Understanding the Velocity and Acceleration of Blocks Linked by Multiple Pulleys

[terryds]

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 v_A = 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..

 

[ehild]

Use the condition that the length of the rope is constant.

 

[terryds]

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 ?

 

[BvU]

It's a given that D is pulled down at 1 m/s

 

[Chestermiller]

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 v_A, what is $dL_{AC}/dt$ equal to?

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

Chet

 

[ehild]

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 a_A=4 m/s², downward. If its velocity is V_A 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 V_A and V_C what is the acceleration if B?

 

[terryds]

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 v_A, what is $dL_{AC}/dt$ equal to?

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

Chet

L_CD = L_DE
If mass A moves down with a velocity of v_A, what is dL_AC/dt equal to? It is v_A, right ?
If pulley C moves downward with a velocity of vC, what is dL_CD/dt equal to? What is dL_DE/dt equal to? It is v_C, right ?

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 a_A=4 m/s², downward. If its velocity is V_A 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 V_A and V_C 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 v_A, right ?
v_B = v_A - v_D, right ?
Please explain the correct answers. I'm really confused when finding the relationship of acceleration.

 

[Chestermiller]

L_CD = L_DE
If mass A moves down with a velocity of v_A, what is dL_AC/dt equal to? It is v_A, right ?
If pulley C moves downward with a velocity of vC, what is dL_CD/dt equal to? What is dL_DE/dt equal to? It is v_C, 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 v_A and v_C)?

Chet

 

[Nathanael]

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.)

 

[terryds]

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 v_A and v_C)?

Chet

L_AC+L_CD+L_DE+L_EB = const
v_A + 2 v_C + v_B = 0
v_B = v_A - 2 v_C
Right ?

 

[terryds]

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 ?

 

[Chestermiller]

L_AC+L_CD+L_DE+L_EB = const
v_A + 2 v_C + v_B = 0
v_B = v_A - 2 v_C
Right ?

Check the sign on v_A.

 

[Nathanael]

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 ?

 

[terryds]

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 ( v_A + v_B ), right ?
Then, how to find the relation between the acceleration ?

 

[terryds]

Check the sign on v_A.

-v_A + 2 v_C - v_B = 0
v_B = 2 v_C - v_A

I think v_A is negative since it goes downward ( the rope lengthens ), v_C is positive since the rope shortens, v_B is negative since the rope lengthens

Is it right ? Hmmm
Or we can just ignore the positive/negative rule
v_A + 2 v_C + v_B = 0
v_B = - (v_A + 2 v_C)

Which one is right ?

 

[Nathanael]

It will get longer by 2x..

Correct.

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.

So, the velocity of pulley D is 0.5 ( v_A + v_B ), right ?

Yes :) this is correct (ignoring directions). Also, V_a does not necessarily equal V_b, but their sum must get shorter by 2 m/s

Then, how to find the relation between the acceleration ?

Differentiate the last equation with respect to time.

 

[terryds]

Correct.Well, EB and AC don't necessarily have to decrease by the same amount.Yes :) this is correct (ignoring directions). Also, V_a does not necessarily equal V_b, but their sum must get shorter by 2 m/sDifferentiate the last equation with respect to time.

v_D = 0.5 ( v_A + v_B )
a_D = 0.5 ( a_A + a_B )
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
a_D = 0.5 ( a_A + a_B )
0 = 0.5 (4 + a_B)
a_B = -4 m/s²

v_D = 0.5 ( v_A + v_B )
-1 = 0.5 ( 4 + v_B )
v_B = -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 ?

 

[Nathanael]

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.

Can you tell me why it is negative ?

It is negative because it is in the opposite direction as block A

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.

 

[terryds]

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 :)