6-22 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m_1&=2.00\ut{kg}\\ m_2&=3.80\ut{kg}\\ \theta&=25.0\degree\\ \mu_1&=0.226\\ \mu_2&=0.113\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\Sigma \vec F = m\vec a,$$ $$ \begin{cases} \Sigma F_{1x} &= m_1 a\\ \Sigma F_{2x} &= m_2 a\\ \Sigma F_{1y} &= 0\\ \Sigma F_{2y} &= 0\\ \end{cases} $$ $$ \begin{cases} m_1 a &= m_1g\sin\theta+F_T-\mu_1F_{N1}\\ m_2 a &= m_2g\sin\theta-F_T-\mu_2F_{N2}\\ 0&= F_{N1}-m_1g\cos\theta\\ 0&= F_{N2}-m_2g\cos\theta\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} F_T&=\(\mu _1-\mu _2\)\frac{ m_1 m_2 }{m_1+m_2}g\cos \theta\\ &=\frac{2147}{14500}g\cos25\degree\\ &\approx 1.316013757149527\ut{N}\\ &\approx 1.32\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} a&=\frac{m_1 \left(\sin \theta-\mu _1 \cos \theta\right)+m_2 \left(\sin \theta-\mu _2 \cos \theta\right)}{m_1+m_2}g\\ &=\left(\sin 25\degree-\frac{4407 \cos 25\degree}{29000}\right)g\\ &\approx 2.793823678372284\ut{m/s^2}\\ &\approx 2.79\ut{m/s^2}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} F_{Tc}&=\(\mu _2-\mu _1\)\frac{ m_2 m_1 }{m_2+m_1}g\cos \theta\\ &=-\frac{2147 }{14500}g\cos 25\degree\\ &=-F_T \end{aligned} $$ $$ \begin{aligned} a_c&=\frac{m_2 \left(\sin \theta-\mu _2 \cos \theta\right)+m_1 \left(\sin \theta-\mu _1 \cos \theta\right)}{m_2+m_1}g\\ &= \left(\sin 25\degree-\frac{4407 \cos 25\degree}{29000}\right)g\\ &=a \end{aligned} $$