Question

A bosun’s chair is a device used by a boatswain to lift himself to the top of the mainsail of a ship. A simplified device consists of a chair, a rope of negligible mass, and a frictionless pulley attached to the top of the mainsail. The rope goes over the pulley, with one end attached to the chair, and the boatswain pulls on the other end, lifting himself upward. The chair and boatswain have a total mass M = 95.7 kg.

a) If the boatswain is pulling himself up at a constant speed, with what magnitude of force must he pull on the rope?

b) If, instead, the boatswain moves in a jerky fashion, accelerating upward with a maximum acceleration of magnitude a = 1.81 m/s², with what maximum magnitude of force must he pull on the rope?

Answer 1

a) let the man and the bosun's chair as a single system. Assume that the tension in each side of the rope be T. Assume that the combined mass of man and the chair is m If the boatswain is pulling himself up at a constant speed, then the magnitude of force is, T +T-mg 0 1n substitute 95.7 kg for mass m and 9.8 m/s for g (95.7 kg) (9.8 m/s') -468.93 N Hence, the magnitude of force he must pull the rope is 468.93 N b) Assume that the upward direction to be positive Ii, instead, the boatswain moves in a jerky tashion, accelerating upward with a maximum acceleration of magnitude a 1.81 m/s2. Then the force is given as, T-(&+a) 95.7 kg (9.8 m/s +1.81 m/s - 555.53 N Hence, the maximim magnitude of force he must pull the rope is 555.53 N