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Imagine there is a painter, weighing $180~\rm lb$, that is working from a bosun's chair hung down the side of a tall building.

enter image description here

Suppose that he pulls down on a fall rope with such a force that he presses against the chair with a force of $100~\rm lb$. You can assume that the chair's weight is $30~\rm lb$.

For finding the acceleration of the painter and the chair, I took into account that the weights of the painter and the chair are $180~\rm lb$ and $30~\rm lb$ respectively. I used this idea to perform the following step:

$$\text {Total mass of the painter and the chair} = \left(\frac{(180 + 30)~\rm lb}{g}\right) $$

He exerts a downward force of $100~\rm lb$ on the chair. His net motion will be upwards.

I think the $100~\rm lb$ force the person exerts on the chair is transferred to the rope he is pulling on. enter image description here

But that is just the string he is pulling on. The diagram shows that only one end of the rope is attached to the bosun chair. That end will have an upwards force of $(100 + 180 + 30)~\rm lb$ (as shown in the picture). This way, one end will have a $y~\rm lb$ force and other a force of $(100 + 180 + 30)~\rm lb$. I don't know if this is possible and I am not totally convinced that the rope is experiencing a force of $100~\rm lb$ due to the painter pulling on it.

How can I properly use Newton's third law to determine the impact of the $100~\rm lb$ downwards force on the overall system (the painter and bosun's chair)?

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closed as off-topic by AccidentalFourierTransform, stafusa, Jon Custer, John Rennie, CR Drost Sep 24 '18 at 19:05

This question appears to be off-topic. The users who voted to close gave this specific reason:

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If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Also, you shouldn't make a list of edits to your question. A list of edits is automatically made that people can look at if needed. $\endgroup$ – Aaron Stevens Sep 24 '18 at 18:20
  • $\begingroup$ How has my question become off-topic again? $\endgroup$ – a_sid Sep 24 '18 at 19:14
  • $\begingroup$ @a_sid It was good of you to post this meta question to ask about that. $\endgroup$ – David Z Sep 24 '18 at 20:02
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    $\begingroup$ @a_sid It might help this question if you don't label things as "my work" as this makes it seem like the question is a "check my work" question. It might also help to put your main question in bold. $\endgroup$ – Aaron Stevens Sep 25 '18 at 10:16
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    $\begingroup$ @a_sid (2 comments up) No, it's not unfair to vote to close without giving additional information. I understand that it may be frustrating for you, but it's a common and legitimate way for things to happen here. $\endgroup$ – David Z Sep 30 '18 at 1:14
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You need to think of the man and the chair separately. This is because if you combine the man and the person into one thing, you get a single equation with Newton's second law with two unknown values (the acceleration and the upwards tension forces).

The man has a tension force acting upwards on him, a normal force upwards on him, and his weight downwards. The chair has tension acting upwards, the same magnitude of a normal force downwards, and it's weight downwards. We can exploit that each of the tension forces are the same (which is something that is incorrect in your diagram)$^*$, as well as the fact that the man and chair will have the same acceleration. Therefore, by Newton's second law,

$$ma=T+N-mg$$ $$Ma=T-N-Mg$$

Where $m$ and $M$ are the masses of the person and the chair respectively.

Since you are given what $m$, $M$, and $N$ are, these are two equations with two unknown values. Note that the 100 lbs corresponds to $N$, not $T$. I will leave the math to you to find what $a$ and $T$ are.

Also, note that the sum of these equations is what we get by treating evening as one system that I was discussing at the beginning of this answer. $$(m+M)a=2T-(m+M)g$$

Notice how we cannot get actual values for $a$ or $T$ with this equation and the given information. We need the previous two equations (or this one and just one of the previous ones) to solve the problem.


Your question title seems to be concerned with more of understanding a cause-and-effect relationship between the forces. I would say what happens is that the man pulls on the rope. This causes two things. 1) It applies an upward force to the man 2) It applies an equal upward force to the chair. Each of these two things have an opposite effect on the normal force between the person and the chair. The first lessens it, and the second makes it larger. The net effect is given in the problem as a 100 lb normal force.

It might be instructive to solve for $N$ and $a$ rather than $T$ and $a$. $$N=\frac{T(m-M)}{m+M}$$ $$a=\frac{2T}{m+M}-g$$

As we can see, the normal force between the man and the chair is directly proportional to the force the man applies to the rope. We can explore this problem thinking of different scenarios. For example, if the man does not pull on the rope ($T=0$), then $N=0$, so the man and chair will fall with an acceleration of $-g$, as expected.

We also see that in order for the man to accelerate upwards he must apply more than half of the total weight of himself and the chair, which makes sense since the whole system is pulled upwards with a force of $2T$. In other words, he gets double out of what he puts in essentially. This also let's us see that if $a>0$, then $N>\frac{g(m-M)}{2}$, which is true in your problem. You can play around with the equations like this and learn other things about the system.


$^*$ If we assume a massless rope, and a massless, frictionless pulley, then the tension throughout the string is uniform. This is why the tension forces acting on the man and the chair are the same.

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  • $\begingroup$ How is the force of tension acting upwards on the painter? Is it because the rope exerts an upwards pointing force on him since he applied a downwards pointing force (of the same magnitude) on the rope? $\endgroup$ – a_sid Sep 24 '18 at 17:43
  • $\begingroup$ @a_sid Yes that is right. It is Newton's third law. Just imagine trying to climb a rope. You pull down on the rope, which makes you move upwards. $\endgroup$ – Aaron Stevens Sep 24 '18 at 18:16
  • $\begingroup$ I don't understand how you are getting N > (g(m-M))/2. $\endgroup$ – a_sid Sep 25 '18 at 4:28
  • $\begingroup$ @a_sid Set $a>0$ and get an expression of the form $T>\tau$. Then use this in the equation for $N$. $\endgroup$ – Aaron Stevens Sep 25 '18 at 6:44
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There are a variety of ways to solve this problem, but I solved it pretty easily by analyzing the painter, and the whole system (painter + chair), giving 2 equations and 2 unknowns.

The painter's force on the chair (100 lbs) $= M_p(g+a)-T$, where $M_p$ is the painter's mass (100 lbs / g), $a$ is the upward acceleration, and $T$ is the rope's tension.

The force on the combined system is $2T = (M_p+M_c)(g+a)$, because twice the rope's tension is pulling up on the system. $M_c$ is the mass of the chair (30 lbs / g).

With a little bit of algebra, it's pretty easy to solve from there.

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    $\begingroup$ Looks familiar :) $\endgroup$ – Aaron Stevens Sep 23 '18 at 21:36
  • $\begingroup$ @AaronStevens You're right! I just found it a little easier to solve using the combined system for the 2'nd equation. But great answer! $\endgroup$ – Stuart Van Horne Sep 26 '18 at 19:11
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Due to Newton's 3rd law, when he exerts 100lbs downwards on the chair, the chair exerts the same upwards on him. That upwards force comes from the string.

So, considering the man+chair as the system, there is an upwards force (string) of 100lbs and a downwards force (gravity) of 210 lbs. It looks like he is accelerating downwards, and not upwards.

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  • $\begingroup$ So, considering the man+chair as the system, there is an upwards force (string) of 100lbs..... How is the force on the string pointing upwards? The question states that the painter pulls down on the fall rope. Shouldn't the force on the string be acting downwards? $\endgroup$ – a_sid Sep 22 '18 at 7:40
  • $\begingroup$ @a_sid He pulls downwards in the rope with a downwards force. The rope conversely pulls upwards in him. Imagine that I am pulling you up from a deep hole; even if you pull downwards in me (with your weight and/or with a pulling force) then I am still pulling upwards in you. $\endgroup$ – Steeven Sep 22 '18 at 9:39
  • $\begingroup$ So you are essentially saying that the rope exerts an equal and opposite force of 100 lbs on the painter. That will cause the 100 lbs force he exerts on the rope to get cancelled out, wouldn't it? Wouldn't that leave tension in the rope (opposing the force of gravity) as the only active force? $\endgroup$ – a_sid Sep 22 '18 at 15:37
  • $\begingroup$ The tension force is not 100 lbs, and the man has two forces acting upwards on him, not one. If you work the problem out you find the the man and chair accelerate upwards. $\endgroup$ – Aaron Stevens Sep 23 '18 at 20:00

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