A helicopter rescue activity at sea is modelled as follows. The helicopter is stationary and a man is suspended from it by means of a vertical, light, inextensible wire that may be raised or lowered, as shown in Fig. 6.1.
\includegraphics{figure_6_1}
- When the man is descending with an acceleration 1.5 m s\(^{-2}\) downwards, how much time does it take for his speed to increase from 0.5 m s\(^{-1}\) downwards to 3.5 m s\(^{-1}\) downwards?
How far does he descend in this time? [4]
The man has a mass of 80 kg. All resistances to motion may be neglected.
- Calculate the tension in the wire when the man is being lowered
- with an acceleration of 1.5 m s\(^{-2}\) downwards,
- with an acceleration of 1.5 m s\(^{-2}\) upwards. [5]
Subsequently, the man is raised and this situation is modelled with a constant resistance of 116 N to his upward motion.
- For safety reasons, the tension in the wire should not exceed 2500 N. What is the maximum acceleration allowed when the man is being raised? [4]
At another stage of the rescue, the man has equipment of mass 10 kg at the bottom of a vertical rope which is hanging from his waist, as shown in Fig. 6.2. The man and his equipment are being raised; the rope is light and inextensible and the tension in it is 80 N.
\includegraphics{figure_6_2}
- Assuming that the resistance to the upward motion of the man is still 116 N and that there is negligible resistance to the motion of the equipment, calculate the tension in the wire. [4]