12-11 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} y&=2.1\ut{m}\\ x&=0.91\ut{m}\\ h&=0.25\ut{m}\\ m&=27\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{cases} p&=y-h\\ q&=h\\ \end{cases} $$ $$ \begin{cases} A&=(0,0)\\ B&=(x,0)\\ C&=(x,y)\\ D&=(0,y)\\ P&=(0,p)\\ Q&=(0,q)\\ \end{cases} $$ $$ \begin{cases} N_{Py}&={mg\over2}\\ N_{Qy}&={mg\over2}\\ \end{cases} $$ $$ \begin{cases} \Sigma F_{x}&=0\\ \Sigma \tau&=0\\ \end{cases} $$ $$ \begin{cases} 0&=N_{Px}+N_{Qx}\\ 0&=-pN_{Px}-qN_{Qx}-{x\over2}mg\\ \end{cases} $$ $$ \begin{cases} N_{Px}&=-{x \over 2(p-q)}mg\\ N_{Qx}&={x \over 2(p-q)}mg\\ \end{cases} $$ $$ \begin{cases} N_{Px}&=-{x \over 2(y-2h)}mg\\ N_{Qx}&={x \over 2(y-2h)}mg\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec N_P &=-{x \over 2(y-2h)}mg\i+{1\over2}mg\j\\ &=-{2457 \over 320}g\i+{27\over2}g\j\\ &=-\br{7.529668453125\i+13.2389775\j}\times10\ut{N}\\ &\approx \br{7.5\i+13\j}\times10\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \vec N_Q &={x \over 2(y-2h)}\i+{1\over2}mg\j\\ &={2457 \over 320}g\i+{27\over2}g\j\\ &=\br{7.529668453125\i+13.2389775\j}\times10\ut{N}\\ &\approx \br{7.5\i+13\j}\times10\ut{N}\\ \end{aligned} $$