College Physics (4th Edition)

Let $T_l$ be the tension in the lower cable. Let $m_l$ be the mass of the lower crate. We can consider the forces of the lower crate: $\sum F = m_l~a$ $T_l-m_l~g = m_l~a$ $T_l = m_l~(a+g)$ $T_l = (100~kg)(1.0~m/s^2+9.80~m/s^2)$ $T_l = 1080~N$ The tension in the lower cable is 1080 N Let $T_u$ be the tension in the upper cable. Let $m_u$ be the mass of the upper crate. We can consider the forces on the system of both crates: $\sum F = (m_l+m_u)~a$ $T_u-m_l~g-m_u~g = (m_l+m_u)~a$ $T_u = (m_l+m_u)~(a+g)$ $T_u = (100~kg+200~kg)(1.0~m/s^2+9.80~m/s^2)$ $T_u = 3240~N$ The tension in the upper cable is 3240 N