50N from left, 50N from right. Tension in string still 50N?

[Ocata]

Let's say I go to the store and purchase a rope. On the package of string, it says that the string can only withstand 100 Newtons before the rope snaps. I go home and test the tension in the rope. I tell a friend to pull on the rope at 99 Newtons while I pull on the rope with 99 Newtons. Did I exceed the recommended tension? If the labeling on the package is correct, will the rope snap because there is some force within the string that causing the rope to experience more than 99 Newtons in the horizontal directions (excluding gravity)? Not sure what stress and strain are yet, but could I be confusing tension with some other concept that I have yet to learn about. I feel that tension in the string should somehow be as strong as the sum of the forces acting on it.

 

[A.T.]

I feel that tension in the string should somehow be as strong as the sum of the forces acting on it.

The sum of forces acting on the string is zero, if you both pull with equal but opposite forces.

 

[rumborak]

EDIT: Never mind.

 

[A.T.]

the tension on the rope is double of the initial scenario however.

The tension is 99N if both pull with 99N. Not sure what is "double" here?

 

[robbertypob]

Here's what I think.

If you are holding one end of the string and standing still, your friend pulling at 99N will exert 99N of force on the string. If you then start pulling at 99N in the opposite direction, the total force on the string is 198N. The forces are equal and opposite so the string won't move anywhere but it may well snap as you are over the 100N limit.

 

[A.T.]

...the total force on the string is 198N. ...but it may well snap as you are over the 100N limit.

Nope.

 

[rumborak]

EDIT: Hmm, I'm confused myself now. After all, the scenario is static.

 

[A.T.]

the scenario Ocata is describing is similar to the pulley setup on that page.

That is how I understand the OP:

From:
http://physics101samimuller.blogspot.de/2014/12/tension-force.html

You have to add both forces to arrive at the tension.

No. The sum of forces is zero, but the tension is 99N

 

[rumborak]

Yeah, I edited my post. Given how the scenario is static, whether a wall is holding it in place, or something else pulling with the same force, comes down to the same thing.

 

[Ocata]

Thank you. If I may, suppose my friend, standing on the left side of the string, continues to pull with a force of 99 N while I, on the right side of the string increase my pull to 100 N. Does the string have 99 Newtons on Tension, 100 Newtons, or a combination? Does my friend on the left feel 99 N of force pulling or 100? Do I feel 100 or 99 N?

 

[A.T.]

If I may, suppose my friend, standing on the left side of the string, continues to pull with a force of 99 N while I, on the right side of the string increase my pull to 100 N. Does the string have 99 Newtons on Tension, 100 Newtons, or a combination?

It will have a tension gradient, going from 100N on your side, to 99N on his side. An the whole thing isn't static anymore, because the forces don't balance.

Does my friend on the left feel 99 N of force pulling or 100? Do I feel 100 or 99 N?

The string pulls on you with same but opposite force as you pull on the string. Same for your friend.

 

[Ocata]

Would the attached image provide an accurate representation of the description you provided?

If so, then what, may I ask, does the rope feel in the middle? Suppose you choose a point exactly in the middle of the rope and measure its tension. Would it be 100 N, 99 N, or 100 N in one direction and 99 in the other direction?

 

[tony873004]

If you pull with 99 N, your friend has no choice but to pull with 99 N. That's Newton's 3rd Law. Your friend could be a doorknob. Tie the other end of the rope to a doorknob. If you pull with 99 N, the doorknob must pull with 99 N.
This reminds me of the tug-of-war question. Which team wins? Most would say the team that pulls hardest. No! According to Newton's 3rd Law, they must both pull with the same force. The team that wins is the team that pushes the ground hardest.

 

[Ocata]

If you pull with 99 N, your friend has no choice but to pull with 99 N. That's Newton's 3rd Law. Your friend could be a doorknob. Tie the other end of the rope to a doorknob. If you pull with 99 N, the doorknob must pull with 99 N.
This reminds me of the tug-of-war question. Which team wins? Most would say the team that pulls hardest. No! According to Newton's 3rd Law, they must both pull with the same force. The team that wins is the team that pushes the ground hardest.

Of course Tony. If the doorknob can only pull with 99 N and I pull with 100 N, the doorknob will have to accelerate at f=ma. That would be 1m/s^2 if the doorknob is 1kg.

And to expand on this using two people to vary the pulling forces,

If we suppose friction on the ground is included in the force of the pulls and the puller on the left pulls with 99 N of force and the puller on the right pulls with 100 N, there will be a net force on the rope of 1 N, right? So the rope would accelerate at F/m = a. Suppose the rope is 1 kg. The rope will accelerate at 1N/1kg = 1 m/s^2 relative to the ground. If the puller on the left, supposing he continues to grasp the rope, will have no choice but to "slide" or accelerate in some way at 1m/s^2 along with the left end of the rope, yes?

 

[Ocata]

These tension forces can be a bit confusing.

 

[A.T.]

If you pull with 99 N, your friend has no choice but to pull with 99 N.

Only for a static situation or if we consider the rope to be mass-less.

 

[A.T.]

Would the attached image provide an accurate representation of the description you provided?

Placing the force arrows is difficult if you don't separate the bodies like shown here in step 3:
http://adaptivemap.ma.psu.edu/websites/structures/analysis_of_frames_and_machines/analysisofframesandmachines.html

Suppose you choose a point exactly in the middle of the rope and measure its tension.

Assuming a uniform mass distribution in the rope it would be 99.5N.

If we suppose friction on the ground is included in the force of the pulls and the puller on the left pulls with 99 N of force and the puller on the right pulls with 100 N, there will be a net force on the rope of 1 N, right? So the rope would accelerate at F/m = a. Suppose the rope is 1 kg. The rope will accelerate at 1N/1kg = 1 m/s^2 relative to the ground.

Yes.

If the puller on the left, supposing he continues to grasp the rope, will have no choice but to "slide" or accelerate in some way at 1m/s^2 along with the left end of the rope, yes?

Initially he could just extend his arm, but beyond that he will fall over or slide if he fails to provide 100N.

 

[NascentOxygen]

Of course Tony. If the doorknob can only pull with 99 N and I pull with 100 N, the doorknob will have to accelerate at f=ma. That would be 1m/s^2 if the doorknob is 1kg.

And to expand on this using two people to vary the pulling forces,

If we suppose friction on the ground is included in the force of the pulls and the puller on the left pulls with 99 N of force and the puller on the right pulls with 100 N, there will be a net force on the rope of 1 N, right? So the rope would accelerate at F/m = a. Suppose the rope is 1 kg. The rope will accelerate at 1N/1kg = 1 m/s^2 relative to the ground. If the puller on the left, supposing he continues to grasp the rope, will have no choice but to "slide" or accelerate in some way at 1m/s^2 along with the left end of the rope, yes?

To determine the acceleration, you need to use the combined masses of the rope and the person/object fastened to its other end since these move as one.

 

[Ocata]

A.T, thanks, I didn't see post #11 last night. It's very informative.

 

[Merlin3189]

If you are holding one end of the string and standing still, your friend pulling at 99N will exert 99N of force on the string. If you then start pulling at 99N in the opposite direction, the total force on the string is 198N...

I think this has been dismissed? Though I find these discussions a bit hard to follow sometimes.

But what I wanted to say was, before the second sentence, before you start pulling on the rope, what happens to it? I guess it will start to uncoil as he moves away from you, as he will have to if he is to apply 99N, but if it's a heavy rough rope perhaps eventually the friction of the ground on the rope will exceed 99N and stop him. (If it's not heavy enough, he pulls out all the rope and continues accelerating into the distance, so that you can never exert your pull because the rope has gone!)

Now when you start pulling on the other end in the opposite direction, it depends how long the rope is. If he had pulled out less than half before the friction stopped him, then the same will happen for you and you will be stopped before you have pulled out half the rope. Each of you will be pulling with 99N against the static friction of your portion of rope and there will be a slack pile of rope remaining in the middle.
If he had pulled out exactly half, then the situation would be the same, but without any slack in the middle: you would still each be pulling against the static friction of half the rope and there would be no tension exactly at the centre. The tension in the rope will be zero at the centre and rise gradually away from the centre to reach a maximum of 99N at each end. ( If the ends of the rope are lifted from the ground, there may be a variation in the rate of change of tension, but the tension at centre and end are unaffected .)

If he had pulled out more than half the rope, then you will start pulling on "his" part of the rope before you stop. The zero tension point will disappear and there will be tension in all parts of the rope, with the minimum position moving towards the centre. You will stop as this minimum reaches the centre and the situation is then symmetric with each of you pulling with 99N, the maximum tension at the ends, reducing gradually thanks to friction where the rope is still in contact with the ground and to gravity where you have lifted the rope from the ground. The tension in the centre will always be less than 99N in this way of looking at it.

If you start taking account of the elasticity of the rope and the mass of you and your friend, then you may get some oscillations which could vary the tension above and below the static values. But then you would have to have some spare rope, so that you can let the rope slip to limit your tug to 99N when it tries to accelerate your mass, and it gets a bit more complicated once you start looking dynamically.

That is rather more than I intended to say, when I wondered what would happen if we paused after the first sentence!

The "total force on the string" seems to be lacking the weight, any forces from the ground and any force due to air pressure. If you add up the sum of the magnitudes of all the forces on the string, I guess it could add up to more than 198N. Maybe "total force" is not a very useful concept?

If you pull with 99 N, your friend has no choice but to pull with 99 N.

Only for a static situation or if we consider the rope to be mass-less.

Well I hope we are sticking with the static situation. But I don't see why "mass-less"? If the rope is heavy, but suspended and static, then the situation is symmetric and the force equal (though not opposite) at each end. If the rope rests at least partially on the floor, then your friend has the option of leaving some of the pulling to friction! But gravity will not preferentially aid him, unless you are higher than he.

As far as "no choice" goes, I think this may be a barber not shaving himself. Unless you (& your friend) are allowed to release the rope, or at any rate let it slide through your hands, I can't see how you can maintain exactly 99N at all times. Is this the a pachyderm in the forum.

 

[A.T.]

Well I hope we are sticking with the static situation.

No we didn't.

 

[Ocata]

Hi Merlin,

Perhaps you went over my head a little bit, lol. Not entirely sure what you mean by "pulling out half of the rope."

Thanks

 

[Merlin3189]

Ok. Well in that case I'd have to say that whether the rope is massless or not, you can't keep to 99N unless you and your friend are also massless.

If we can have a light inextensible string attached to two opposing 99N forces with no mass and no other forces, then the original question gets even more trivial.
If so, I'd take a different tack in explaining tension - what is it, where does it come from, why is total tension not meaningful, etc.

 

[Ocata]

Suppose I tied the rope to a box of 1,000,000 kg (M) and, while I experience enough friction to remain static, the box is on a frictionless surface.

If I pull the rope with 1N, and keep pulling on the rope so that it remains 100% taut, the box will accelerate toward me at F/M = a = 1/1,000,000 = .000001m/s^2. It will accelerate very slowly, but it will still accelerate. I find it amazing that no matter how much mass an object has, it will still accelerate when acted upon by another force.

However, when a box of just 100kg (m) is sitting on a table with some friction, it won't budge when applying 100N (or 100x the force that I applied on the gigantic mass in the previous scenario).

Does this difference in the way the two boxes interact with my pulling force mean that the 100kg box (m) is pulling against my 100N pull with more opposing force than the 1,000,000 kg box (M)?

Does the M kg box even have a resisting/opposing force? If I continue applying the small force of 1N to the M kg box, it will continue to accelerate, so there must not be any force to match my force and balance out the net force to 0, right?

So does mass itself does not provide a pulling force at all? Box M may feel to me like it's pulling against my pull, but it really isn't since it's actually accelerating very slowly - even though I can't detect it?

Scenario 1: If I try to evaluate all the opposing forces to find the net force, suppose I'm pulling 100N on the rope, the rope is pulling 100N on me, the other end of the rope attached to Box M is pulling on it with 100N and the box M is pulling on the rope with 100N, is this correct? If so, then why does the box M in the frictionless state not oppose my pull with 100N? It seems all the forces should cancel each other out for a net force of 0. The rope and I should be a net force of 0 and the rope and the box M should be another net force. Two net forces of 0 should mean no acceleration, I thought.

Scenario 2: Now, if I pull the box m (with enough friction to remain static), I still count a net force of 0. My 100 N applied to the rope, the rope applying 100N on me, the rope applying 100 N on the box, the box applying 100 N on the rope, the surface applying 100 N on the block, the block applying 100 N on the surface. 3 net forces of zero.

In both scenarios, all of the force pairs seem to balance each other out to a net force of 0, unless I'm over counting or undercounting a force somewhere..

 

[Merlin3189]

Ocata: OP has the friend start pulling on the string (sorry I picked up rope from another post without noticing) and I wondered what would happen. If the string were laid out in a straight line, then (assuming it to be inextensible, but not light) it would immediately start to accelerate like a solid body. Assuming a small delay before you try to pull on your end, then it might be out of your reach before you could catch hold.
So I thought, just have it be long and loosely piled in a heap so that even while your friend was racing away, your end would stay still for a second or so.

If perchance the string is light (or lightish) then applying 99N to it is going to accelerate it very quickly - a 2m string of 100N working load strength would be less than 0.1kg and so would accelerate away at about 1000m/sec/sec or more. Pausing for even 0.1 sec before grabbing the string to exert your pull, the string is at least 5m away from you and traveling at 100m/sec! Even if you were running before your friend started pulling, you'd not catch it.
Perhaps your friend starts by pulling the string towards you and you catch the end as it flies past. If you are massless, then you're simply going to reduce the rope's acceleration to zero and move with your friend and the rope at whatever speed they have reached. If you have mass, and as AT wants we look at dynamics, then it all gets more complex. If the string were inextensible then the problem becomes impossible: one of you has to have infinite acceleration. So now we have to consider elasticity and length of the string and although the problem may be tractable, we are almost certainly going to break the constant 99N rule of the original problem.The problem with the OP question is the difficulty of applying a single force of 99N to a string, unless that string is a long, heavy rope. I thought that using a long, heavy rope might make the problem more tractable.
Additionally allowing it to slide along a rough floor provides damping to the spring mass system and maybe allows us to gloss over the dynamic problems and look at the steady state results..

 

[Merlin3189]

Does this ... mean that the 100kg box (m) is pulling against my 100N pull with more opposing force than the 1,000,000 kg box (M)?

Is either box pulling? Maybe they just sit there and you are pulling them.

Does the M kg box even have a resisting/opposing force?

Probably not. The smaller box has a friction force from the table pushing in the opposite way to you and pushing as hard as you.

So does mass itself does not provide a pulling force at all?

Except for gravity.

Box M may feel to me like it's pulling against my pull, but it really isn't since it's actually accelerating very slowly - even though I can't detect it?

Now if you're pulling something, does it matter whether it is moving or not, accelerating or not, don't you still feel the same? What you feel is you pulling. If there were no resistance, you would not BE pulling - you'd be moving your hand (maybe holding a floppy bit of string.) You can only pull or push if there is something to pull or push against.
Mass is something you can pull or push and you cause acceleration. If an object were light and had zero mass, in the absence of other forces, you could not really push it - if you tried, as soon as you touched it, it would shoot away from you with infinite speed from the tiniest imperceptible force you had applied as you contacted it. If it were only lightish, then you could apply a small force, but it would soon accelerate away before you felt you were giving it a proper push.
Friction is what generally provides an "other force" , though it could be another object behind it as well.
If you push your 100kg box (which IMO is not small!) it does not accelerate away, because the table is pushing against via friction. For your big block, you ruled that out, so even though it is more massive, it does, as you say, accelerate, but so slowly that you can apply considerable force for some time before it moves away from you..

 

[Ocata]

Thanks Merlin, I realize I should be as specific as I can when asking questions about this topic.

Suppose I have a 100kg block resting in space with zero velocity (compared to the strings I will attach to it). Suppose I attach a string to the left and right end and each string is a million miles long so that the block can continue accelerating without running out of string for a period of time. Suppose the strings are attached to a reel or wind up device that can wind up the million mile string at faster and faster rates until the strings are accelerating at near the speed of light. And suppose the reel device is anchored in space such that no amount of force can cause the devices themselves to accelerate. Given these initial conditions:

Suppose I pull one end if the string( the right side) while you let the left side run loose such that the block accelerates toward me at rate of 9.8m/s^2. Since f=ma = 100kg*9.8m/s^2 = 980N, if I had a spring measuring device in the string, it would measure 980N.

Now, if you introduced a matching force of 980N pulling from the left side, the net force on the block would be zero, so the block would continue traveling at a constant velocity in my direction.

Question 1:
Does my measuring device still read 980N? If you had a measuring device on your side, would it also read 980N ?

Question 2:
If the block was made of metal, except for a small slice of styrofoam in the middle (such that if the block were to split, it would split down the middle), then would the block be any more likely to rip apart while accelerating toward me at 9.8m/s^2 than if it were experiencing a net force of zero with your force of 980N and my force of 980N applied to it?

Through our discussion so far, I'd guess that applying a force of 980N in one direction and causing acceleration in the block puts the same amount of force on the block as if the 980N were being applied to both sides. Thus the block is exactly equally likely to rip apart at the styrofoam slice.

Okay, so starting the scenario over with the same initial conditions. Suppose the block is resting at zero velocity and we both start pulling on the rope with the exact force such that we both want to accelerate the block toward each other at 9.8 meters per second. Do I apply a force of 980N towards me and you apply a force of 980N toward you. The net force is zero. My measuring device reads 980 N and your device reads 980N.

Now suppose all of a sudden you apply more force such that I begin to accelerate toward you at 2m/s^2. According to F=ma, the force needed to apply an acceleration of 2m/s^2 to a 100kg block is 200N.

Question 3:
So the force you are now applying is 980N+200N = 1180N, yes?

Question 4:
So, if your measuring device attached to the string would measure 1180N, what does my measuring device read? 980N or 1180N?

 

[Merlin3189]

Suppose I have a 100kg block resting in space with zero velocity ... I attach a string to the left and right end ... the strings are attached to a... device that can wind up the ... string at faster and faster rates until the strings are accelerating at near the speed of light. And suppose the reel device is anchored in space such that no amount of force can cause the devices themselves to accelerate.

This is getting dangerous! I believe in relativity, but I've never mastered the sums.

Given these initial conditions:

Suppose I pull ... the right side while you let the left side run loose such that the block accelerates toward me at rate of 9.8m/s^2. Since f=ma = 100kg*9.8m/s^2 = 980N, if I had a spring measuring device in the string, it would measure 980N.

Correct.

Now you introduced a matching force of 980N pulling from the left side, the net force on the block would be zero, so the block would continue traveling at a constant velocity in my direction..

Correct.

Question 1:
Does my measuring device still read 980N? If you had a measuring device on your side, would it also read 980N ?.

Correct.

Question 2:
If the block was made of metal, except for a small slice of styrofoam in the middle (such that if the block were to split, it would split down the middle), then would the block be any more likely to rip apart while accelerating toward me at 9.8m/s^2 than if it were experiencing a net force of zero with your force of 980N and my force of 980N applied to it?.

Ok. Imagine the block is two blocks, joined by your spring force meter.
When just you are accelerating it to the right. The left block of 50kg is accelerating at 1g, so there is a force on it of 490N to the R. this is provided by the link between the blocks, the spring device, which therefore registers 490N. The right hand block similarly is 500kg accelerating at 1g, so net force on it is 490N. This is the result of 980N on your string opposed by the 490N of the spring device.
Now when we both pull: I pull the left hand block with 980N. But it does not accelerate, so there must be a balancing force of 980N from the linking spring. Similarly you are pulling the RH block with 980N which is balanced by the 980N from the spring and there is no acceleration. In this case the two halves are more likely to split: the force in the spring joining them is 980N, while when it was accelerating the force was only 490N.

Through our discussion so far, I'd guess that applying a force of 980N in one direction and causing acceleration in the block puts the same amount of force on the block as if the 980N were being applied to both sides. Thus the block is exactly equally likely to rip apart at the styrofoam slice..

So I think that is wrong.

Okay, so starting the scenario over with the same initial conditions. Suppose the block is resting at zero velocity and we both start pulling on the rope with the exact force such that we both want to accelerate the block toward each other at 9.8 meters per second. Do I apply a force of 980N towards me and you apply a force of 980N toward you. The net force is zero. My measuring device reads 980 N and your device reads 980N..

Correct.

Now suppose all of a sudden you apply more force such that I begin to accelerate toward you at 2m/s^2. According to F=ma, the force needed to apply an acceleration of 2m/s^2 to a 100kg block is 200N.

Question 3:
So the force you are now applying is 980N+200N = 1180N, yes? .

Correct. You are still pulling 980N, so I must pull 200N harder than you.

Question 4:
So, if your measuring device attached to the string would measure 1180N, what does my measuring device read? 980N or 1180N?

980N
You are still pulling with 980N. The block is experiencing a net force of 200N to the left, that is 980N to the right as shown on your meter and 1180N to the left as shown on mine.

 

[Ocata]

Thank you, this is really starting to make sense.

So if I have a 1000kg block and I apply 1000N, it will accelerate at 1m/s^2. If I divide the block into two 500kg blocks attached by a string or spring force meter, then applying the same force of 1000N to the string I am pulling will distribute the force equally over both blocks. F=ma = (500kg +500kg)(1m/s^2) = 1000N. Each block will have a force of 500N. F=ma = 500kg(1m/s^2)= 500N.

I'm starting to see that the amount of force over the two blocks would vary depending on the string being massless or not. If the rope connecting the two blocks contains a mass, then the force of 1000N is distributed equally with the dividing line in the middle of the connecting rope. So if the rope has a mass of 10kg, then applying 1000N to the string on my side would cause a total acceleration of f/m = 1000/1010= .99m/s^2. The block on the right still has a mass of 500kg, so the force on it is 500(.99)=495N. Since the block on the left has a mass of 500kg, the force on it is 495N as well, which leaves 10N for the force of the rope. 495N + 495N + 10N = 1000N. Yet if I calculate the force of the rope directly, I get F=ma= 10kg(.99)=9.9N. Why the discrepancy? Is the force on the rope 10N or 9.9N?If I put a spring force meter in the middle of the connecting rope, I am dividing the mass of the whole set equally. (500kg +5kg)+(500kg+5kg) = 1010kg. The spring force meter in the middle reads 505kg(.99)= 499.95N. And (2)499.95N = 999.9 N. But I started out applying 1000 N, why when I add up the forces applied to each part, does it return a result of 999.9N?

If we take away the rope connecting the two blocks and replacing it with a spring force meter and adding a spring force meter in front of the right block as you described, then since each of the 500kg blocks would generate an acceleration of 1m/s^2 with 1000N applied, then each block would have a force of f=ma= 500kg(1m/s^2) = 500N. If I understand correctly, the force meter in between the blocks would read 500N due to the force on the left block. The force on the right hand block is 500N as well, but does the force meter on the right side of the right hand block read 500N or 1000N?

 

[A.T.]

Why the discrepancy? Is the force on the rope 10N or 9.9N?...But I started out applying 1000 N, why when I add up the forces applied to each part, does it return a result of 999.9N?

Use symbolic math instead of numerical values, to check for equivalence.

The force on the right hand block is 500N as well, but does the force meter on the right side of the right hand block read 500N or 1000N?

Draw a diagram with force vectors. What is the net force on each block?

 

[Ocata]

Use symbolic math instead of numerical values, to check for equivalence.

Hi A.T., I just looked up "check for equivalence" a the results returned abstract algebra solutions. I apologize, but I am just in calc 3. I don't yet know anything about abstract algebra. Is it possible to understand what's going on here without abstract algebra?

Draw a diagram with force vectors. What is the net force on each block?

I'm pulling with 1000N to the right of the right hand block and the left hand block is "pulling" on the right Han block with 500N, yes? So the net force on the right hand block is 500N to the right. However, I just learned to be careful getting confused between net force and tension as they are not the same thing. So it shouldn't be 500N. It should read 1000N since the force meter is connected to my pull and I happen to be pulling at 1000N?

 

[Merlin3189]

Draw a diagram with force vectors. What is the net force on each block?

Agree. I've not shown this because I don't have any easy software for creating diagrams, but my desk is littered with scraps of paper and I always do sketches for myself.

... will distribute the force equally over both blocks. F=ma = (500kg +500kg)(1m/s^2) = 1000N. Each block will have a force of 500N. F=ma = 500kg(1m/s^2)= 500N.

Yes. I don't like the theorem about "distribute the force equally" even though it may be true, because F=ma seems to tell the whole story and the distribution theorem is just a shortcut which saves you thinking about Newton's basis.

... the connecting rope ... has a mass of 10kg, then applying 1000N to the string on my side would cause a total acceleration of f/m = 1000/1010= .99m/s^2.
block on the right ... force on it is 500(.99)=495N.
block on the left force on it is 495N as well, which leaves 10N for the force of the rope. 495N + 495N + 10N = 1000N.
Yet if I calculate the force of the rope directly, I get F=ma= 10kg(.99)=9.9N. Why the discrepancy? Is the force on the rope 10N or 9.9N?

Rounding errors. (Perhaps it would be good to follow A.T. and use algebra.)
Accn = 1000/1010 = 0.9900990099
F_R= F_L = 500 x Accn = 495.0495
1000 - 495.0495 -495.0495 =9.90099
F_rope=10 x Accn = 9.90099
All agree when you work out carefully. (Or algebraically.) Your original calc of accn was only done to 2sf, so at the end of further calcs you can't be confident of more than 1sf

If I put a spring force meter in the middle of the connecting rope, I am dividing the mass of the whole set equally. (500kg +5kg)+(500kg+5kg) = 1010kg. The spring force meter in the middle reads 505kg(.99)= 499.95N. And (2)499.95N = 999.9 N. But I started out applying 1000 N, why when I add up the forces applied to each part, does it return a result of 999.9N?

Same again.

If we take away the rope connecting the two blocks and replacing it with a spring force meter and adding a spring force meter in front of the right block as you described, then since each of the 500kg blocks would generate an acceleration of 1m/s^2 with 1000N applied, then each block would have a force of f=ma= 500kg(1m/s^2) = 500N. If I understand correctly, the force meter in between the blocks would read 500N due to the force on the left block. The force on the right hand block is 500N as well, but does the force meter on the right side of the right hand block read 500N or 1000N?

The NET force on the RH block is 500N, made up of 1000N to the R from the string and 500N to the L from the other half block. So the force meter on the RH string reads 1000N.

 

[Ocata]

This is making so much more sense. Thank you for making me realize the concrete impact of calculating the entire set of decimal places.

Suppose I divide the one 1000kg blocks into ten 100kg blocks (numbered 1 through 10 with the 1st on being on the far left and the 10th one being on the far right) and I pull with 2000N on the RH side of the 10th block. The entire acceleration will equal 2000N/1000kg = a = 2m/s^2 and each box will have a force of 100kg(2m/s^2) = 200N. The force meter connected to the string on which I am pulling will read 2000N. If a force meter is connected between each 100kg block, will each of the 9 force meters read 200N for a total of 1800N or will each force meter read a consecutively smaller number? In thinking about your explanation earlier for why the block is more likely to rip apart with an opposing force, I'm going to guess that the force meters will read consecutively smaller numbers. If so, would they read like this: F = ma => 100kg(9)(2), 100kg(8)(2), 100kg(7)(2), ..., 100kg(3)(2), 100kg(2)(2), 100kg(1)(2)

So if you were to offset the Styrofoam division in the original 1000kg block to the far right within the block and pulled the RH string to the R with 1000N, it would be much more likely to rip apart than if you were to offset the Styrofoam division to the far left within the block, yes?

 

[Tom_K]

Suppose I pull one end if the string( the right side) while you let the left side run loose such that the block accelerates toward me at rate of 9.8m/s^2. Since f=ma = 100kg*9.8m/s^2 = 980N, if I had a spring measuring device in the string, it would measure 980N.

Now, if you introduced a matching force of 980N pulling from the left side, the net force on the block would be zero, so the block would continue traveling at a constant velocity in my direction.

Not necessarily. The block was never moving at a constant velocity; it was accelerating towards you at 9.8 m/s^2 with a net force acting on it so it definitely would not continue moving at a constant velocity.

When you introduce the matching force on the left side, there will be a discontinuity in the acceleration (jerk) the net force will suddenly drop to zero, the acceleration drops to zero but the velocity is not easy to determine.

The Atwood[/PLAIN] machine was made just for this type of experiment. Suppose you have two equal masses of 10 kg each that are connected by a massless, frictionless string and the string is run over a frictionless pulley. That is an ideal Atwood machine. Have the mass on the left resting on the ground with a bit of slack and you support the mass on the right in your hand. Now let the mass on the right drop. It will fall with an acceleration of 9.8 m/s^2 until the slack is all taken up. Now the force of gravity suddenly has to work against the two masses pulling in opposite directions. Isn’t this similar to the situation you described, with matching forces of 98 N on both ends of the string?

Do you think the right mass will continue falling with constant acceleration, or will it jerk to an abrupt stop or slow to a constant velocity? If a constant velocity, what direction will the two mass system go?

What do you think the tension in the string will be?

 

[A.T.]

the acceleration drops to zero but the velocity is not easy to determine.

If acceleration drops to zero, then velocity stays at the value it had when the balancing force was introduced.

 

[A.T.]

Thank you for making me realize the concrete impact of calculating the entire set of decimal places.

You cannot always calculate the entire set of decimal places, as they are often infinite. The actual lesson from this is not use numerical values, but symbols.

 

[Tom_K]

If acceleration drops to zero, then velocity stays at the value it had when the balancing force was introduced.

In the case of the two 10 kg masses on the Atwood machine, say mass1 is allowed to fall from rest for one second before the string becomes taut. Mass1 will have an instantaneous velocity of 9.8 m/s at the instant the slack runs out, and mass2 will still be at rest on the floor. In order for mass1 to keep moving at that same velocity, mass2 would need to be accelerated from rest to 9.8 m/s in an instant.

Is that possible?

I think that if the acceleration were to be brought down to zero in a continuous manner, such as slowly applying a brake, then the two masses would move at whatever the instantaneous velocity mass1 has when acceleration reaches 0. But that would entail a continuous slowing of mass1 so that its instantaneous velocity is also very near zero when mass2 starts moving.

In the case that I described, where both masses are subjected to an impulse force, this is a jerk system, where acceleration is brought to zero after a discontinuity, and in that case the velocity is unpredictable. The jerk introduces an element of chaotic motion into the situation. The system may move slowly in the same direction as mass1 was going, or it could come to a sudden halt and even move back down on the side of mass2. I think it is likely to oscillate back and forth at a very small amplitude and period (as a pendulum) until air resistance brings it to a stop.

 

[A.T.]

Mass1 will have an instantaneous velocity of 9.8 m/s at the instant the slack runs out, and mass2 will still be at rest on the floor.

That has nothing to do with Ocata's scenario you quoted in post #34, where there is only one body to analyze (the block), not two bodies with two different velocities. The scenario tells us all the forces on the block and its mass, and that's all we need to know to compute its acceleration.

 

[Tom_K]

That has nothing to do with Ocata's scenario you quoted in post #34, where there is only one body to analyze (the block), not two bodies with two different velocities. The scenario tells us all the forces on the block and its mass, and that's all we need to know to compute its acceleration.

How is it different? Let the center of the string on the Atwood machine have some mass then and let that be exactly centered at the top of the pulley when the string goes taut. The other two masses are there only to apply the force and the counter force. You can let them each be 100 kg so that the force is 980 N exactly as in the original situation.

In fact, all this will do is add an equal increment of mass to each of the other masses and the situation is still exactly the same.

Now do you think the velocity of the mass at the top dead center of the pulley is predictable when the string goes taut?

 

[A.T.]

How is it different?

All the forces on the block are already given. We don't need to worry about some string going taut etc.

 

[Ocata]

Error: In post #33, first sentence of second paragraph, I made the word block plural by mistake.

The following sentence:

Suppose I divide the one 1000kg blocks into ten 100kg blocks (numbered 1 through 10 with the 1st on being on the far left and the 10th one being on the far right) and I pull with 2000N on the RH side of the 10th block. The entire acceleration will equal 2000N/1000kg = a = 2m/s^2 and each box will have a force of

should read:

"Suppose I divide the one 1000kg **block** into ten 100kg blocks (numbered 1 through 10 with the 1st on being on the far left and the 10th one being on the far right) and I pull with 2000N on the RH side of the 10th block. The entire acceleration will equal 2000N/1000kg = a = 2m/s^2 and each box will have a force of..."

My apologies. I will illustrate this as soon as I get home from work. Hopefully by 830am eastern time.

 

[A.T.]

I'm going to guess that the force meters will read consecutively smaller numbers.

Yes, you have a tension gradient if you accelerate a massive extended body by pulling at one end.

 

[Bassyboy]

I think this could easily be solved by using a torque wench if you have a setting of 99 Newton metres and put another toque wrench exactly the same they would both click off at 99 Newton facing each other right? So if the rope is in the middle of a tugger war with let's say a car and both had the same torque (no horsepower) at all string would not brake but if you added horsepower the string would snap! It's kinda hard to explain

 

[Merlin3189]

Thank you for making me realize the concrete impact of calculating the entire set of decimal places.

As A.T. says, you can't calculate "the entire set of decimal places".But you have to use enough. If you only use 2 significant figures in your calculation, then you end up with 2sf or less, so you can't then complain if you get 9.9 instead of 10. Even if you had calculated to 6sf, you might have ended up with 9.99999 instead of 10, but you should not think that the results differ in either case - whether you got 9.9 and 10 working to 2sf or 9.99999 and 10 working to 6sf, you can't tell that the answers are not the same.
Although A.T. suggests you avoid this issue when working symbolically, accuracy, errors and significant figures are important topics you should try to understand because you will have to do numerical calculations as well. It is not just that sometimes the sums don't work out perfectly as decimals (like 1000/1010), but you can never measure anything exactly and even constants like g are not known exactly.
This is a bit of a sore point for me since coming to PF, where, at least in the HW section, everybody (nearly) seems to think the answer is always an "equation" which you can put into your calculator or computer and up pops the answer. Calculators & computers do give numbers (lots of them), but you have to turn those into sensible answers.

... divide the 1000kg block into ten 100kg blocks... and pull with 2000N on the RH ... acceleration will equal 2000N/1000kg = a = 2m/s^2 and each box will have a NET force of 100kg(2m/s^2) = 200N. (eg. the 3rd box will be subject to two forces, 400N to the left and 600N to the right -> net 200N to right.)
The force meter connected to the string on which I am pulling will read 2000N.
If a force meter is connected between each 100kg block, will each of the 9 force meters read 200N for a total of 1800N (NO)
or will each force meter read a consecutively smaller number? like this: F = ma => 100kg(9)(2), 100kg(8)(2), 100kg(7)(2), ..., 100kg(3)(2), 100kg(2)(2), 100kg(1)(2) (YES)

So if you were to offset the Styrofoam division in the original 1000kg block to the far right within the block and pulled the RH string to the R with 1000N, it would be much more likely to rip apart than if you were to offset the Styrofoam division to the far left within the block, yes? (YES)

You could see this easily on a diagram.

 

[Ocata]

Thank you Merlin,

I was actually working on a diagram since I got home and just completed it and the numbers work out exactly as your diagram depicts. I even scanned it in already and was ready to upload it as a reference to what I was attempting to describe, but your diagram represents exactly the scenario I was trying to describe. What I've internalized in the process so far is getting a feel for why the tension forces are equal along the string. And it is, of course, exactly because the string is massless, which, through your and A.T.'s guidance, helped me to grasp. If the string had mass then each block would still "contain" the same force value of 200N because the mass of the blocks would not have changed. However, the force vectors created by the tension on the RHS of each block would take into account the mass of all the blocks up until that point + the mass of all the strings up until that point.

For instance the tension after the 4th block would be:
(B1(kg)+String1(kg) + B2(kg)+String2(kg) + B3(kg) +String3(kg)+ B4(kg))(F/(mass of all blocks and strings)
= (B1(kg)+String1(kg) + B2(kg)+String2(kg) + B3(kg) +String3(kg)+ B4(kg))(2m/s^2) = Tension(N) on RHS of Block 4.

And the tension on the LHS of the 4th block would be:
(B1(kg)+String1(kg) + B2(kg)+String2(kg) + B3(kg) +String3(kg)))(F/(mass of all blocks and strings)
= (B1(kg)+String1(kg) + B2(kg)+String2(kg) + B3(kg) +String3(kg))= Tension(N) on LHS of Block 4.

The difference of the two forces would be Block4(kg).

Am I describing the idea correctly?

 

[A.T.]

The difference of the two forces would be Block4(kg).

Times acceleration

Am I describing the idea correctly?

Yes, the general idea is: You can cut the system into bodies in different ways, and apply Newtons Laws to them.

 

[Ocata]

Merlin and A.T., Thank you.