A pulley system with two pulleys and two suspended masses

[member 731016]

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!

 

[kuruman]

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

 

[member 731016]

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!

 

[BvU]

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.

$\$

 

[member 731016]

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!

 

[erobz]

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!

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.

 

[BvU]

is there way to prove this thought?

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

$\$

 

[member 731016]

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!

 

[member 731016]

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

$\$

Thank you for your reply @BvU!

 

[erobz]

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.

 

[member 731016]

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?

Many thanks!

 

[member 731016]

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

Many thanks!

 

[erobz]

Thank you for your reply @erobz !

Is the ok for the variables?
View attachment 322798
Many thanks!

Yeah, that will work.

 

[member 731016]

Yeah, that will work.

Thank you for your reply @erobz !

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

Many thanks!

 

[erobz]

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.

 

[member 731016]

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!

 

[erobz]

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.

 

[member 731016]

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!

 

[member 731016]

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!

 

[erobz]

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.

 

[member 731016]

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!

 

[erobz]

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 $z \to z+ \Delta z$.

 

[member 731016]

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!

 

[member 731016]

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!

 

[erobz]

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 $z + \Delta z$.

 

[member 731016]

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!

 

[member 731016]

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!

 

[erobz]

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.

 

[member 731016]

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!

 

[member 731016]

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

Oh thank you for your reply @erobz !

 

[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!