A pulley system with two pulleys and two suspended masses

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Homework Help Overview

The discussion revolves around a pulley system involving two pulleys and two suspended masses. Participants explore the mechanics of the system, particularly the movement of the blocks and the relationship between the lengths of the rope sections.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question how the direction of movement for blocks A and B is determined without given masses. They also inquire about the relationship between the distances moved by the blocks and the lengths of the rope.

Discussion Status

Some participants have provided insights into the mechanics of the system, suggesting that the movement of the blocks can be understood through the configuration of the rope. Others are exploring the implications of their assumptions and seeking further clarification on the relationships between the variables involved.

Contextual Notes

There are discussions about the equality of the masses and the implications of that assumption on the movement of the blocks. Participants also mention the need for clarity on the lengths of the rope sections and how they relate to the overall system dynamics.

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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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The problem states that the masses are equal and the solution is based on that.
 
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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!
 
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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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!
 
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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Callumnc1 said:
is there way to prove this thought?

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

##\ ##
 
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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!
 
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 !
 

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