# 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...
ChiralSuperfields
Homework Statement
Pls see below
Relevant Equations
Pls see below
For this problem,

The solution is,

However, how do they know the block B will move up and block A will move down? The masses of each are not given so could be the other way round if ##m_b > m_a##?

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

Many thanks!

The problem states that the masses are equal and the solution is based on that.

ChiralSuperfields
kuruman said:
The problem states that the masses are equal and the solution is based on that.

Sorry I did not see that! I will try the problem again.

Many thanks!

Callumnc1 said:
Also how do they know that if block B moves up by 1cm, block A will move down by 2cm?
That is a matter of 'counting': in the picture you see three lengths of cord. Think of the situation that block A is all the way down: by then A has dropped by two lengths, whereas B has gone up by only one length.

##\ ##

ChiralSuperfields and Lnewqban
BvU said:
That is a matter of 'counting': in the picture you see three lengths of cord. Think of the situation that block A is all the way down: by then A has dropped by two lengths, whereas B has gone up by only one length.

##\ ##

I think I might see what you mean from intuition, is there way to prove this thought?

Many thanks!

Callumnc1 said:

I think I might see what you mean from intuition, is there way to prove this thought?

Many thanks!

Write the total length of the inextensible rope ##L##, In terms of all the other lengths of each section. There will be a couple more steps, but that is where you start.

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ChiralSuperfields
Callumnc1 said:
is there way to prove this thought?

Yes ! The overwhelming beauty of physics: The experiment decides !

##\ ##

ChiralSuperfields
erobz said:
View attachment 322795

Write the total length of the inextensible rope ##L##, In terms of all the other lengths of each section. There will be a couple more steps, but that is where you start.

## L = l + s + 2.5 \times purple~semicircle##

Many thanks!

BvU said:
Yes ! The overwhelming beauty of physics: The experiment decides !

##\ ##

Callumnc1 said:

## L = l + s + 2.5 \times purple~semicircle##

Many thanks!
Sorry, I shouldn't have drawn that top section purple, it could be any length. I'll update the image.

1) just assign a random variable to each length that is not already labeled.
2) you are missing some length of rope in that sum.

ChiralSuperfields
erobz said:
Sorry, I shouldn't have drawn that top section purple, it could be any length.

1) just assign a random variable to each length that is not already labled
2) you are missing some length of rope in that sum.

Is the ok for the variables?

Many thanks!

From the diagram it looks like ##c = b##, but would you like me to leave it like that @erobz ?

Many thanks!

erobz
Callumnc1 said:

Is the ok for the variables?
View attachment 322798
Many thanks!
Yeah, that will work.

ChiralSuperfields
erobz said:
Yeah, that will work.

##L = a + s + b + c + l##

Many thanks!

Callumnc1 said:

##L = a + s + b + c + l##

Many thanks!
ok, but you are still missing a length of rope there.

ChiralSuperfields
erobz said:
ok, but you are still missing a length of rope there.

Sorry should be ##L = a + 2s + b + c + l## and assuming that the pulleys are the same size as the rope is taut, then ##c = b ## so simplifying gives ##L = a + 2s + 2b + l##

Many thanks!

erobz
Callumnc1 said:

Sorry should be ##L = a + 2s + b + c + l##

Many thanks!
Great! Now identify the only variables that change in that equation when the blocks move.

ChiralSuperfields
erobz said:
Great! Now identify the only variables that change in that equation when the blocks move.

The variables that change are ##l## and ##s##

Many thanks!

erobz
Callumnc1 said:

The variables that change are ##l## and ##s##

Many thanks!
I guess they are functions of time then?

Many thanks!

Callumnc1 said:
I guess they are functions of time then?

Many thanks!
Don't worry about that just yet.

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...

Write the new equation.

ChiralSuperfields
erobz said:
Don't worry about that just yet.

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...

The change for each length is

##\Delta l##
##\Delta s##

Which in the original equation is

##L = a + 2\Delta s + 2b + \Delta l##Many thanks!

erobz
Callumnc1 said:

The change for each length is

##\Delta l##
##\Delta s##

Many thanks!
So what is the new equation? Rember a variable that changes goes from ##z \to z+ \Delta z##.

ChiralSuperfields

I this this equation,

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

Can also be expressed as##L_i = L_f##
## a + 2s_i + 2b + l_i = a + 2s_f + 2b + l_f##
## 2s_i + l_i =2s_f+ l_f##

Many thanks!

erobz said:
So what is the new equation? Rember a variable that changes goes from ##z \to z+ \Delta z##.

Do you mean

##z_i = z_f + \Delta z##?

If you want I could edit post #23 to use that notation. This is giving me conservation of string vibes I remember sometime talking about it on Discord.

Many thanks!

erobz
Callumnc1 said:

Do you mean

##z_i = z_f + \Delta z##?

If you want I could edit post #23 to use that notation. This is giving me conservation of string vibes I remember sometime talking about it on Discord.

Many thanks!
you don't need the subscripts. Its just saying you take what was labeled ##z## in the original eq. and replace it with ##z + \Delta z##.

ChiralSuperfields
I think @erobz that,

## L_i = 2s_i + l_i =2s_f+ l_f = L_f##

is really just a conservation of string statement, correct?

Many thanks!

erobz
erobz said:
you don't need the subscripts. Its just saying you take what was labeled ##z## in the original eq. and replace it with ##z + \Delta z##.

I'm not too familiar with that notation. I will change to that notation.

Many thanks!

Callumnc1 said:

I'm not too familiar with that notation. I will change to that notation.

Many thanks!
I believe it's the standard we approach stuff in Calculus.

ChiralSuperfields
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!

erobz said:
I believe it's the standard we approach stuff in Calculus.

Here it is @erobz ,

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

Many thanks!

Callumnc1 said:
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##

ChiralSuperfields
erobz said:
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##

So an equivalent notation is

##z ≈ z + dz##

Callumnc1 said:
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##

ChiralSuperfields
erobz said:
Ok, now equate the two equations through the variable ##L## and cancel the terms. What are you left with?

Which two equations sorry?

Many thanks!

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