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