# Multiple Pulleys Problem

## Homework Statement

http://www.sumoware.com/images/temp/xziltxnrjlckqjqs.png [Broken]

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..

Last edited by a moderator:

ehild
Homework Helper
Use the condition that the length of the rope is constant.

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

Chestermiller
Mentor
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

ehild
Homework Helper
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?

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

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 ?

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.

Chestermiller
Mentor
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

Nathanael
Homework Helper
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.)

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 ?

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
Mentor
LAC+LCD+LDE+LEB = const
vA + 2 vC + vB = 0
vB = vA - 2 vC
Right ?
Check the sign on vA.

Nathanael
Homework Helper
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 ?

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

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 ?

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 ?

Nathanael
Homework Helper
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 ( 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

Then, how to find the relation between the acceleration ?
Differentiate the last equation with respect to time.

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/s

Differentiate 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)

Nathanael
Homework Helper
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)
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 A

This 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 :)