What is the velocity of the block pulled by strings

[randomgamernerd]

Homework Statement

:[/B]In the arrangement shown in the figure, the two ends P and Q of the light unstrechable string move downward with uniform speed u as shown. The upward speed of the mass m is v. Then which of the following is correct:
A) v cos60°=u
B) 2v cos60=u
C) v =2ucos60
D) v= u cos60

Homework Equations

: [/B]we need to take components and equate I guess

The Attempt at a Solution

since the strings are massless, hence velocity of string is u throughout.
so each string has a velocity ucos60° in the direction of normal.
Hence 2ucos60°=v
apparently this is wrong i don't know why.
I have attached two pictures...
one of the question and one of my attempt.
Please bear with my clumsy handwriting.
Thanks in advance

 

[tnich]

Homework Statement

:[/B]In the arrangement shown in the figure, the two ends P and Q of the light unstrechable string move downward with uniform speed u as shown. The upward speed of the mass m is v. Then which of the following is correct:
A) v cos60°=u
B) 2v cos60=u
C) v =2ucos60
D) v= u cos60

Homework Equations

: [/B]we need to take components and equate I guess

The Attempt at a Solution

since the strings are massless, hence velocity of string is u throughout.
so each string has a velocity ucos60° in the direction of normal.
Hence 2ucos60°=v
apparently this is wrong i don't know why.
I have attached two pictures...
one of the question and one of my attempt.
Please bear with my clumsy handwriting.
Thanks in advance

So you are saying that the velocity of the mass is twice the normal velocity of the strings pulling on it? You are adding the normal velocities of the strings together?

 

[randomgamernerd]

So you are saying that the velocity of the mass is twice the normal velocity of the strings pulling on it? You are adding the normal velocities of the strings together?

yah

 

[tnich]

yah

So the mass is traveling twice as fast as the strings pulling on it? How would that work?
Imagine that the angle is 0.

 

[haruspex]

each string has a velocity ucos60° in the direction of normal.

That is true for the bit of string right up close to the pulley, but the rest of it moves differently.
The piece up close to the mass is getting closer to the pulley at rate u, but it also has a component normal to that.

 

[randomgamernerd]

So the mass is traveling twice as fast as the strings pulling on it? How would that work?
Imagine that the angle is 0.

that is because velocities of the strings are adding up

 

[randomgamernerd]

That is true for the bit of string right up close to the pulley, but the rest of it moves differently.
The piece up close to the mass is getting closer to the pulley at rate u, but it also has a component normal to that.

didn't get you..can you please rephrase a bit?

 

[tnich]

that is because velocities of the strings are adding up

Take the case where the angle of the strings is 0° and they are hanging vertically with the mass at the ends. Consider one of the strings, traveling upwards at a velocity v = u cos(0) = u. So the end of the string connected to the mass is moving upward at velocity v = u, while the mass is moving upward at velocity 2u. But since the end of the string is connected to the mass and the mass is moving upward at velocity 2u, the end of the string must be be moving upward at velocity 2u. How can the end of string move at velocity u and velocity 2u at the same time?

 

[haruspex]

didn't get you..can you please rephrase a bit?

If you look at a part of the string about to reach a pulley it is moving (mostly) towards the pulley at speed u. Hence its vertical component of movement is u cos θ, where θis the angle of the string to the vertical.
But if you look at a part of the string right next to the mass it is moving upwards, so somewhat towards the pulley, again at speed u, but also tangentially. If its overall movement is upwards, and its component towards the pulley is u, what is that upward speed? What is the tangential component?

 

[ehild]

since the strings are massless, hence velocity of string is u throughout.
so each string has a velocity ucos60° in the direction of normal.
Hence 2ucos60°=v
apparently this is wrong i don't know why.

The length of the string does not change. if the vertical pieces get longer by uΔt the piece L gets shorter by uΔt, and the mass rises by Δy. Apply the Law of Cosines to get Δy from the yellow triangle (as Δt is small you can ignore the second-order terms). v=Δy/Δt.

 

[randomgamernerd]

If you look at a part of the string about to reach a pulley it is moving (mostly) towards the pulley at speed u. Hence its vertical component of movement is u cos θ, where θis the angle of the string to the vertical.
But if you look at a part of the string right next to the mass it is moving upwards, so somewhat towards the pulley, again at speed u, but also tangentially. If its overall movement is upwards, and its component towards the pulley is u, what is that upward speed? What is the tangential component?

can you explain with a diagram[emoji28]

 

[randomgamernerd]

The length of the string does not change. if the vertical pieces get longer by uΔt the piece L gets shorter by uΔt, and the mass rises by Δy. Apply the Law of Cosines to get Δy from the yellow triangle (as Δt is small you can ignore the second-order terms). v=Δy/Δt.

View attachment 225181

yes, i get this part. But my question is if both the strings are changing by the same length, each causes a ∆y rise in the mass, so that gets added up and in total, the mass rises by 2∆y

 

[randomgamernerd]

ok, i am posting an imaginary situation which is roughly similar to this question, i want you guys just to confirm if I’m on the right track. Please bear with my poor handwriting.

so according to you people, change in height of the mass 5 Kg is equal to change in length of the 2kg mass. Right?? I guess I was confusing the mass with some moving massless pulley.Had it been a pulley instead of the block, then the displacement lf pulley would have been twice the change in length of each half of the string going round it..right?

 

[ehild]

yes, i get this part. But my question is if both the strings are changing by the same length, each causes a ∆y rise in the mass, so that gets added up and in total, the mass rises by 2∆y

No, the strings do not cause ∆y rise each. ∆y is the rise of the mass, due to the shortening of the strings. It is simple geometry.
See the symmetric figure: Given a, b, and theta, determine the distance d.

 

[randomgamernerd]

No, the strings do not cause ∆y rise each. ∆y is the rise of the mass, due to the shortening of the strings. It is simple geometry.
See the symmetric figure: Given a, b, and theta, determine the distance d.
View attachment 225262

i think i get it...can you please go through my new post(#13)

 

[ehild]

i think i get it...can you please go through my new post(#13)

The situation in your Post #13 is different from the original problem. The two hanging masses should be equal.
If they are, the constant length of both strings ensures that as much one piece of string gets longer as much the other piece gets shorter.
If there are different masses attached to the mass in the middle, the symmetry breaks, it will go sideways.

 

[randomgamernerd]

The situation in your Post #13 is different from the original problem. The two hanging masses should be equal.
If they are, the constant length of both strings ensures that as much one piece of string gets longer as much the other piece gets shorter.
If there are different masses attached to the mass in the middle, the symmetry breaks, it will go sideways.

View attachment 225272

no, i meant a situation where the strings remain parallel, not at any angle with the normal..

 

[ehild]

no, i meant a situation where the strings remain parallel, not at any angle with the normal..

If the two masses are not equal either the strings do not stay parallel, of the block between them will turn.

 

[randomgamernerd]

If the two masses are not equal either the strings do not stay parallel, of the block between them will turn.

I am neglecting its rotational behavior at present for simplicity of understanding, like let's consider it point mass for the present discussion, but the strings are close enough so that they remain parallel.

 

[ehild]

I am neglecting its rotational behavior at present for simplicity of understanding, like let's consider it point mass for the present discussion, but the strings are close enough so that they remain parallel.

Then one string becomes slack.

 

[randomgamernerd]

Then one string becomes slack.

okay...
so what about the acceleration of the block connected to the slack string?

 

[haruspex]

can you explain with a diagram[emoji28]

Try to draw your own diagram from this description:

Suppose a section of string is passing up and over a pulley at speed u.
That means every part of the string section must be getting closer to the pulley at that rate.
If we take some point on this string we can resolve its velocity into two components: one along the string and one normal to it.
E.g. with u=0 it could still happen that the string is revolving around the pulley for some reason.
We can call these components the radial and tangential velocities.

At any instant, the tangential velocity varies along the string, being maximum where it contacts the mass. Let the tangential speed there be v and the angle the string makes to the vertical be θ.
We know that the end of the string contacting the mass moves vertically, so the horizontal components of the tangential and radial velocities cancel there:
u sin(θ) = v cos(θ)
In the context of the question, both the radial velocity, u, and the tangential velocity contribute to the rising speed of the mass.
The contribution from u is u cos(θ) and from the tangential speed v sin(θ) = u sin(θ)tan(θ). So the upward speed of the mass must be
u cos(θ)+u sin(θ)tan(θ) = u(cos²(θ)+sin²(θ))/cos(θ) = u sec(θ).

Alternatively, we could have got there much faster by looking at it the other way around. If the mass is getting nearer to the pulley at rate u then the component of its motion that is towards the pulley must be u: u = w cos(θ).
But I went through the other analysis to show why your method went wrong: you left out the tangential motion of the string.

 

[randomgamernerd]

Try to draw your own diagram from this description:

Suppose a section of string is passing up and over a pulley at speed u.
That means every part of the string section must be getting closer to the pulley at that rate.
If we take some point on this string we can resolve its velocity into two components: one along the string and one normal to it.
E.g. with u=0 it could still happen that the string is revolving around the pulley for some reason.
We can call these components the radial and tangential velocities.

At any instant, the tangential velocity varies along the string, being maximum where it contacts the mass. Let the tangential speed there be v and the angle the string makes to the vertical be θ.
We know that the end of the string contacting the mass moves vertically, so the horizontal components of the tangential and radial velocities cancel there:
u sin(θ) = v cos(θ)
In the context of the question, both the radial velocity, u, and the tangential velocity contribute to the rising speed of the mass.
The contribution from u is u cos(θ) and from the tangential speed v sin(θ) = u sin(θ)tan(θ). So the upward speed of the mass must be
u cos(θ)+u sin(θ)tan(θ) = u(cos²(θ)+sin²(θ))/cos(θ) = u sec(θ).

Alternatively, we could have got there much faster by looking at it the other way around. If the mass is getting nearer to the pulley at rate u then the component of its motion that is towards the pulley must be u: u = w cos(θ).
But I went through the other analysis to show why your method went wrong: you left out the tangential motion of the string.

Is this what you mean?
(I have drawn only one of the pulleys)

 

[haruspex]

View attachment 225320
Is this what you mean?
(I have drawn only one of the pulleys)

Yes, but you see how difficult it is to express it all in one diagram. You show three arrows from the block but no clue as to what they mean.

Might be better not to show the block at all. Label the arrow towards the left pulley as u (the string end's radial component of velocity) the arrow to the right as v, the tangential component, and put a doubled arrow on the vertical to indicate the resultant velocity.

 

[randomgamernerd]

Yes, but you see how difficult it is to express it all in one diagram. You show three arrows from the block but no clue as to what they mean.

Might be better not to show the block at all. Label the arrow towards the left pulley as u (the string end's radial component of velocity) the arrow to the right as v, the tangential component, and put a doubled arrow on the vertical to indicate the resultant velocity.

okay, finally I get the physics behind what's happening.. Thanks a lot.

 

[ehild]

okay...
so what about the acceleration of the block connected to the slack string?

Sorry, I answered hastily.
Slack means zero tension. Let be m2>>m3 >m1. If you release the system, m2 can go down and m3 up so fast, that the piece CE of the blue string becomes slack, and that causes m1 accelerate downward with g. But soon the pieces CE and DE become of equal length again, shorter by the same amount as AB and FG get longer.
The length of a string does not change, so ΔAB+ΔCE= 0 and ΔDE+ΔFG=0, the magnitudes of the acceleration are the same, but the tensions are different in the strings.

If m3 is connected to a third pulley, there is one string only, and ΔAB+ΔCE+ΔDH+ΔFG=0 and ΔCE=ΔDH, so ΔCE=(ΔAB+ΔFG)/2

 

[randomgamernerd]

Sorry, I answered hastily.
Slack means zero tension. Let be m2>>m3 >m1. If you release the system, m2 can go down and m3 up so fast, that the piece CE of the blue string becomes slack, and that causes m1 accelerate downward with g. But soon the pieces CE and DE become of equal length again, shorter by the same amount as AB and FG get longer.
The length of a string does not change, so ΔAB+ΔCE= 0 and ΔDE+ΔFG=0, the magnitudes of the acceleration are the same, but the tensions are different in the strings.

View attachment 225319

If m3 is connected to a third pulley, there is one string only, and ΔAB+ΔCE+ΔDH+ΔFG=0 and ΔCE=ΔDH, so ΔCE=(ΔAB+ΔFG)/2

View attachment 225321

thanks , now i get it..