# Homework Help: Multiple Pulleys Problem

1. Feb 9, 2015

### terryds

1. The problem statement, all variables and given/known data
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

2. Relevant equations
W = ΔKE
V(t)^2 = V0^2 + 2 a Δs

3. The attempt at a solution

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: May 7, 2017
2. Feb 9, 2015

### ehild

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

3. Feb 9, 2015

### terryds

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 ?

4. Feb 9, 2015

### BvU

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

5. Feb 9, 2015

### Staff: 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

6. Feb 9, 2015

### 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 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?

7. Feb 9, 2015

### terryds

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 ?

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.

8. Feb 9, 2015

### Staff: Mentor

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

9. Feb 9, 2015

### Nathanael

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. Feb 9, 2015

### terryds

LAC+LCD+LDE+LEB = const
vA + 2 vC + vB = 0
vB = vA - 2 vC
Right ?

11. Feb 9, 2015

### terryds

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. Feb 9, 2015

### Staff: Mentor

Check the sign on vA.

13. Feb 9, 2015

### Nathanael

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 ?

14. Feb 9, 2015

### terryds

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. Feb 9, 2015

### terryds

-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. Feb 9, 2015

### Nathanael

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.

17. Feb 9, 2015

### terryds

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)

18. Feb 9, 2015

### Nathanael

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.

19. Feb 9, 2015

### terryds

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