# Why is the net force on block A twice that on block B?

1. Jul 9, 2015

### Kyle.Nemeth

www.webassign.net/serpop/p7-47.gif

In the figure that I've posted in the link above, it is assumed that both blocks are initially held at rest and then released. Why does block A move with twice the speed of block B?

2. Jul 9, 2015

### Staff: Mentor

It's because of the Law of Conservation of String. (a deep and subtle result of String Theory )

Suppose block B moves upwards a distance d. What happens to the lengths of the middle and right-hand sections of the string? What has to happen to the left-hand section of the string in order to keep the total length constant?

Last edited: Jul 9, 2015
3. Jul 9, 2015

### gmax137

Forget about speed for a minute. If you pull block A down one centimeter, what happens to block B? It goes up, right? How far does it go up?

edit - jtbell beat me to the punch

4. Jul 9, 2015

### Kyle.Nemeth

I'm still not quite sure.. to me, it seems that if I were to pull block A down by 1 cm, then the middle string will want to pull up by 1 cm.

5. Jul 9, 2015

### Staff: Mentor

If the middle string shortens by 1 cm, doesn't the right-hand string also have to shorten by 1 cm? What would happen to the total length of the string if that were true?

6. Jul 9, 2015

### Kyle.Nemeth

Hmm. So if I were to imagine that the middle and right string were instead one string with a block attached, then this string would move up by 1 cm upon pulling block A down by 1 cm.

If this new string were bent around a pulley and attached to a platform as in the figure, such that its length is the same, then upon pulling block A down by 1 cm, the middle string must move up 1/2 cm and also for the right string?

7. Jul 9, 2015

### andresB

Asumming the pulleys have no mass (so the tension is the same across all the string), then there is only one tension pulling block A upwards while there are two pulling block B.

8. Jul 9, 2015

### Kyle.Nemeth

I understand that, and I actually tried using Newton's second law to show what the acceleration of block A would be. I modeled each block separately as a particle under a net force in the vertical direction. The horizontal components of force are all zero so,

For block A,

T = m(g-aA)

where aA is the acceleration of block A

For block B,

T = ½m(aB+g)

Where aB is the acceleration of block B

This yields,

aB = g-2aA

and I'm not sure whether that is correct. I certainly know that I haven't been able to take any information from it...

9. Jul 10, 2015

### Staff: Mentor

Assuming that the two masses are equal, this is correct. This gives you one equation for the two unknowns aA and aB. You can get a second equation from the geometrical condition that the total length of string is constant. You're close to it in your second statement in post #6. This gives you two equations in two unknowns.

10. Jul 10, 2015

### Thom_Silva

I got to the same result as you! . And it is strange, if the speed A at any given moment is twice as the speed at b. Then aA=2aB. I also don't get it.... It seemed such an easy problem. We are probably wrong on something. From the calculations we also get that (g-aA)/(g+aB)= 1/2. I don't know if that is relevant, but it is interesting

11. Jul 10, 2015

### Staff: Mentor

Yep.
This is also valid. It's actually the same as Kyle's last equation from post #8:
which you can verify by multiplying out your equation to get rid of the fraction on the left side. So you have two equations in the two unknowns aA and aB. What's the first thing that you should think when you see two equations in two unknowns?

(or three equations in three unknowns, or four equations in four unknowns, etc.)

Last edited: Jul 10, 2015
12. Jul 10, 2015

### Kyle.Nemeth

So we now have a system of equations and can solve for both accelerations and compare them, right?..

I'm still wrapping my head around the string conservation idea too.. The only way it makes sense to me is by imagining that the middle and right string instead is just one string that has been "divided" into two strings. If the left string is pulled down by 1 cm, then the combination of the movement of the middle and right string must be 1 cm also. Is this where I'm wrong?

Edit:

Since Thom's equation is just a rearrangement of my equation, wouldn't we just end up with 1 = 1?

Edit #2:

I realize that you may be referring to the equation

aA = 2aB

In the context of solving the system

And I was reffering to

(g-aA) / (g+aB) = ½

My apologies for the confusion. I'm trying to show that

aA = 2aB

Which is why I'm trying to understand the string conservation idea.

Last edited: Jul 10, 2015
13. Jul 10, 2015

### Thom_Silva

Two equations two unknowns should be easy to solve right ? But in this case i couldn't. Can you show us the result for aA and aB? The system of equations ends up undetermined in my case. Help !

14. Jul 10, 2015

### Kyle.Nemeth

Very nice

15. Jul 10, 2015

### gmax137

You have said this a few times. It confuses me, because the middle and right strings ARE just one string. There is only one string, it is attached to A on one end and the ceiling on the other end.

16. Jul 10, 2015

### Staff: Mentor

The second and third equations in my post #11 are indeed rearrangements of each other, so solving them together simply gives 1 = 1 as Kyle noted. Oops.

Try solving the first and second equations together, or the first and third. If you get stuck, show what you did.

17. Jul 10, 2015

### Thom_Silva

We need one more equation to solve this....

18. Jul 10, 2015

### Staff: Mentor

Here's a simple way to derive the acceleration constraint from the "conservation of string". (Although just playing with a piece of string should be enough for you to figure it out.)

Let $x_a$ be the distance below the ceiling of mass a; let $x_b$ be the distance below the ceiling of mass b. Using those distances, express the length of the string:
$L = x_a + 2x_b$.

Now take the time derivative of both sides. (Since the length doesn't change, its derivative is zero.) Do it once to get the velocities: $0 = v_a + 2v_b$; Do it again to get the acceleration: $0 = a_a + 2a_b$.

That tells you that $a_a = -2a_b$. (If all you care about is the magnitude, then $a_a = 2a_b$.)

Make sense?

19. Jul 10, 2015

### Thom_Silva

Good explanation Doc, But can that be true as well as aB = g-2aA ?

20. Jul 10, 2015

### Staff: Mentor

Yes. This allows you to find the values of aB and aA, by solving the two equations together.

21. Jul 10, 2015

### Staff: Mentor

To recap: Your two equations are:

(1) aA = 2aB

and either

(2a) aB = g-2aA

or

(2b) (g-aA)/(g+aB)= 1/2

which are equivalent.

aA and aB are unknown. g is known. Two equations in two unknowns. Why do you think we need one more equation?

22. Jul 10, 2015

### Thom_Silva

Adding aA = 2aB its solvable. But that seems like cheating, i would like to solve the equations to prove that in fact, aA= 2aB. But the explanation of Doc proves it to me. But still if anyone knows another way to get to the solution without assuming that aA = 2aB from the first place, i would like to know. Thank you anyway :)

23. Jul 10, 2015

### Staff: Mentor

You can't. That condition is a geometrical constraint that arises from using a fixed-length string (no stretching or compression). It comes in addition to the equations that you derive from applying Newton's Second Law to the two masses.

If the string were a rubber band, instead, you would not be able to use that condition.

24. Jul 10, 2015

### Kyle.Nemeth

Doc, your derivation makes perfect sense, thank you.

I think the trouble I'm having is visualizing this string conservation principle in my head in regards to the figure I posted along with the thread.

My train of thought was that when block A is pulled down by 1 cm, the middle string must move up by 1 cm. I think I realized that I was assuming the pulley attached to B was stationary. Is this correct?....

25. Jul 10, 2015

### Staff: Mentor

If the pulley attached to B were held stationary (e.g. also attached to the ceiling), then you would not be able to pull down block A at all. (unless the string was actually a rubber band)