S\\ &\approx 20.38012105272343\ut{J}\\ &\approx 20.4\ut{J}\\ \end{aligned} $$