A pulley system with two pulleys and two suspended masses

[member 731016]

Here it is @erobz ,

$L = a + 2s + \Delta 2s+ 2b + l + \Delta l$

Many thanks!

 

[erobz]

Hi @erobz ,

Is the reason why you used arrows for this $z \to z+ \Delta z$ because if you use equal sign then

$z = z+ \Delta z$
$z = z+ z_f - z_i$
$z_i = z_f$

Oh which is conservation of the quantity z anyway!

Many thanks!

Its just shorthand math notation for saying $z$ goes to $z + \Delta z$. The variable $z$ wouldn't equal $z + \Delta z$ except in the case that $\Delta z \to 0$

 

[member 731016]

Its just shorthand math notation for saying $z$ goes to $z + \Delta z$. The variable $z$ wouldn't equal $z + \Delta z$ except in the case that $\Delta z \to 0$

Thank you for your reply @erobz !

So an equivalent notation is

$z ≈ z + dz$

 

[erobz]

Here it is @erobz ,

$L = a + 2s + \Delta 2s+ 2b + l + \Delta l$

Many thanks!

Ok, now equate the two equations through the variable $L$ and cancel the terms. What are you left with?

Edit: keep the $\Delta$ next to the $s$

 

[member 731016]

Ok, now equate the two equations through the variable $L$ and cancel the terms. What are you left with?

Thank you for your reply @erobz!

Which two equations sorry?

Many thanks!

 

[erobz]

Which two equations sorry?

Many thanks!

The first one we derived with that one. they both equal $L$. You are attempting to figure out how the changes in each variable relate to each other.

 

[member 731016]

The first one we derived with that one. they both equal $L$.

Thank you for your reply @erobz !

That must be $L = a + 2s + 2b + l$ and $L = a + 2s + \Delta 2s+ 2b + l + \Delta l$

So
$a + 2s + 2b + l = a + 2s + \Delta 2s+ 2b + l + \Delta l$
$0 = \Delta 2s + \Delta l$
$-\Delta 2s = \Delta l$
$2(s_f - s_i) = l_f - l_i$
Many thanks!

 

[erobz]

Thank you for your reply @erobz !

That must be $L = a + 2s + 2b + l$ and $L = a + 2s + \Delta 2s+ 2b + l + \Delta l$

So
$a + 2s + 2b + l = a + 2s + \Delta 2s+ 2b + l + \Delta l$
$0 = \Delta 2s + \Delta l$

Many thanks!

Move one of them to the other side of the zero. And if you want to know how far pulley B moves in terms of a how far pulley A moves solve for $\Delta s$.

 

[erobz]

Again. Keep the $\Delta$ touching the $s$. They are stuck together. Joined at the hip! don't put that 2 in between them.

 

[member 731016]

Again. Keep the $\Delta$ touching the $s$. They are stuck together. Joined at the hip! don't put that 2 in between them.

Thank you for your reply!

Sorry @erobz missed that message,

$-2\Delta s = \Delta l$

Many thanks!

 

[member 731016]

Thank you for your help @erobz !

Just a side question, how were we allowed to add the $\Delta$ to $s$ and $l$ then set them equal to each other?

Many thanks!

 

[erobz]

Thank you for your help @erobz !

Just a side question, how were we allowed to add the $\Delta$ to $s$ and $l$ then set them equal to each other?

Many thanks!

Because the change is unsigned. It could be a positive change. It could be a negative change. It may appear like were adding to both lengths but is turns out we weren't. Notice the equation dictates a positive change in the variable $l$ ( $\Delta l > 0$) demands and negative change in $s$ ( $\Delta s < 0$), and visa-versa.

 

[member 731016]

Because the change is unsigned. It could be a positive change. It could be a negative change. It may appear like were adding to both lengths but is turns out we weren't. Notice the equation dictates a positive change in the variable $l$ ( $\Delta l > 0$) demands and negative change in $s$ ( $\Delta s < 0$), and visa-versa.

Thank you for your reply @erobz !

I think we can only add the $\Delta$ to the variable that is changing if know that there is a constraint for example the string length is conserved, correct?

Many thanks!

 

[erobz]

Thank you for your reply @erobz !

I think we can only add the $\Delta$ to the variable that is changing if know that there is a constraint for example the string length is conserved, correct?

Many thanks!

No. You can add it to anything. The only constraint needed is that you have a relationship between some variables. It's simply saying we have this relationship between all these variables, lets pick some or, all that can change and change them a bit from where they currently are valued at, and see how they respond to each other. How it all pans out or the end goal is specific to each situation.

The main motivation is derivatives and differential equations, where you are looking for the change in one variable w.r.t a change in another variable. You've got time to learn all this. You don't have to figure all this out yesterday.

 

[member 731016]

No. You can add it to anything. The only constraint needed is that you have a relationship between some variables. It's simply saying we have this relationship between all these variables, lets pick some or, all that can change and change them a bit from where they currently are valued at, and see how they respond to each other. How it all pans out or the end goal is specific to each situation.

The main motivation is derivatives and differential equations, where you are looking for the change in one variable w.r.t a change in another variable. You've got time to learn all this. You don't have to figure all this out yesterday.

Thank you for your help @erobz ! :)

 

[haruspex]

$L = a + 2\Delta s + 2b + \Delta l$

No, the deltas are the increases in length. The original lengths should still be in the equation. You should have just replaced $s$ with $s+\Delta s$, etc.

 

[BvU]

I am somewhat flabbergasted at the length () of this thread!
If you do the experiment with a real piece of rope things will be clear quickly.

If you still insist on a mental exercise, remove all unessentials: reduce the pulleys to points, etc

$\$

 

[erobz]

No, the deltas are the increases in length. The original lengths should still be in the equation. You should have just replaced $s$ with $s+\Delta s$, etc.

It was getting late! I guess my mind was just inserting symbols that weren’t there.

Edit: Actually at a later post (#31) than what you must have been quoting they did get the proper result.

 

[Lnewqban]

However, how do they know the block B will move up and block A will move down?
...
Also how do they know that if block B moves up by 1cm, block A will move down by 2cm?

Do you now understand how do they know those things?

If not, please see:
https://www.physicsforums.com/threa...2-dynamics-problem-involving-a-pulley.991370/

 

[member 731016]

No, the deltas are the increases in length. The original lengths should still be in the equation. You should have just replaced $s$ with $s+\Delta s$, etc.

Thank you for your reply @haruspex ! I will try that sometime.

Many thanks!

 

[member 731016]

I am somewhat flabbergasted at the length () of this thread!
If you do the experiment with a real piece of rope things will be clear quickly.

If you still insist on a mental exercise, remove all unessentials: reduce the pulleys to points, etc

$\$

Thank you for your reply @BvU!

Sorry, what do you mean reduce the pulley to points?

Many thanks!

 

[member 731016]

It was getting late! I guess my mind was just inserting symbols that weren’t there.

Edit: Actually at a later post (#31) than what you must have been quoting they did get the proper result.

Thank you for your reply @erobz !

 

[member 731016]

Do you now understand how do they know those things?

If not, please see:
https://www.physicsforums.com/threa...2-dynamics-problem-involving-a-pulley.991370/

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View attachment 322811

Thank you for sharing that thread @Lnewqban !

 

[BvU]

Sorry, what do you mean reduce the pulley to points?

You can get far with just a few pins or paperclips and a piece of wire, real or imagined ..

$\$

 

[member 731016]

You can get far with just a few pins or paperclips and a piece of wire, real or imagined ..

$\$

Ok thank you @BvU !