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

[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?

 

[jtbell]

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?

 

[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

 

[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.

 

[jtbell]

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?

 

[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?

 

[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.

 

[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-a_A)

where a_A is the acceleration of block A

For block B,

T = ½m(a_B+g)

Where a_B is the acceleration of block B

This yields,

a_B = g-2a_A

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

 

[jtbell]

a_B = g-2a_A

Assuming that the two masses are equal, this is correct. This gives you one equation for the two unknowns a_A and a_B. 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.

 

[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

 

[jtbell]

a_A=2a_B

Yep.

(g-a_A)/(g+a_B)= 1/2

This is also valid. It's actually the same as Kyle's last equation from post #8:

a_B = g-2a_A

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 a_A and a_B. 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.)

 

[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

a_A = 2a_B

In the context of solving the system

And I was referring to

(g-a_A) / (g+a_B) = ½

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

a_A = 2a_B

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

 

[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 !

 

[Kyle.Nemeth]

(a deep and subtle result of String Theory )

Very nice

 

[gmax137]

The only way it makes sense to me is by imagining that the middle and right string instead is just one string...

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.

 

[jtbell]

The system of equations ends up undetermined in my case. Help !

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.

 

[Thom_Silva]

We need one more equation to solve this...

 

[Doc Al]

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.

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?

 

[Thom_Silva]

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

 

[jtbell]

But can that be true as well as aB = g-2aA

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

 

[jtbell]

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

We need one more equation to solve this...

To recap: Your two equations are:

(1) a_A = 2a_B

and either

(2a) a_B = g-2a_A

or

(2b) (g-a_A)/(g+a_B)= 1/2

which are equivalent.

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

 

[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 :)

 

[jtbell]

i would like to solve the equations to prove that in fact, aA= 2aB

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.

 

[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?...

 

[jtbell]

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)

 

[Kyle.Nemeth]

Yes, since it is attached to the ceiling. Would it be true if instead of being attached to the ceiling, it were attached to a roll of string assuming the the roll of string had enough string to provide?

 

[jtbell]

Yes, but then of course the "conservation of string" principle would not apply. And if there were no friction, I suspect the string would simply spool off the roll until it reaches its end.

 

[Kyle.Nemeth]

I understand... The point I am trying to make for this case Is that, if the right string were displaced by 1 cm, then the middle and left string would displace by 1 cm too... This helps me realize that if pulley B were not stationary, than this is not the case. From there, I'm trying to understand how the displacement of the middle and right strings happen..

I thank you for your patience jtbell, you have been a great help to me.

 

[jtbell]

(returning to the original situation) If you pull mass A down by 1 cm, the left string lengthens by 1 cm. Therefore the total of the middle and right strings together must shorten by 1 cm.

As the pulley on mass B moves upwards, the middle and right strings must shorten by equal amounts. Therefore they must each shorten by 1/2 cm, so that their total shortens by 1 cm.

 

[Kyle.Nemeth]

I understand now. Thank you jtbell. I think I had that correct in the second statement of post #6 (unless there was a subtlety in my statement that deemed me technically incorrect).

I apologize for dragging out the question. I am grateful for all the help I've received from the physics community!