1. **Problem statement:** We have three boxes stacked in a lift accelerating upwards at $0.5\ \text{m/s}^2$. The masses are $m_1=20\ \text{kg}$ (bottom), $m_2=10\ \text{kg}$ (middle), and $m_3=5\ \text{kg}$ (top). The tension in the lift cable is $T=4620\ \text{N}$. Gravity $g=10\ \text{m/s}^2$. We need to find: (a) mass of the lift $M$, (b) reaction between floor and 1st box $R_1$, (c) reaction between 1st and 2nd box $R_2$, (d) reaction between 2nd and 3rd box $R_3$. 2. **Formulas and concepts:** - Total upward force = tension $T$. - Total downward force = weight of lift + weights of boxes. - Newton's 2nd law: $T - (M + m_1 + m_2 + m_3)g = (M + m_1 + m_2 + m_3)a$. - Reaction forces are normal forces between surfaces supporting weight and acceleration. 3. **Calculate total mass and acceleration forces:** Let total mass $= M + 20 + 10 + 5 = M + 35$. Using Newton's 2nd law for the whole system: $$T - (M + 35)g = (M + 35)a$$ Substitute values: $$4620 - (M + 35) \times 10 = (M + 35) \times 0.5$$ 4. **Solve for $M$:** $$4620 = 10(M + 35) + 0.5(M + 35) = 10.5(M + 35)$$ $$M + 35 = \frac{4620}{10.5} = 440$$ $$M = 440 - 35 = 405\ \text{kg}$$ 5. **Reaction between floor and 1st box $R_1$:** $R_1$ supports boxes 1, 2, 3 and their acceleration: $$R_1 = (m_1 + m_2 + m_3)(g + a) = 35 \times (10 + 0.5) = 35 \times 10.5 = 367.5\ \text{N}$$ 6. **Reaction between 1st and 2nd box $R_2$:** $R_2$ supports boxes 2 and 3: $$R_2 = (m_2 + m_3)(g + a) = 15 \times 10.5 = 157.5\ \text{N}$$ 7. **Reaction between 2nd and 3rd box $R_3$:** $R_3$ supports box 3 only: $$R_3 = m_3(g + a) = 5 \times 10.5 = 52.5\ \text{N}$$ **Final answers:** (a) Mass of lift $M = 405\ \text{kg}$ (b) Reaction floor and 1st box $R_1 = 367.5\ \text{N}$ (c) Reaction 1st and 2nd box $R_2 = 157.5\ \text{N}$ (d) Reaction 2nd and 3rd box $R_3 = 52.5\ \text{N}$