# Cord tension & how to calculate force magnitude of a cord & pulley

## Homework Statement

A man sitting in a bosun's chair that dangles from a massless rope, which runs over a massless, frictionless pulley and back down to the man's hand. the combined mass of the man and chair is 103.0 kg. With what force magnitude must the man pull on the rope if he is to rise (a) with a constant velocity)

and

(c) if the rope on the right extends to the ground and is pulled by a coworker, with what force magnitude must the co-worker pull for the man to rise with constant velocity?

a picture showing the particular case:
http://puu.sh/14XOQ [Broken]

f = m*a

## The Attempt at a Solution

To find the force magnitude I used newton's second law:
F_res = m * a
where the resulting force being:
T-m*g - T being the cord tension, and since he is rising it is larger than f(g) = m*g

The velocity is constant, and the acceleration is therefore 0, making the equation:

T = (m*g)

but that is not correct, for some reason, which I do not quite understand, the cord tension is 2T, making the actual solution

T = (m*g)/2

In the case where the co-worker is pulling the cord, the cord tension is just T, making the magnitude required to lift the person in the bosun's chair double that of himself lifting.

So while I did get the proper results, I was hoping somebody could shed some light on why one must calculate the cord tension as such i.e. himself pulling equals 2T and just T if somebody else does.

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Doc Al
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So while I did get the proper results, I was hoping somebody could shed some light on why one must calculate the cord tension as such i.e. himself pulling equals 2T and just T if somebody else does.
Picture an imaginary box around the man and his chair. How many strands of rope extend from that box? Remember that each strand of rope exerts the full tension.

That's the cleverness of the bosun's chair.

As such: http://puu.sh/14Ya6 [Broken] ?

if yes; to continue with that line of thinking, if one was to picture an imaginary box around the co-worker standing on the ground below him, only one strand extends from that box, thus T = 1?

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Doc Al
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As such: http://puu.sh/14Ya6 [Broken] ?
Right.
if yes; to continue with that line of thinking, if one was to picture an imaginary box around the co-worker standing on the ground below him, only one strand extends from that box,
Right.
thus T = 1?
Not sure what you mean by T = 1. I'd say that the total upward force in that case = T. As opposed to the bosun's chair, which would have the upward force = 2T.

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Not sure what you mean by T = 1. I'd say that the total upward force in that case = T.

That is also what I meant, poorly formulated by me. Thanks for the replies, much appreciated.

Chestermiller
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I don't think you got the idea of what Doc Al was saying. You've heard of free body diagrams, correct? Draw a free body diagram showing all the forces acting on the combination of guy and chair (not the guy on the ground) for each of the two cases.

in case (a) there's a downwards force equal to f(g) and a upward force of T (cord tension) on the bosun's chair since it is connected to a rope. But since the man's hands are also "attached" to the rope, there is another upwards of force of T making the sum of upwards forces 2T. (my translation of the diagram into text)

in case (c) the man's hands are not "attached" to the rope, and therefore the upwards force is just T. Which I assume is the explanation.. sort of.

the keywords in your reply to me was "combination of guy and chair".. I initially thought that as in case (a) where the man pulling himself up resulted in another upwards force T, likewise I assumed in case (c) that the guy on ground pulling would result in another upwards force T... but since the rope is only attached at one place, namely the bosun's chair, the upwards force is just T.

do I have it entirely wrong?

Chestermiller
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No. You have it right. You doped it out correctly. Good job. The importance of using free body diagrams cannot be overemphasized.