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anonymous
 5 years ago
physics question, suppose two people pull on rope. person A is 50 kg and pulls towards the right (+) with 30 N force. person b who is 40 kg pulls towards the left with a 20 N force. assume the rope is massless. I dont get why the tension in the rope is the same magnitude. For the rope the forces on it are F1  F2 = 0*a. but F1 and F2 should be the 30N and 20N , so it should be not zero but 10 N to the right for tension. something doesnt make sense here
anonymous
 5 years ago
physics question, suppose two people pull on rope. person A is 50 kg and pulls towards the right (+) with 30 N force. person b who is 40 kg pulls towards the left with a 20 N force. assume the rope is massless. I dont get why the tension in the rope is the same magnitude. For the rope the forces on it are F1  F2 = 0*a. but F1 and F2 should be the 30N and 20N , so it should be not zero but 10 N to the right for tension. something doesnt make sense here

This Question is Closed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are suggesting that at one part of the rope, there is 10 newtons of tension, and at the other end there is 0 tension (i.e. slack)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not sure why you're looking at the forces on the rope. You're assuming it's massless, so you can't really talk about the rope's acceleration.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i was using my book. it says the forces on the rope are F1  F2 = 0. isnt tension the force on the rope?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are you being asked to find?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A rope is one element. It can't have variable force along it. So whatever the tension is in the rope, it has to be the same all along the rope.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Here is the full paragraph. "Suppose two people pull the ends of a rope with oppositely directed forces F1 and F2 (in bold). The rope also exerts forces F1 and F2 on the people (in bold)From the second law, we know F1  F2 = ma , where m is the mass of the rope and a is its acceleration. If that acceleration is zero , or if the mass of the rope is so small that we can idealize the rope as having zero mass, then F1  F2 = 0. In this special situation the forces exerted by the rope on the two people are equal in magntitude

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well im wondering, what is the force on the rope , if it says F1  F2 = 0, then F1 = F2, but 30 N does not equal 20 N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What it's actually saying is that it's impossible for person A to be pulling with 30N of force and person B to be pulling with 20N of force and for there to be zero acceleration. But you know that already since: $$F=ma$$ If F were not 0, you'd have an acceleration.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in my example, one guy pulls 30 N right and the other guy pulls 20 N left

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't see where you got 30N and 20N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In your example, there would be acceleration towards the guy pulling with 30N.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There would be a net force of 10N which would result in an acceleration in the direction of the guy pulling with 30N force.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, but you are contradicting the hypothisis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my question is, what is the force on the rope? and why does it say the force exerted by the rope on the two people is equal in magnitude. and which force is the tension. so there are 3 questions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They are saying that If the rope is not accelerating, we know the forces on both ends are equal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you get the forces at both ends are equal, and what is this force? the tension?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and which force is it, what is exerting on what

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you said the forces at both ends are equal. what do you mean by that. the force that the rope exerts on the two people are equal?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, the force on the rope.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but there are different forces, one is 30 N and one is 20 N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, you made up that example, where the conclusion is false, and trying to make a claim about the implication.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the conclusion is false?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you said the force on the rope at both ends is equal. clearly this is false, it is 20 N and 30 N , they are not equal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the acceleration is not 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, youre saying the net force on both ends is equal in magnitude?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but my rope is massless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm saying IF the acceleration is 0, then the NET force on the rope is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why does the book say F1F2 = 0, so F1 = F2,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't talk about force acting on massless things.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It says F1F2 is 0 BECAUSE it claimed already that the acceleration is 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then F1 and F2 are zero?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is tension the F1 and F2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If a =0 then \[\sum F = F1F2 = 0 \implies F1 = F2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But if the forces do not sum to 0 than the acceleration of the rope will not be 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but in my example, what is F1 and F2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is the acceleration equal to 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nomo, if you cant type , you should restart.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0www.openstudy.com start over

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A review of massless rope mechanics: A rope carries a tension. This tension is the force felt at the endpoints of the rope by whatever the rope is attached to. So, if you pull against a rope caught tied to a wall:  > 30N the tension in the rope is 30N. But, by action/reaction, the wall is also pulling on the rope. And since there is no acceleration, it is pulling with an equal and opposite force of 30N. Now let's look at your example. You now have two people, pulling on the rope. One is pulling harder than the other. Without doing the math, it's obvious that whoever is pulling harder is going to move the other person. That's exactly what happens: 20N <> 30N > everything moves this way So your next question is, what's the tension in the rope? Well, the tension in the rope has to be 20N, because it's the force the weaker puller is pulling with. The rest of the force of the stronger puller, 10N, goes into acceleration. So: 1. The tension in the rope is 20N. 2. The net external force on the rope is 10N. Hope this helps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get a different answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.030 N  T = 50kg * a 20 N + T = 40 kg * a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Heh. Cantorset, you are in fact correct.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Those are the equations of motion, and I should have written them out before posting.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 10 N = 90 kg *a , so a = 1/9 m*s^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one sec, my browser is crashing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.024.444 N is the tension

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The problem with this exercise is that in the absence of any frictional forces, you 're going to have that both people accelerate toward each other which will remove the tension.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what is F1 and F2, i still dont see

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your acceleration is not constant.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0F1 and F2 are the tension forces?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0F1 is the force of one person pulling, F2 is the force of the other.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but thats false, because 30  20 is not 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The equations of motion that cantorset wrote are valid at the instant when both people start pushing, as are the results. However, polpak is correctly pointing out that this is a dynamic problem, not a static problem. Odds are you'll never actually find a problem like this in a problem set, because there is too much information missing to be able to give a complete answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nomo, so do you agree that 24.444 is the tension?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In the instant when they both pull, (and in the absence of any friction) The tension is 24.444

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is the acceleration given to be 0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0"Suppose two people pull the ends of a rope with oppositely directed forces F1 and F2 (in bold). The rope also exerts forces F1 and F2 on the people (in bold)From the second law, we know F1  F2 = ma , where m is the mass of the rope and a is its acceleration. If that acceleration is zero , or if the mass of the rope is so small that we can idealize the rope as having zero mass, then F1  F2 = 0. In this special situation the forces exerted by the rope on the two people are equal in magnitude, and the rope can be thought of as simply transmitting a force from one person to the otherl the force at any point in the rope is referred to as tension. It is the same everywhere in the rope only if the rope is unaccelerated or if is idealized as massless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, but the rope is given to be massless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So the acceleration is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But that's not a problem, it's a lesson.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You CANNOT have forces acting on massless things.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Your example is a problem, and it's a problem not covered by this lesson, since it clearly stipulates that the acceleration is 0; in your case, it isn't.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, so the book is contradicting itself?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are trying to contradict the book

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But not making a reasoned argument.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, you're contradicting the book. You are constructing a situation where the acceleration cannot be zero, and trying to analyze the problem as if it could.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , i know the acceleration is not zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So then the book doesn't apply anymore.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then the forces cannot be the same.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets look at the problem again. the forces on the rope is F1  F2 = ma , the mass of the rope is neglible. so we have F1  F2 = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Therefore the F1 = F2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you contradict by saying \[F1 \ne F2\] then \[ma \ne 0.\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but didnt we stipulate earlier that F1 = 30 N and F2 = 20 N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You cannot say that ma = 0 if f1 not equal f2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The sum of the forces IS ma.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the sum is nonzero ma is nonzero. If ma is 0 the sum is 0. If the sum is 0 ma is 0. If ma is nonzero the sum is nonzero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but in my problem, i started with 30 N and 20 N , and they applied the forces at either end, correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can *only* assume the rope is massless *IF* F1=F2. Otherwise, the discussion in the book doesn't apply.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So while there are realworld situations where F1 is not equal to F2, you can't analyze them using what the book says. It *only* applies for the special case where F1=F2.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but didnt we assume that the two guys are applying two different forces

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We did. But then you tried to use what the book told you. You can't do that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If forces are nonzero then ma is nonzero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Pick one or the other, but you can't just pick both to be different.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok let me try again, one sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0person A is 50 kg and pulls towards the right (+) with 30 N force. person b who is 40 kg pulls towards the left with a 20 N force. assume the rope is massless.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You cannot say the sum of the forces is not zero and ma is 0. You cannot say that ma is not zero but the sum of the forces is 0. Because the sum of the forces IS ma. If you pick one you are also choosing the other.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can't assume the rope is massless.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they do that in problems though, i thought

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They do that when the forces are equal.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are trying to do something like: Let \(1 = 0 \implies 2+1 = 2\) And then complaining because you know 2+1 should be 3.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont understand , what is wrong with my problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have ma = 0, but the sum of the forces is not 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If it is not accelerating then there is no net force on the object.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Things do not accelerate in the absence of a net force.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Things with net forces acting on them MUST accelerate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what exactly is the contradiction in the question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nothing wrong with assuming the rope is massless person A is 50 kg and pulls towards the right (+) with 30 N force. person b who is 40 kg pulls towards the left with a 20 N force. assume the rope is massless.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You try to claim \(m = 0 \implies ma=0\) But \[\sum F = ma\] And you pick two forces also that don't sum to 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We are telling you THERE IS SOMETHING WRONG with that assumption.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what is the force on the rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because what happens in reality is that ma is not 0, You will get a tension, the tension will pull the two people together

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh you said there is no force on the rope, since its massless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not saying anything. You are choosing two sides of the equation that are nonequal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, you cant have tension with a massless rope?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Which do you want, a rope that doesn't accelerate? or a net force?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ive seen many tension problems where we assume the mass of the rope is negligible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Was the rope accelerating?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one sec, let me get a problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0An important property of massless ropes is that the total force on the rope must be zero at all times. To prove this, we go back to Newton's Second Law. If a net force acts upon a massless rope, it would cause infinite acceleration, as a = F/m , and the mass of a massless rope is 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://www.sparknotes.com/physics/dynamics/newtonapplications/section3.rhtml

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right. So if you assume the rope is massless you have to have it not accelerating. If it is not accelerating the net force must be 0. If the net force is zero, each force acting on the rope must be cancelled by another foce.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so youre saying, if you have a 40 kg guy and a 50 kg guy and they are pulling on a rope in opposite directions, and the rope is massless, then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then the force exerted by the rope on the two people must be equal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but then the two people, what about the force of the two people exerting on the rope , or the forces on the rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The net force they exert on the rope must be 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but they can have different forces as well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but then by newtons third law, then that would force them to have the same force

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They cannot be exerting different forces.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If they were then the sum would not be 0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets say we have a pulley and two masses

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one mass is 50 kg, the other mass is 30 kg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0m1 = 50 kg, m2 = 30 kg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they go up and down, like vertically

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0simple pulley. the force on m1 are 9.8 *50  T , where + is down direction

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the force on m2 is T  30 *9.8

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the problem says the mass of the cable and the pulley is neglible

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that means don't try to use it for finding the mass

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0, wait, you said then the acceleration is zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the pulley is accelerating, i mean the cable is accelerating

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You just keep talking in circles. What exactly is the question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the force on the cable should be zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since it is massless, but there is force on the cable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats what the problem states

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It says ignore the mass, not that it's massless.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In all these cases you are going to assume that the mass of the rope is ignorable, and that the tension in the rope is uniform.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think it says the mass is zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's telling you to ignore it because otherwise it becomes impossible to solve with your current tools. IT CANNOT BE 0 everything (even light) has mass. Forces cannot act on objects with no mass.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it says let it be zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not going to continue this further. If you want to argue further you can contact the authors, or your professor.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the rope has mass m, then Newton’s second law applied to the rope gives T  T'= ma. If the mass m is taken to be zero, however, as in the upcoming examples, then T = T'.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the acceleration of a massless object with a force of 1 newton acting on it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here i will show you a link , one sec http://www.wadsworthmedia.com/marketing/sample_chapters/0495106194_ch04.pdf page 70

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't disagree with what they state there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0All they are saying with the 'massless rope' thing is that the tension in both ends is the same. It's a trick used to simplify your problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in reality that cannot happen.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But if you violate the precepts of the 'trick' then you cannot use the rules they infer from it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but if the tension is the same, then it wont move

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm not talking about the external force on the rope.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the tension is the internal force.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The point is that it is not slack, it does not break, and it is not elastic.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can have it move, but you cannot try to find it's acceleration in the absence of other objects because it has no mass.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's a tool, it is not real.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It is a way to magically give direction to forces acting on objects.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is a diagram of a pulley http://www.sparknotes.com/physics/dynamics/newtonapplications/section3.rhtml

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it says specifically the pulley and cable are massless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you have a question, or are you just going to keep repeating the same thing and pasting examples from the book.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have read the book. I don't need to see it again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok if the pullies and cable are massless then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how can it have acceleration

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm done. Until you actually read what I have written I have no reason to continue to write.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so your saying the cable experiences no acceleration, ok i agree

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well if it is massless then it experiences no acceleration. the cable i mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm saying massless cables do not goddamn exist. If you want to talk about them then you have to obey very specific rules.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well actually kevlar ropes exist which can carry so much weight, that in comparison to the weight it is massless

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They are a figment of your imagination. A tool to make it easy for you to solve these problems without a hell of a lot of integrals.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok , so then my original problem was impossible, having a massless rope and 20 N left and 30 N right, force

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0On a rope by itself, yes.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Because F = ma therefore F/m = a and m = 0 so F/m is not defined.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the accelleration is not 0, it's exploding.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The only time it is ok to talk about external forces on massless objects is when those forces cancel out (a = 0)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok then can you think of an example where we want a massless rope?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The pulley example is fine.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0As long as you talk about the a of the masses, not the a of the rope.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0All you need to consider about the rope is that the tension is uniform.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the point of having a massless rope is to make or force the tension to be the same throughout , it simplifies the problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0even though intuitively, there are different forces on the ends of the rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so you might think there are two different tensions on the ends of the rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like in my example, there would be a tension of 20 N opposite to the guy pulling left 20N , and a tension of 30 opposite to the guy pulling to the right 30 N ... well i guess in this case we need to know the mass of the rope to determine the tension err, i mean .... the tension does not hold because it is accelerating , so there isnt tension , in the case where there is mass in the rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0look at the pully example. You have that 50*9.8  T = (50+30)a 30*9.8 + T = (50+30)a So we can find T, and we can find a because we assume that T and a are the same (one block accelerates up while the other accelerates down. We know the rope is accelerating at the same rate as the blocks. So there is a net force on the rope but the tension is uniform. Also the rope is not massless, we just ignore it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0supposing that in my example the mass of the rope is 1 kg. so a guy is pulling 20 N to the left and 30 N to the right someone else pulls.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the tension in the rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you're making up impossible scenarios again.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just to make it simple , 1 kg rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you assuming the rope stays taut?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh i see what you mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you assuming that the guys are standing on the ground and don't fall over?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does the ground have friction?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or with a pulley example , even

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its much more complicated

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok pulley, 50 kg moving down, 40 kg moving up. the cable is 1 kg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is the tension in the cable,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this will be my last example, sorry for keeping ya

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i was wondering if that made sense

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also ignore gravity for the cable,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do you calculate the tension of the rope , if it has mass 1 kg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The same was you would if it had no mass.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so the forces are 50* 9.8  40 * 9.8 , the forces on the cable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.050* 9.8  40 * 9.8 = 1kg * a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the net force on the cable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you do that the rope will fly away.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Stop obsessing about the stupid rope. It doesn't work like that in reality and you don't have the tools to look at it that way. Look at the masses.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We MUST assume that they all accelerate at the same rate. Otherwise we have a dynamics system and you don't have the PDE's to solve that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The blocks, and the ropes all move with the same a.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so dont treat the rope independently of the masses

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, because if you have a 40N force acting on a 1kg rope it will go shooting off into space.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And if the mass of the rope is smaller, it will fly away faster.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They just want you to ignore the rope, not make any calculations based on its mass or acceleration.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh actually its (5040)* 9.8=98 Newtons force on the cable, ok . bad way of looking at it, also for my other example, two guys 30 N and 20 N right and left respectively, rope is 1 kg. the tension is still 24.4444, doing it the right way. dont treat the cable seperately

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok got it. ignore the rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how come its not obvious the tension is 24.444

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the other guy assumed the tension was 20N.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so with my other rope example, it would be wrong to do (30  20) = 1kg * a , 10N/1kg = a , or 10 m/s^2. it will fly off , ok i see

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it would shoot off in the direction of the stronger person

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so this is why we dont deal with the rope directly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you say something like Two men are having a tug of war. One man (weighs 50kg) is pulling with a force of 20N, the other weighing 40kg is pulling with a force of 10N. What is the acceleration? and the tension in the rope.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Again you must assume the acceleration is a constant for the whole 2 men system

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If it werent then the tension in the rope would change

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 10 N in the right direction?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do 10 N = ( 40 + 50) * 1kg ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0errr, 10 N = (40 + 50) kg * a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually that's still a bad example because we have to assume some friction under their feet or they would just go toward each other. (and lose tension in the rope)

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0ok i don't know anything about physics but it sounds like cantorset is trying to use a theorem. And it sounds like you guys are saying just because the conclusion is true doesn't mean the is. p>q is not equivalent to q>p or are you guys talking about somtehing different. You might have already figure this out so sorry if you have. if you don't have any clue what i'm talking about ignore this post

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so a = 1/9 or .1111 m/s^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, that's what I was complaining about earlier myininaya. But I think we worked through it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i ignored the tension

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in fact i ignored the rope altogether

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I like the pully problem better ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This one still has too many unknowns.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can do 20 N  T = 50 * a T  10 N = 40 * a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh right, we solved for a , so 20 N  T = 50 * 1/9

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sure, that will tell you how fast they fly together.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And how tight the rope was at the instant they both pulled (before they start moving)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh right, so we would have friction , and...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Though it's not quite right because you did the forces wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes this only works for an instant at the beginning before they actually start moving

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Two men are having a tug of war. One man (weighs 50kg) is pulling with a force of 20N, the other weighing 40kg is pulling with a force of 10N. 20 N  T = 50 * a (force on bigger guy) T  10 N = 40 * a (force on smaller guy )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The net acceleration of the system will be 0, and you'll do stuff like this when you get to center of mass, but the system you have there doesn't correctly model what happens.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you dont agree the tension is 14.444 N , in the idealized case

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what do you mean net acceleration is zero. oh you mean that the people are staying in their places and not moving (so the string is taut)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Imagine for a moment that it's an astronaut(75kg) in space, pulling (20N) on a cord attached to a free floating 50 kg backpack. The astronaut pulls the rope. The rope pulls the astronaut.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why not 2 different astronauts in space

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'd rather start with this one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so whats the question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is tension in rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What is the acceleration of the backpack?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isnt it just 20 N = 50kg * a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And what is the acceleration of the astronaut?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we never said the astronaut is accelerating

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hes just pulling, so maybe zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he's floating in space, pulling on a cord attached to a 50kg backpack.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he's definitely gonna accelerate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because the bookbag pulls back on him ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is he exerting a force?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0he is pulling on a rope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then the rope must also be pulling on him

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right with equal and opposite force

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in the opposite direction to bookbag

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right, so they start flying toward eachother.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And the tension in the rope?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(for one instant anyway)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.020 N  T = 75 * a , and a = 20 /75

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You are listing the forces on the astronaut?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but there is tension in that force, so im solving for the tension

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What forces are on the astronaut?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we know astronaut is accelerating 20/75 m * s^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0duh, equal and opposite force

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now lets do the same thing, but with 2 astronauts.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait, whats the tension?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Name the forces on the astronaut. (don't bother with the number)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the forces on the astronaut are 20 N ....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes. And what is exerting that force?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ummm, just the tension ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the tension comes from the equal and opposite force

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so the tension is , instantaneously 20 N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we dont even have to solve for that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So now lets analyze two astronauts. If one is just floating and the other pulls, we have the same situation as we do in the backpack case right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so lets do both pulls >

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Will the tension in the rope be greater, the same, or less if both astronauts pull?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well if they both pull 20 N say

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Imagine only one pulls.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0They both fly together right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0like our bookbag case

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Will they fly together just as fast if they both pull, or will it be slower, or faster?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0without thinking about it, faster >?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it depends on the pulls, are they equal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0even if they are equal.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If they accelerate faster than they would if just one pulled, what can we say about the tension?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then the tension must be greater

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so lets say that they both pull 20 N force

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.020 N  T = 75 * a . and

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0they are both pulling on each other 20 N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, they are both pulling on the rope with 20N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the m1 = 75 kg, and m2 = 50 kg , the astronauts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but i will examine the forces on the astronauts, and glean out the tension

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0True, but the force on the astronaut from the rope is not 20N.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If it were, they would accelerate just as if only one was pulling.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 20 N is from the opposite reaction ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0theres only tension then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i keep thinking of the pulley example

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes For astronaut 1, the only force on them is the tension.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right theres no gravity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if we assume the tension is the same throughout, dunno

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the rope doesn't fly away.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how to calculate tension

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well we add the forces and the reaction forces

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 2 * 20 + 20/75 + 50/75 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it's easier than that.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you familiar with hanging scales?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you know like hanging scales in a grocery store?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what was wrong with my analysis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok well it tells how much the thing in the basket weighs by the tension pulling down.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If I put a 1kg mellon in the basket the scale will read 1kg. If I push down on the scale with a 9.8N of force what will the scale read?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02kg actually. but that's ok. Point is my force is added to the tension from the mellon

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean if you push down with the melon in the scale, oh

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, 1 kg is getting pushed down with 9.8 N of force

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and if you push down 9.8 N and keep the force there, its like adding another kg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so how does that help us with the tension when two astronauts are pulling on each other 20 N

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If I accelerate the scale upwards at a rate of 9.8m/s^2 the scale will also read 2kg

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how can you accelerate a scale upwards?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we're talking about a scale in a store. it doesnt register upward forces

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh you mean downwards, same thing, ok

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0upwards meaning pushing towards the ground, ok

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No I mean upwards because accelerating the scale up will change the tension in the chain

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no I mean blasting the whole apparatus into the sky.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and the melon is inside the scale ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I'm just confusing you further here. The point with the astronauts is this. If they both pull, the tension in the rope will be the sum of their individual tensions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and its not 20 + 20 + 20/75 + 20 / 50 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, it's 20 + 20 = 40

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what about those reaction forces

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the newtons third law pairs

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that effects the astronauts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's no good without pictures

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the astronauts are accelerating, one is accelerating 20/75 and other is accelerating 20/50 m/s^2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.040/75 and 40/50, because the reaction to tension

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm, and why do you add the tensions to get 40
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