A pulley system with two pulleys and two suspended masses

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...
  • #1
ChiralSuperfields
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Homework Statement
Pls see below
Relevant Equations
Pls see below
For this problem,
1677189160333.png

The solution is,
1677189178930.png

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!
 
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  • #2
The problem states that the masses are equal and the solution is based on that.
 
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  • #3
kuruman said:
The problem states that the masses are equal and the solution is based on that.
Thank you for your reply @kuruman !

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

Many thanks!
 
  • #4
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.

##\ ##
 
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  • #5
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.

##\ ##
Thank you for your reply @BvU!

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

Many thanks!
 
  • #6
Callumnc1 said:
Thank you for your reply @BvU!

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

Many thanks!
1677200458048.png


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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  • #7
Callumnc1 said:
is there way to prove this thought?

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

##\ ##
 
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  • #8
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.
Thank you for your reply @erobz ! That diagram is quite helpful :)

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

Many thanks!
 
  • #9
BvU said:
Yes ! The overwhelming beauty of physics: The experiment decides !

##\ ##
Thank you for your reply @BvU!
 
  • #10
Callumnc1 said:
Thank you for your reply @erobz ! That diagram is quite helpful :)

## 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.
 
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  • #11
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.
Thank you for your reply @erobz !

Is the ok for the variables?
1677200509314.png

Many thanks!
 
  • #12
From the diagram it looks like ##c = b##, but would you like me to leave it like that @erobz ?

Many thanks!
 
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  • #13
Callumnc1 said:
Thank you for your reply @erobz !

Is the ok for the variables?
View attachment 322798
Many thanks!
Yeah, that will work.
 
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  • #14
erobz said:
Yeah, that will work.
Thank you for your reply @erobz !

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

Many thanks!
 
  • #15
Callumnc1 said:
Thank you for your reply @erobz !

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

Many thanks!
ok, but you are still missing a length of rope there.
 
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  • #16
erobz said:
ok, but you are still missing a length of rope there.
Thank you for your reply @erobz !

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!
 
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  • #17
Callumnc1 said:
Thank you for your reply @erobz !

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.
 
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  • #18
erobz said:
Great! Now identify the only variables that change in that equation when the blocks move.
Thank you for your reply @erobz !

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

Many thanks!
 
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  • #19
Callumnc1 said:
Thank you for your reply @erobz !

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

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

Many thanks!
 
  • #20
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.
 
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  • #21
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...
Thank you for your reply @erobz !

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!
 
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  • #22
Callumnc1 said:
Thank you for your reply @erobz !

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##.
 
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  • #23
Thank you for your reply @erobz !

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!
 
  • #24
erobz said:
So what is the new equation? Rember a variable that changes goes from ##z \to z+ \Delta z##.
Thank you for your reply @erobz!

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!
 
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  • #25
Callumnc1 said:
Thank you for your reply @erobz!

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##.
 
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  • #26
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!
 
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  • #27
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##.
Oh ok thank you for your reply @erobz !

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

Many thanks!
 
  • #28
Callumnc1 said:
Oh ok thank you for your reply @erobz !

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.
 
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  • #29
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!
 
  • #30
erobz said:
I believe it's the standard we approach stuff in Calculus.
Oh thank you for your reply @erobz !
 
  • #31
Here it is @erobz ,

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

Many thanks!
 
  • #32
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##
 
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  • #33
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##
Thank you for your reply @erobz !

So an equivalent notation is

##z ≈ z + dz##
 
  • #34
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##
 
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  • #35
erobz said:
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!
 

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