11-35 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec r&=3.0\i-2.0\j+4.0\k\ut{m}\\ \vec F_1&=3.0\i-4.0\j+5.0\k\ut{N}\\ \vec F_2&=-3.0\i-4.0\j+5.0\k\ut{N}\\ \vec P&=3.0\i+2.0\j+4.0\k\ut{m} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec \tau_a&=\vec r\times\vec F_1\\ &=6.0\i-3.0\j-6.0\k\ut{N\cdot m} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec \tau_b&=\vec r\times\vec F_2\\ &=6.0\i-27\j-18\k\ut{N\cdot m} \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \vec \tau_c&=\vec r\times(\vec F_1+\vec F_2)\\ &=12\i-3.0\times10\j-24\k\ut{N\cdot m} \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \vec \tau_d&=(\vec r-\vec P)\times(\vec F_1+\vec F_2)\\ &=-40\i\ut{N\cdot m} \end{aligned} $$