Understanding the Velocity and Acceleration of Blocks Linked by Multiple Pulleys

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Homework Help Overview

The discussion revolves around the dynamics of two blocks linked by a system of pulleys, specifically focusing on the velocities and accelerations of the blocks as one block is pulled down at a constant velocity. The problem involves analyzing the relationships between the movements of the blocks and the pulleys, as well as applying principles of mechanics to determine the unknown quantities.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the accelerations and velocities of blocks A and B, questioning whether their accelerations are equal in magnitude.
  • Some participants suggest using the constant length of the rope as a constraint to derive relationships between the movements of the blocks and pulleys.
  • There is discussion about the implications of external forces acting on the system and how they affect energy conservation principles.
  • Questions arise regarding the correct interpretation of the signs for velocities and accelerations, particularly in relation to the direction of movement.

Discussion Status

The discussion is ongoing, with participants providing insights and raising questions about the relationships between the variables involved. Some guidance has been offered regarding the use of rope length constraints and the implications of pulley movements, but there is still uncertainty about the correct relationships and calculations.

Contextual Notes

Participants are working under the assumption that the system is influenced by the constant downward pull of pulley D, which is affecting the dynamics of blocks A and B. There is also a focus on the need to clarify the directionality of the velocities and accelerations involved in the problem.

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Homework Statement


http://www.sumoware.com/images/temp/xziltxnrjlckqjqs.png

Two blocks ( A and B ) are linked by rope through three identical pulleys C, D, and E.
Pulley C and E are in stationary, while D is pulled down so its constant velocity 1 m/s
If vA = 4m/s when A goes down 2 m, determine the velocity and acceleration also the direction of block B when A goes 2 m down

Choices :
a) 4 m/s and 1 m/s^2 downward
b) 4 m/s and 2 m/s^2 downward
c) 4 m/s and 2 m/s^2 upward
d) 6 m/s and 4 m/s^2 downward
e) 6 m/s and 4 m/s^2 upward

Homework Equations


W = ΔKE
V(t)^2 = V0^2 + 2 a Δs

The Attempt at a Solution


[/B]
W = ΔKE
ma s = (1/2) m v^2
a (2) = (1/2) (4)^2
a = 4 m/s^2

So, the acceleration of block A is 4 m/s^2 downward
But, what is the block B acceleration and its velocity when block A goes 2m down ?

I need to find the relationship between block A acceleration and block B acceleration.
Their accelerations are same in magnitude, right ? (Since the rope is free)
So, block B acceleration is 4 m/s^2 downward, right ?
But, what about block B velocity ?
Using the equation
V(t)^2 = V0^2 + 2 a Δs
V(t) = √(2aΔs) = √16 = 4 m/s

But, my answer is not in the choices..
Please help me..
 
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Use the condition that the length of the rope is constant.
 
ehild said:
You can not apply conservation of energy, as some external force does work on the system when pulling down pulley D with constant velocity.
Use the condition that the length of the rope is constant.

Hmm..
If A goes 2m down, the rope C->A increases 2m in length
So, rope D->C decreases 2m in length
Then, rope E->D decreases 2m in length
And, rope E->B increases 2m in length
Right ?
Then, how to find the acceleration ?
 
It's a given that D is pulled down at 1 m/s
 
Following ehild's suggestion, you have ##L_{AC}+L_{CD}+L_{DE}+L_{EB}=Const.##

From the figure, how is ##L_{CD}## (always) related to ##L_{DE}##?

If mass A moves down with a velocity of vA, what is ##dL_{AC}/dt## equal to?

If pulley C moves downward with a velocity of vC, what is ##dL_{CD}/dt## equal to? What is ##dL_{DE}/dt## equal to?

Chet
 
As D moves downward with 1 m/s , the length of the piece AC+BE shortens at rate ... ?
A moves downward with uniform acceleration. You got aA=4 m/s2, downward. If its velocity is VA downward, the piece of rope AC lengthens at this rate. The piece EB has to shorten at rate ...?
From the velocity of B in terms of VA and VC what is the acceleration if B?
 
BvU said:
It's a given that D is pulled down at 1 m/s

Chestermiller said:
Following ehild's suggestion, you have ##L_{AC}+L_{CD}+L_{DE}+L_{EB}=Const.##

From the figure, how is ##L_{CD}## (always) related to ##L_{DE}##?

If mass A moves down with a velocity of vA, what is ##dL_{AC}/dt## equal to?

If pulley C moves downward with a velocity of vC, what is ##dL_{CD}/dt## equal to? What is ##dL_{DE}/dt## equal to?

Chet

LCD = LDE
If mass A moves down with a velocity of vA, what is dLAC/dt equal to? It is vA, right ?
If pulley C moves downward with a velocity of vC, what is dLCD/dt equal to? What is dLDE/dt equal to? It is vC, right ?

ehild said:
As D moves downward with 1 m/s , the length of the piece AC+BE shortens at rate ... ?
A moves downward with uniform acceleration. You got aA=4 m/s2, downward. If its velocity is VA downward, the piece of rope AC lengthens at this rate. The piece EB has to shorten at rate ...?
From the velocity of B in terms of VA and VC what is the acceleration if B?

The length piece AC+BE shortens at rate 0.5 m/s
The piece EB has to shorten at rate vA, right ?
vB = vA - vD, right ?
Please explain the correct answers. I'm really confused when finding the relationship of acceleration.
 
terryds said:
LCD = LDE
If mass A moves down with a velocity of vA, what is dLAC/dt equal to? It is vA, right ?
If pulley C moves downward with a velocity of vC, what is dLCD/dt equal to? What is dLDE/dt equal to? It is vC, right ?
Very good. Now, if you take the time derivative of this equation ##L_{AC}+L_{CD}+L_{DE}+L_{EB}=Const.## , and use the previous results, what do you get for the downward velocity of mass B, namely ##v_B=dL_{EB}/dt## (in terms of vA and vC)?

Chet
 
terryds said:
The length piece AC+BE shortens at rate 0.5 m/s
This is not right. If we pull the pulley D down by a certain distance, by how much does the rope in the middle get longer?
(This extra length comes from AC+BE being shortened.)
 
  • #10
Chestermiller said:
Very good. Now, if you take the time derivative of this equation ##L_{AC}+L_{CD}+L_{DE}+L_{EB}=Const.## , and use the previous results, what do you get for the downward velocity of mass B, namely ##v_B=dL_{EB}/dt## (in terms of vA and vC)?

Chet

LAC+LCD+LDE+LEB = const
vA + 2 vC + vB = 0
vB = vA - 2 vC
Right ?
 
  • #11
Nathanael said:
This is not right. If we pull the pulley D down by a certain distance, by how much does the rope in the middle get longer?
(This extra length comes from AC+BE being shortened.)
Hmm.. It must go down in the same distance as the distance the pulley D goes down.
So, its rate is also 1 m/s, right ?
 
  • #12
terryds said:
LAC+LCD+LDE+LEB = const
vA + 2 vC + vB = 0
vB = vA - 2 vC
Right ?
Check the sign on vA.
 
  • #13
terryds said:
Hmm.. It must go down in the same distance as the distance the pulley D goes down.
So, its rate is also 1 m/s, right ?
Take a look at this picture I made (it's harder to explain with words).

If the pulley goes down by x then the rope gets longer by ?

pulley.png
 
  • #14
Nathanael said:
Take a look at this picture I made (it's harder to explain with words).

If the pulley goes down by x then the rope gets longer by ?

View attachment 78839

It will get longer by 2x.. Thanks for your nice illustration..
Then the piece EB decreases x in length and AB also decreases x in length, right ?
So, the velocity of pulley D is 0.5 ( vA + vB ), right ?
Then, how to find the relation between the acceleration ?
 
  • #15
Chestermiller said:
Check the sign on vA.
-vA + 2 vC - vB = 0
vB = 2 vC - vA

I think vA is negative since it goes downward ( the rope lengthens ), vC is positive since the rope shortens, vB is negative since the rope lengthens

Is it right ? Hmmm
Or we can just ignore the positive/negative rule
vA + 2 vC + vB = 0
vB = - (vA + 2 vC)

Which one is right ?
 
  • #16
terryds said:
It will get longer by 2x..
Correct.

terryds said:
Then the piece EB decreases x in length and AB also decreases x in length, right ?
Well, EB and AC don't necessarily have to decrease by the same amount.

terryds said:
So, the velocity of pulley D is 0.5 ( vA + vB ), right ?
Yes :) this is correct (ignoring directions). Also, Va does not necessarily equal Vb, but their sum must get shorter by 2 m/s

terryds said:
Then, how to find the relation between the acceleration ?
Differentiate the last equation with respect to time.
 
  • #17
Nathanael said:
Correct.Well, EB and AC don't necessarily have to decrease by the same amount.Yes :) this is correct (ignoring directions). Also, Va does not necessarily equal Vb, but their sum must get shorter by 2 m/sDifferentiate the last equation with respect to time.

vD = 0.5 ( vA + vB )
aD = 0.5 ( aA + aB )
Right ?

Then, using the work-kinetic energy formula, I've found that acceleration of A is 4 m/s2
I will use negative sign for up direction and positive for down direction
aD = 0.5 ( aA + aB )
0 = 0.5 (4 + aB)
aB = -4 m/s2

vD = 0.5 ( vA + vB )
-1 = 0.5 ( 4 + vB )
vB = -6 m/s2

So, the acceleration is 4m/s2 upward ? and the velocity is 6 m/s upward ?

Can you tell me why it is negative ?
I imagine the scene and I think that block B will go in the same direction as block A, right ? (The block A goes down, so the middle pulley goes up, then block B goes down)
So, could you please tell me the mistake I made ?
 
  • #18
terryds said:
So, the acceleration is 4m/s2 upward ? and the velocity is 6 m/s upward ?
If the block A started with a velocity 4 m/s downwards and an acceleration 4 m/s2 downwards, then yes your answer is correct.

terryds said:
Can you tell me why it is negative ?
It is negative because it is in the opposite direction as block A

terryds said:
I imagine the scene and I think that block B will go in the same direction as block A, right ? (The block A goes down, so the middle pulley goes up, then block B goes down)
So, could you please tell me the mistake I made ?
This is not true. The middle pulley is always going down at 1m/s, that is unchangeable.
 
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  • #19
Nathanael said:
If the block A started with a velocity 4 m/s downwards and an acceleration 4 m/s2 downwards, then yes your answer is correct.It is negative because it is in the opposite direction as block AThis is not true. The middle pulley is always going down at 1m/s, that is unchangeable.

Yeah.. I forgot that the middle pulley is going down..
Thanks for your nice help :)
 

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