1. **State the problem:** We have a lift with a total mass of 1200 kg (lift + person) accelerating upwards at $\frac{7g}{10}$ m/s², where $g$ is the acceleration due to gravity. We need to find: a) The tension in the lift cable. b) The mass $M$ of the person given the reaction force on the person is $63g$ N. --- 2. **Given data:** - Total mass $m = 1200$ kg - Acceleration $a = \frac{7g}{10}$ m/s² - Gravitational acceleration $g$ (standard gravity) - Reaction force on person $R = 63g$ N --- ### Part (a): Find the tension $T$ in the lift cable 3. The forces acting on the lift system are: - Weight downward: $W = mg$ - Tension upward: $T$ Using Newton's second law upward: $$T - mg = ma$$ Rearranged: $$T = m(g + a)$$ Substitute $a = \frac{7g}{10}$: $$T = 1200 \times \left(g + \frac{7g}{10}\right) = 1200 \times \frac{10g + 7g}{10} = 1200 \times \frac{17g}{10} = 2040g$$ So the tension is $T = 2040g$ N. --- ### Part (b): Find the mass $M$ of the person given the reaction force $R = 63g$ N 4. The reaction force $R$ is the normal force exerted by the floor on the person. The person experiences: - Weight downward: $Mg$ - Normal reaction upward: $R$ Using Newton's second law for the person accelerating upward at $a = \frac{7g}{10}$: $$R - Mg = Ma$$ Rearranged: $$R = M(g + a)$$ Substitute $R = 63g$ and $a = \frac{7g}{10}$: $$63g = M \left(g + \frac{7g}{10}\right) = M \times \frac{17g}{10}$$ Divide both sides by $g$: $$63 = M \times \frac{17}{10}$$ Solve for $M$: $$M = \frac{63 \times 10}{17} = \frac{630}{17} \approx 37.06$$ So the mass of the person is approximately 37.06 kg. --- **Final answers:** - a) Tension in the cable: $T = 2040g$ N - b) Mass of the person: $M \approx 37.06$ kg