# Work done by a chain

1. May 8, 2017

### yecko

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

http://i.imgur.com/TdPWVgr.png

2. Relevant equations
WD=mgh

3. The attempt at a solution
mass per length=0.8kg/m
mgh= ∫(0,10)(0.8g(10-y))dy=[8gy-0.4y^2g](0,10)=80g-40g=40g
whats wrong with my calculation?
thanks

2. May 8, 2017

### BvU

What is 10-y in your equation ?

3. May 8, 2017

### yecko

the height the chain need to travel for each horizontal slice

4. May 8, 2017

### BvU

I thought you were integrating the force needed to do the work ....

But as for the height: then what are the limits of your integration ?

5. May 8, 2017

### yecko

0 to10
So what is your way to do so?
thanks

6. May 8, 2017

### BvU

No. Half the chain can be left as is.
comes down to the same thing.
Note that not everything has to be dragged up all the way to the roof.... so you have to revise the 10-y

7. May 8, 2017

### kuruman

Question: What is the change in potential energy of the center of mass?

8. May 8, 2017

### haruspex

I suggest you are making an assumption about how the bottom of the chain is to be lifted up. You perhaps are imagining standing on the roof, hauling manually, no equipment. I can think of two alternatives that would lead to the lower answer.

9. May 8, 2017

### yecko

Half of the chain means the center has zero change?
Does that mean the limit of y is 0 to 5
And the h=2(5-y)?

10. May 8, 2017

### haruspex

Are you trying to do it by integration for the algebraic exercise? There is a rather simpler approach.

11. May 8, 2017

### kuruman

If not by integration, the simpler approach referred to by @haruspex would be to consider that the external agent that lifts the chain, does work against gravity equal to the change in gravitational potential energy of the center of mass.

12. May 9, 2017

### yecko

Oh yes! 0.8*5*5g=20g
Right!!!!

But what if i need to integrate it? For instance the number of the question is not that perfect?
By mgh, whats wrong with these numbers?

13. May 9, 2017

### BvU

10-2y is correct. The first link has to be hauled up 10 m, the last one can stay in position. The one at e.g. y = 4 m goes up only 2 m.

14. May 9, 2017

### yecko

Thanks
intergrate mgh=0.8g(10-2y) from 0 to 5 =20g
Finally got it!