5-41 할리데이 11판 솔루션 일반물리학

$$\ab{a,b}$$ $$ \begin{cases} m_1&=3.2\ut{kg}\\ m_2&=5.4\ut{kg}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\Sigma \vec F = 0 \Harr \Delta \vec v=0,$$ $$ \begin{cases} \Sigma F_1 &= 0\\ \Sigma F_2 &= 0\\ \end{cases} $$ $$ \begin{cases} T_1-m_1g-T_2 &= 0\\ T_2-m_2g &= 0\\ \end{cases} $$ $$ \begin{cases} T_1 &= (m_1+m_2)g\\ T_2 &= m_2 g\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} T_2 &= m_2 g\\ &= 52.95591\ut{N}\\ &\approx 53\ut{N}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} T_1 &= (m_1+m_2)g\\ &= 84.33719\ut{N}\\ &\approx 84\ut{N}\\ \end{aligned} $$ $$\ab{c,d}$$ $$ \begin{cases} m_3&=4.8\ut{kg}\\ m_5&=5.5\ut{kg}\\ T_3&=199\ut{N} \end{cases} $$ $$ \begin{cases} \Sigma F_3 &= 0\\ \Sigma F_4 &= 0\\ \Sigma F_5 &= 0\\ \end{cases} $$ $$ \begin{cases} T_3-m_3g-T_4 &= 0\\ T_4-m_4g-T_5 &= 0\\ T_5-m_5g &= 0\\ \end{cases} $$ $$ \begin{cases} T_3 &= (m_3+m_4+m_5)g\\ T_4 &= (m_4+m_5)g\\ T_5 &= m_5g\\ \end{cases} $$ $$\ab{c}$$ $$ \begin{aligned} T_5&=\frac{11}{2}g\\ &=53.936575\ut{N}\\ &\approx 54\ut{N}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} T_4&=199-\frac{24}{5}g\\ &=151.92808\ut{N}\\ &\approx 1.5\times10^2\ut{N}\\ \end{aligned} $$