# A pulley system with two pulleys and two suspended masses

• ChiralSuperfields
In summary: Write the new equation.So ##l## and ##s## are the only length that can change. Apply a change to each of the variables.For instance if a varianle was ##z## we would apply a change by adding ##\Delta z## to it. etc...
Callumnc1 said:
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.

ChiralSuperfields
erobz said:
The first one we derived with that one. they both equal ##L##.

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
Callumnc1 said:

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##.

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

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

Sorry @erobz missed that message,

## -2\Delta s = \Delta l##

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!

Callumnc1 said:
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.

ChiralSuperfields
erobz said:
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.

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!

Callumnc1 said:

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.

Last edited:
ChiralSuperfields
erobz said:
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 ! :)

erobz
Callumnc1 said:
##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.

ChiralSuperfields
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

##\ ##

ChiralSuperfields and Lnewqban
haruspex said:
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.

Last edited:
ChiralSuperfields
Callumnc1 said:
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?

https://www.physicsforums.com/threa...2-dynamics-problem-involving-a-pulley.991370/

ChiralSuperfields
haruspex said:
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!

BvU said:
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

##\ ##

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

Many thanks!

erobz said:
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.

Callumnc1 said:
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 ..

##\ ##

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

##\ ##
Ok thank you @BvU !

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