anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
You 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
  • anonymous
I'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
  • anonymous
hi

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anonymous
  • anonymous
well i was using my book. it says the forces on the rope are F1 - F2 = 0. isnt tension the force on the rope?
anonymous
  • anonymous
What are you being asked to find?
anonymous
  • anonymous
A 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
  • anonymous
Here 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
  • anonymous
I agree with that.
anonymous
  • anonymous
well 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
  • anonymous
What 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
  • anonymous
in my example, one guy pulls 30 N right and the other guy pulls 20 N left
anonymous
  • anonymous
I don't see where you got 30N and 20N
anonymous
  • anonymous
i made up an example
anonymous
  • anonymous
In your example, there would be acceleration towards the guy pulling with 30N.
anonymous
  • anonymous
you mean 10 N
anonymous
  • anonymous
There would be a net force of 10N which would result in an acceleration in the direction of the guy pulling with 30N force.
anonymous
  • anonymous
Yes, but you are contradicting the hypothisis
anonymous
  • anonymous
my 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
  • anonymous
They are saying that If the rope is not accelerating, we know the forces on both ends are equal
anonymous
  • anonymous
how do you get the forces at both ends are equal, and what is this force? the tension?
anonymous
  • anonymous
and which force is it, what is exerting on what
anonymous
  • anonymous
you 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
  • anonymous
No, the force on the rope.
anonymous
  • anonymous
but there are different forces, one is 30 N and one is 20 N
anonymous
  • anonymous
which we stipulated
anonymous
  • anonymous
Yes, you made up that example, where the conclusion is false, and trying to make a claim about the implication.
anonymous
  • anonymous
the conclusion is false?
anonymous
  • anonymous
you 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
  • anonymous
So the acceleration is not 0
anonymous
  • anonymous
wait, youre saying the net force on both ends is equal in magnitude?
anonymous
  • anonymous
but my rope is massless
anonymous
  • anonymous
I'm saying IF the acceleration is 0, then the NET force on the rope is 0.
anonymous
  • anonymous
why does the book say F1-F2 = 0, so F1 = F2,
anonymous
  • anonymous
You can't talk about force acting on massless things.
anonymous
  • anonymous
what is F1 and F2 ?
anonymous
  • anonymous
It says F1-F2 is 0 BECAUSE it claimed already that the acceleration is 0
anonymous
  • anonymous
so then F1 and F2 are zero?
anonymous
  • anonymous
No.
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
is tension the F1 and F2
anonymous
  • anonymous
\[\sum F = ma\]
anonymous
  • anonymous
If a =0 then \[\sum F = F1-F2 = 0 \implies F1 = F2\]
anonymous
  • anonymous
But if the forces do not sum to 0 than the acceleration of the rope will not be 0.
anonymous
  • anonymous
yes but in my example, what is F1 and F2
anonymous
  • anonymous
is the acceleration equal to 0?
anonymous
  • anonymous
nomo, if you cant type , you should restart.
anonymous
  • anonymous
www.openstudy.com start over
anonymous
  • anonymous
A 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
  • anonymous
lets see
anonymous
  • anonymous
i get a different answer
anonymous
  • anonymous
30 N - T = 50kg * a -20 N + T = 40 kg * a
anonymous
  • anonymous
Heh. Cantorset, you are in fact correct.
anonymous
  • anonymous
Those are the equations of motion, and I should have written them out before posting.
anonymous
  • anonymous
so 10 N = 90 kg *a , so a = 1/9 m*s^-2
anonymous
  • anonymous
Yes, I agree.
anonymous
  • anonymous
So what is T?
anonymous
  • anonymous
one sec, my browser is crashing
anonymous
  • anonymous
weird, one sec
anonymous
  • anonymous
plug back in
anonymous
  • anonymous
so 30 - 50/9 N = T
anonymous
  • anonymous
24.444 N is the tension
anonymous
  • anonymous
so natho was wrong
anonymous
  • anonymous
The 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
  • anonymous
so what is F1 and F2, i still dont see
anonymous
  • anonymous
So your acceleration is not constant.
anonymous
  • anonymous
F1 and F2 are the tension forces?
anonymous
  • anonymous
F1 is the force of one person pulling, F2 is the force of the other.
anonymous
  • anonymous
but thats false, because 30 - 20 is not 0
anonymous
  • anonymous
Who said it was = 0?
anonymous
  • anonymous
is the a = 0?
anonymous
  • anonymous
The 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
  • anonymous
nomo, so do you agree that 24.444 is the tension?
anonymous
  • anonymous
pol, the book answer
anonymous
  • anonymous
In the instant when they both pull, (and in the absence of any friction) The tension is 24.444
anonymous
  • anonymous
Is the acceleration given to be 0?
anonymous
  • anonymous
"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
  • anonymous
No, but the rope is given to be massless
anonymous
  • anonymous
So the acceleration is 0.
anonymous
  • anonymous
But that's not a problem, it's a lesson.
anonymous
  • anonymous
You CANNOT have forces acting on massless things.
anonymous
  • anonymous
Your 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
  • anonymous
hmmm, so the book is contradicting itself?
anonymous
  • anonymous
No.
anonymous
  • anonymous
You are trying to contradict the book
anonymous
  • anonymous
But not making a reasoned argument.
anonymous
  • anonymous
No, 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
  • anonymous
as if it were 0.
anonymous
  • anonymous
ok , i know the acceleration is not zero
anonymous
  • anonymous
So then the book doesn't apply anymore.
anonymous
  • anonymous
Then the forces cannot be the same.
anonymous
  • anonymous
ok 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
  • anonymous
Therefore the F1 = F2
anonymous
  • anonymous
If you contradict by saying \[F1 \ne F2\] then \[ma \ne 0.\]
anonymous
  • anonymous
the end.
anonymous
  • anonymous
but didnt we stipulate earlier that F1 = 30 N and F2 = 20 N
anonymous
  • anonymous
You cannot say that ma = 0 if f1 not equal f2
anonymous
  • anonymous
The sum of the forces IS ma.
anonymous
  • anonymous
ok, hmmmm
anonymous
  • anonymous
If the sum is non-zero ma is non-zero. If ma is 0 the sum is 0. If the sum is 0 ma is 0. If ma is non-zero the sum is non-zero.
anonymous
  • anonymous
but in my problem, i started with 30 N and 20 N , and they applied the forces at either end, correct?
anonymous
  • anonymous
You can *only* assume the rope is massless *IF* F1=F2. Otherwise, the discussion in the book doesn't apply.
anonymous
  • anonymous
So while there are real-world 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
  • anonymous
but didnt we assume that the two guys are applying two different forces
anonymous
  • anonymous
We did. But then you tried to use what the book told you. You can't do that.
anonymous
  • anonymous
If forces are non-zero then ma is non-zero.
anonymous
  • anonymous
Pick one or the other, but you can't just pick both to be different.
anonymous
  • anonymous
ok let me try again, one sec
anonymous
  • anonymous
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
  • anonymous
You can't do that.
anonymous
  • anonymous
You 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
  • anonymous
You can't assume the rope is massless.
anonymous
  • anonymous
they do that in problems though, i thought
anonymous
  • anonymous
No.
anonymous
  • anonymous
They do that when the forces are equal.
anonymous
  • anonymous
You 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
  • anonymous
i dont understand , what is wrong with my problem
anonymous
  • anonymous
You have ma = 0, but the sum of the forces is not 0.
anonymous
  • anonymous
You cannot do that.
anonymous
  • anonymous
If it is not accelerating then there is no net force on the object.
anonymous
  • anonymous
Things do not accelerate in the absence of a net force.
anonymous
  • anonymous
im confused
anonymous
  • anonymous
Things with net forces acting on them MUST accelerate.
anonymous
  • anonymous
so what exactly is the contradiction in the question
anonymous
  • anonymous
nothing 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
  • anonymous
You 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
  • anonymous
We are telling you THERE IS SOMETHING WRONG with that assumption.
anonymous
  • anonymous
ok
anonymous
  • anonymous
so what is the force on the rope
anonymous
  • anonymous
Because what happens in reality is that ma is not 0, You will get a tension, the tension will pull the two people together
anonymous
  • anonymous
oh you said there is no force on the rope, since its massless
anonymous
  • anonymous
I'm not saying anything. You are choosing two sides of the equation that are non-equal
anonymous
  • anonymous
wait, you cant have tension with a massless rope?
anonymous
  • anonymous
Which do you want, a rope that doesn't accelerate? or a net force?
anonymous
  • anonymous
Ive seen many tension problems where we assume the mass of the rope is negligible
anonymous
  • anonymous
Was the rope accelerating?
anonymous
  • anonymous
i dont know
anonymous
  • anonymous
I do.
anonymous
  • anonymous
one sec, let me get a problem
anonymous
  • anonymous
Good.
anonymous
  • anonymous
An 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
  • anonymous
http://www.sparknotes.com/physics/dynamics/newtonapplications/section3.rhtml
anonymous
  • anonymous
Right. 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
  • anonymous
so 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
  • anonymous
then the force exerted by the rope on the two people must be equal
anonymous
  • anonymous
yes
anonymous
  • anonymous
but then the two people, what about the force of the two people exerting on the rope , or the forces on the rope
anonymous
  • anonymous
The net force they exert on the rope must be 0.
anonymous
  • anonymous
but they can have different forces as well
anonymous
  • anonymous
Yes
anonymous
  • anonymous
but then by newtons third law, then that would force them to have the same force
anonymous
  • anonymous
the reaction force
anonymous
  • anonymous
Wait, no
anonymous
  • anonymous
They cannot be exerting different forces.
anonymous
  • anonymous
If they were then the sum would not be 0.
anonymous
  • anonymous
ok lets say we have a pulley and two masses
anonymous
  • anonymous
Ok.
anonymous
  • anonymous
one mass is 50 kg, the other mass is 30 kg
anonymous
  • anonymous
m1 = 50 kg, m2 = 30 kg
anonymous
  • anonymous
they go up and down, like vertically
anonymous
  • anonymous
yes
anonymous
  • anonymous
simple pulley. the force on m1 are 9.8 *50 - T , where + is down direction
anonymous
  • anonymous
Certainly.
anonymous
  • anonymous
the force on m2 is T - 30 *9.8
anonymous
  • anonymous
but the problem says the mass of the cable and the pulley is neglible
anonymous
  • anonymous
that means don't try to use it for finding the mass
anonymous
  • anonymous
, wait, you said then the acceleration is zero
anonymous
  • anonymous
but the pulley is accelerating, i mean the cable is accelerating
anonymous
  • anonymous
You just keep talking in circles. What exactly is the question?
anonymous
  • anonymous
the force on the cable should be zero
anonymous
  • anonymous
since it is massless, but there is force on the cable
anonymous
  • anonymous
It's not massless.
anonymous
  • anonymous
thats what the problem states
anonymous
  • anonymous
It says ignore the mass, not that it's massless.
anonymous
  • anonymous
In 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
  • anonymous
i think it says the mass is zero
anonymous
  • anonymous
It'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
  • anonymous
it says let it be zero
anonymous
  • anonymous
I'm not going to continue this further. If you want to argue further you can contact the authors, or your professor.
anonymous
  • anonymous
If 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
  • anonymous
What is the acceleration of a massless object with a force of 1 newton acting on it?
anonymous
  • anonymous
here i will show you a link , one sec http://www.wadsworthmedia.com/marketing/sample_chapters/0495106194_ch04.pdf page 70
anonymous
  • anonymous
I don't disagree with what they state there.
anonymous
  • anonymous
All 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
  • anonymous
in reality that cannot happen.
anonymous
  • anonymous
But if you violate the precepts of the 'trick' then you cannot use the rules they infer from it.
anonymous
  • anonymous
but if the tension is the same, then it wont move
anonymous
  • anonymous
I'm not talking about the external force on the rope.
anonymous
  • anonymous
the tension is the internal force.
anonymous
  • anonymous
The point is that it is not slack, it does not break, and it is not elastic.
anonymous
  • anonymous
You 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
  • anonymous
It's a tool, it is not real.
anonymous
  • anonymous
It is a way to magically give direction to forces acting on objects.
anonymous
  • anonymous
here is a diagram of a pulley http://www.sparknotes.com/physics/dynamics/newtonapplications/section3.rhtml
anonymous
  • anonymous
it says specifically the pulley and cable are massless
anonymous
  • anonymous
Do you have a question, or are you just going to keep repeating the same thing and pasting examples from the book.
anonymous
  • anonymous
I have read the book. I don't need to see it again.
anonymous
  • anonymous
ok if the pullies and cable are massless then
anonymous
  • anonymous
how can it have acceleration
anonymous
  • anonymous
I'm done. Until you actually read what I have written I have no reason to continue to write.
anonymous
  • anonymous
so your saying the cable experiences no acceleration, ok i agree
anonymous
  • anonymous
I'm not saying that.
anonymous
  • anonymous
well if it is massless then it experiences no acceleration. the cable i mean
anonymous
  • anonymous
I'm saying massless cables do not goddamn exist. If you want to talk about them then you have to obey very specific rules.
anonymous
  • anonymous
right, its idealized
anonymous
  • anonymous
well actually kevlar ropes exist which can carry so much weight, that in comparison to the weight it is massless
anonymous
  • anonymous
They 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
  • anonymous
ok , so then my original problem was impossible, having a massless rope and 20 N left and 30 N right, force
anonymous
  • anonymous
On a rope by itself, yes.
anonymous
  • anonymous
Because F = ma therefore F/m = a and m = 0 so F/m is not defined.
anonymous
  • anonymous
right i agree
anonymous
  • anonymous
the accelleration is not 0, it's exploding.
anonymous
  • anonymous
The only time it is ok to talk about external forces on massless objects is when those forces cancel out (a = 0)
anonymous
  • anonymous
ok then can you think of an example where we want a massless rope?
anonymous
  • anonymous
The pulley example is fine.
anonymous
  • anonymous
As long as you talk about the a of the masses, not the a of the rope.
anonymous
  • anonymous
ok
anonymous
  • anonymous
All you need to consider about the rope is that the tension is uniform.
anonymous
  • anonymous
so the point of having a massless rope is to make or force the tension to be the same throughout , it simplifies the problem
anonymous
  • anonymous
yes
anonymous
  • anonymous
even though intuitively, there are different forces on the ends of the rope
anonymous
  • anonymous
so you might think there are two different tensions on the ends of the rope
anonymous
  • anonymous
like 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
  • anonymous
look 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
  • anonymous
supposing 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
  • anonymous
what is the tension in the rope
anonymous
  • anonymous
you're making up impossible scenarios again.
anonymous
  • anonymous
just to make it simple , 1 kg rope
anonymous
  • anonymous
Are you assuming the rope stays taut?
anonymous
  • anonymous
yes
anonymous
  • anonymous
oh i see what you mean
anonymous
  • anonymous
Are you assuming that the guys are standing on the ground and don't fall over?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Does the ground have friction?
anonymous
  • anonymous
or with a pulley example , even
anonymous
  • anonymous
yes
anonymous
  • anonymous
so its much more complicated
anonymous
  • anonymous
ok pulley, 50 kg moving down, 40 kg moving up. the cable is 1 kg
anonymous
  • anonymous
what is the tension in the cable,
anonymous
  • anonymous
this will be my last example, sorry for keeping ya
anonymous
  • anonymous
i was wondering if that made sense
anonymous
  • anonymous
Yes, that's fine.
anonymous
  • anonymous
also ignore gravity for the cable,
anonymous
  • anonymous
yes
anonymous
  • anonymous
how do you calculate the tension of the rope , if it has mass 1 kg
anonymous
  • anonymous
The same was you would if it had no mass.
anonymous
  • anonymous
ok so the forces are 50* 9.8 - 40 * 9.8 , the forces on the cable
anonymous
  • anonymous
No.
anonymous
  • anonymous
50* 9.8 - 40 * 9.8 = 1kg * a
anonymous
  • anonymous
Stop that
anonymous
  • anonymous
the net force on the cable
anonymous
  • anonymous
No.
anonymous
  • anonymous
If you do that the rope will fly away.
anonymous
  • anonymous
Stop 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
  • anonymous
We 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
  • anonymous
The blocks, and the ropes all move with the same a.
anonymous
  • anonymous
ok
anonymous
  • anonymous
ok so dont treat the rope independently of the masses
anonymous
  • anonymous
Yes, because if you have a 40N force acting on a 1kg rope it will go shooting off into space.
anonymous
  • anonymous
And if the mass of the rope is smaller, it will fly away faster.
anonymous
  • anonymous
They just want you to ignore the rope, not make any calculations based on its mass or acceleration.
anonymous
  • anonymous
oh actually its (50-40)* 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
  • anonymous
ok got it. ignore the rope
anonymous
  • anonymous
Yes, exactly.
anonymous
  • anonymous
how come its not obvious the tension is 24.444
anonymous
  • anonymous
the other guy assumed the tension was 20N.
anonymous
  • anonymous
so 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
  • anonymous
it would shoot off in the direction of the stronger person
anonymous
  • anonymous
right.
anonymous
  • anonymous
interesting
anonymous
  • anonymous
so this is why we dont deal with the rope directly
anonymous
  • anonymous
If 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
  • anonymous
ok good question,
anonymous
  • anonymous
Again you must assume the acceleration is a constant for the whole 2 men system
anonymous
  • anonymous
If it werent then the tension in the rope would change
anonymous
  • anonymous
so 10 N in the right direction?
anonymous
  • anonymous
you can do 10 N = ( 40 + 50) * 1kg ?
anonymous
  • anonymous
errr, 10 N = (40 + 50) kg * a
anonymous
  • anonymous
Actually 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
  • myininaya
ok 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
  • anonymous
so a = 1/9 or .1111 m/s^2
anonymous
  • anonymous
yeah, that's what I was complaining about earlier myininaya. But I think we worked through it.
anonymous
  • anonymous
but i ignored the tension
anonymous
  • anonymous
in fact i ignored the rope altogether
anonymous
  • anonymous
I like the pully problem better ;)
anonymous
  • anonymous
This one still has too many unknowns.
anonymous
  • anonymous
you can do 20 N - T = 50 * a T - 10 N = 40 * a
anonymous
  • anonymous
oh right, we solved for a , so 20 N - T = 50 * 1/9
anonymous
  • anonymous
T = 14.4444
anonymous
  • anonymous
Sure, that will tell you how fast they fly together.
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
And how tight the rope was at the instant they both pulled (before they start moving)
anonymous
  • anonymous
oh right, so we would have friction , and...
anonymous
  • anonymous
Though it's not quite right because you did the forces wrong
anonymous
  • anonymous
yes this only works for an instant at the beginning before they actually start moving
anonymous
  • anonymous
i did ?
anonymous
  • anonymous
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. 20 N - T = 50 * a (force on bigger guy) T - 10 N = 40 * a (force on smaller guy )
anonymous
  • anonymous
The 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
  • anonymous
you dont agree the tension is 14.444 N , in the idealized case
anonymous
  • anonymous
what 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
  • anonymous
Imagine 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
  • anonymous
This will work.
anonymous
  • anonymous
why not 2 different astronauts in space
anonymous
  • anonymous
I'd rather start with this one.
anonymous
  • anonymous
ok so whats the question?
anonymous
  • anonymous
what is tension in rope
anonymous
  • anonymous
What is the acceleration of the backpack?
anonymous
  • anonymous
isnt it just 20 N = 50kg * a
anonymous
  • anonymous
20 /50 m s^-2
anonymous
  • anonymous
yep.
anonymous
  • anonymous
ok good
anonymous
  • anonymous
And what is the acceleration of the astronaut?
anonymous
  • anonymous
we never said the astronaut is accelerating
anonymous
  • anonymous
hes just pulling, so maybe zero
anonymous
  • anonymous
he's floating in space, pulling on a cord attached to a 50kg backpack.
anonymous
  • anonymous
he's definitely gonna accelerate.
anonymous
  • anonymous
because the bookbag pulls back on him ?
anonymous
  • anonymous
is he exerting a force?
anonymous
  • anonymous
he is pulling on a rope
anonymous
  • anonymous
then the rope must also be pulling on him
anonymous
  • anonymous
right with equal and opposite force
anonymous
  • anonymous
so 20/75 m*s^-2
anonymous
  • anonymous
in the opposite direction to bookbag
anonymous
  • anonymous
20 N = 75 kg * a
anonymous
  • anonymous
Right, so they start flying toward eachother.
anonymous
  • anonymous
right
anonymous
  • anonymous
And the tension in the rope?
anonymous
  • anonymous
hmmm
anonymous
  • anonymous
(for one instant anyway)
anonymous
  • anonymous
oh
anonymous
  • anonymous
one sec
anonymous
  • anonymous
20 N - T =
anonymous
  • anonymous
20 N - T = 75 * a , and a = 20 /75
anonymous
  • anonymous
so T = 40 N ?
anonymous
  • anonymous
Hold up.
anonymous
  • anonymous
You are listing the forces on the astronaut?
anonymous
  • anonymous
yes, but there is tension in that force, so im solving for the tension
anonymous
  • anonymous
What forces are on the astronaut?
anonymous
  • anonymous
oh ,
anonymous
  • anonymous
lets see
anonymous
  • anonymous
we know astronaut is accelerating 20/75 m * s^-2
anonymous
  • anonymous
20 N
anonymous
  • anonymous
duh, equal and opposite force
anonymous
  • anonymous
right.
anonymous
  • anonymous
Now lets do the same thing, but with 2 astronauts.
anonymous
  • anonymous
so 20 N - T = 75* a
anonymous
  • anonymous
wait, whats the tension?
anonymous
  • anonymous
No.
anonymous
  • anonymous
Name the forces on the astronaut. (don't bother with the number)
anonymous
  • anonymous
the forces on the astronaut are 20 N ....
anonymous
  • anonymous
yes. And what is exerting that force?
anonymous
  • anonymous
ummm, just the tension ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and the tension comes from the equal and opposite force
anonymous
  • anonymous
yes
anonymous
  • anonymous
ok so the tension is , instantaneously 20 N
anonymous
  • anonymous
yes
anonymous
  • anonymous
we dont even have to solve for that
anonymous
  • anonymous
right.
anonymous
  • anonymous
So 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
  • anonymous
ok 2 astronauts
anonymous
  • anonymous
yes
anonymous
  • anonymous
so lets do both pulls >
anonymous
  • anonymous
Will the tension in the rope be greater, the same, or less if both astronauts pull?
anonymous
  • anonymous
i dont know
anonymous
  • anonymous
hmmm, less ?
anonymous
  • anonymous
well if they both pull 20 N say
anonymous
  • anonymous
Hold up.
anonymous
  • anonymous
Imagine only one pulls.
anonymous
  • anonymous
ok
anonymous
  • anonymous
They both fly together right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
like our bookbag case
anonymous
  • anonymous
Will they fly together just as fast if they both pull, or will it be slower, or faster?
anonymous
  • anonymous
dont know
anonymous
  • anonymous
without thinking about it, faster >?
anonymous
  • anonymous
it depends on the pulls, are they equal
anonymous
  • anonymous
even if they are equal.
anonymous
  • anonymous
oh much faster
anonymous
  • anonymous
indeed
anonymous
  • anonymous
If they accelerate faster than they would if just one pulled, what can we say about the tension?
anonymous
  • anonymous
then the tension must be greater
anonymous
  • anonymous
Also correct.
anonymous
  • anonymous
so lets say that they both pull 20 N force
anonymous
  • anonymous
ok.
anonymous
  • anonymous
20 N - T = 75 * a . and
anonymous
  • anonymous
hold on
anonymous
  • anonymous
What is that 20N?
anonymous
  • anonymous
they are both pulling on each other 20 N
anonymous
  • anonymous
No, they are both pulling on the rope with 20N
anonymous
  • anonymous
so the m1 = 75 kg, and m2 = 50 kg , the astronauts
anonymous
  • anonymous
yes but i will examine the forces on the astronauts, and glean out the tension
anonymous
  • anonymous
indirectly
anonymous
  • anonymous
True, but the force on the astronaut from the rope is not 20N.
anonymous
  • anonymous
If it were, they would accelerate just as if only one was pulling.
anonymous
  • anonymous
the 20 N is from the opposite reaction ,
anonymous
  • anonymous
ohhh
anonymous
  • anonymous
theres only tension then
anonymous
  • anonymous
i keep thinking of the pulley example
anonymous
  • anonymous
yes For astronaut 1, the only force on them is the tension.
anonymous
  • anonymous
right theres no gravity
anonymous
  • anonymous
so if we assume the tension is the same throughout, dunno
anonymous
  • anonymous
the rope doesn't fly away.
anonymous
  • anonymous
how to calculate tension
anonymous
  • anonymous
well we add the forces and the reaction forces
anonymous
  • anonymous
yes.
anonymous
  • anonymous
so 2 * 20 + 20/75 + 50/75 ?
anonymous
  • anonymous
no it's easier than that.
anonymous
  • anonymous
Are you familiar with hanging scales?
anonymous
  • anonymous
so i did it wrong >
anonymous
  • anonymous
yeah
anonymous
  • anonymous
you know like hanging scales in a grocery store?
anonymous
  • anonymous
yes
anonymous
  • anonymous
what was wrong with my analysis
anonymous
  • anonymous
Ok well it tells how much the thing in the basket weighs by the tension pulling down.
anonymous
  • anonymous
right
anonymous
  • anonymous
If 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
  • anonymous
9.8 kg ?
anonymous
  • anonymous
2kg actually. but that's ok. Point is my force is added to the tension from the mellon
anonymous
  • anonymous
you mean if you push down with the melon in the scale, oh
anonymous
  • anonymous
right, 1 kg is getting pushed down with 9.8 N of force
anonymous
  • anonymous
and if you push down 9.8 N and keep the force there, its like adding another kg
anonymous
  • anonymous
right.
anonymous
  • anonymous
so how does that help us with the tension when two astronauts are pulling on each other 20 N
anonymous
  • anonymous
If I accelerate the scale upwards at a rate of 9.8m/s^2 the scale will also read 2kg
anonymous
  • anonymous
how can you accelerate a scale upwards?
anonymous
  • anonymous
a rocket? my arms?
anonymous
  • anonymous
a pully?
anonymous
  • anonymous
we're talking about a scale in a store. it doesnt register upward forces
anonymous
  • anonymous
oh you mean downwards, same thing, ok
anonymous
  • anonymous
upwards meaning pushing towards the ground, ok
anonymous
  • anonymous