# Bosuns Chair acceleration problem

## Homework Statement

1. A boatswain's chair (or bosun's chair) is a device which is used to suspend a person from a rope to perform work aloft. It has a rigid seat and can be rigged, as shown in Fig.1, to allow the operator to haul himself upwards. [Naturally, in practice, it is wise to ensure that the cable cannot be released accidentally!]. If the operator’s mass is 70 kg and he exerts a downward pull of 200N on the cable, what is his acceleration?

F=ma

## The Attempt at a Solution

Unfortunately have no idea how to go about it :(

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Doc Al
Mentor
Use Newton's 2nd law. Start by drawing a free body diagram. Hint: Treat the operator and his chair as one unit.

I've tried that. The weight of the operator and his chair 686.7N meaning that the Tension in the two ropes holding him up should be 343.35N. Correct?

Doc Al
Mentor
No, not correct.

The weight of the operator and his chair 686.7N
OK.

meaning that the Tension in the two ropes holding him up should be 343.35N.
You are told that he pulls the rope with 200 N of force, so the tension is given. (You assumed equilibrium, for some reason. Don't do that.)

So the tension in each of the ropes, as there is 2, would be 100N. Would that therefore mean the unbalanced force would be 686.7-200= 486.7N?

haruspex
Homework Helper
Gold Member
I don't see a diagram. What's the pulley arrangement?

Doc Al
Mentor
So the tension in each of the ropes, as there is 2, would be 100N.
No. The force that he pulls on the rope is 200 N, thus, from Newton's 3rd law, the force the rope exerts on his him (which is the tension in the rope) must be equal.

I don't see a diagram. What's the pulley arrangement?
Yes, a diagram would be nice. But I assume it's the standard Bosun's chair arrangement.