# With what force magnitude must a man pull on the rope if he is to rise with a constant velocity? [closed]

I didn't understand why the man should pull with a force of 2T to rise at constant velocity. First,okay he should exercise a tension force T equal to his weight plus the the weight of the chair to stay at rest. but why he should apply a force=2T to constant motion?

## closed as off-topic by Gert, AccidentalFourierTransform, Chris♦, Kyle Kanos, stafusaFeb 12 '18 at 11:12

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• Can you add more details? At present, it's impossible to tell what you're asking. – probably_someone Feb 11 '18 at 19:10
• You have a basic communication problem in that you have never said what $T$ is. Assuming it is tension, then you seem to be assuming that the tension equals the person's weight. Why do you think that? – dmckee Feb 11 '18 at 19:13
• sorry i forgot to upload a photo – Daniel Kamand Feb 11 '18 at 19:13
• I am assuming this because the weight is by definition the force that you should apply to a body to keep it at rest or something like this – Daniel Kamand Feb 11 '18 at 19:15
• Potentiallu useful search term "bosun's chair" – dmckee Feb 11 '18 at 19:16

The answer is: Newton's 1st law.

Firstly, we must realize that the rope is in fact grabbing the man twice.

• If he only held on to it with his hands and it wasn't tied to the chair, then only one $T$ was holding him up.

• If the rope was only tied to his chair and he wasn't holding on with his hands, then again only one $T$ would be holding him up.

But the rope is both tied to the chair and he is holding on to it. This corresponds to two rope ends holding him up. They share the load they are carrying. The fact that those two rope ends belong to the same rope makes no difference.

Therefore, $2T$ is holding/pulling upwards. And the weight $w$ (total weight of man-plus-chair) pulls downwards as always.

Secondly, we must realize that for the velocity to be constant, the acceleration is zero. This means that Newton's 1st law applies.

$$\sum F=0$$

The sum of all forces balances out to zero. Let's plug in those forces:

$$2T-w=0\qquad \Leftrightarrow \qquad w=2T$$

So, the weight equals two-times-the-tension $2T$, not one-time-the-tension $T$ as you initially thought. It might not seem intuitive at first, but remember that tension is a variable force; it can change to fit. The value of $T$ is not the same in those two cases. If two-times-the-tension $2T$ is balancing out the weight $w$, then that tension $T$ is smaller than if one-time-the-tension $T$ was balancing out the weight $w$.

Tension is a variable force, and that might be the tricky part to realize here. Having two rope ends to carry the load means that they share the load, and the tension is thus only half of what it would have been if one rope end carried the whole load alone.

Assuming a massless, frictionless pulley and rope, and tension $T$ in the rope, the force exerted by the rope on the man+chair is always $2T$, because there are two attachment points, both pulling upward with $T$. It can never be anything else. If $2T$ is greater than the weight of the man+chair, he accelerates upward. If less, he accelerates downward. If he is sitting still, or moving up or down at a constant rate, they are the same: $2T$ is the weight of the man+chair.

• To clarify: $T$ is not equal to the weight of the man and chair, which is what the OP thought. It is equal to half their weight. – probably_someone Feb 12 '18 at 6:21
• @probably_someone I guess I maybe answered the wrong question. OP wanted to know why the tension is not equal to the weight? – Ben51 Feb 12 '18 at 6:24
• I'm not quite sure myself, just wanted to cover all the bases in case that was the primary difficulty. – probably_someone Feb 12 '18 at 6:27