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

[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.