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

[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.