Calculating Work Done to Pull Chain Upward: A 2m Chain of Mass 4kg

[Yashbhatt]

Homework Statement

A 2 m long chain of mass 4 kg rests on a table such that 60 cm of it hangs vertically downwards from the table. If the chain has uniform mass distribution, calculate the work done in pulling the entire chain upwards. Ignore the frictional force.

Homework Equations

W = U = mgh, W = Fdcosθ

The Attempt at a Solution

First, I calculated the mass of the 60 cm section of the chain using ratios. Then, I calculated the difference in potential energy of the 60 cm section when it was hanging initially and when it was entirely pulled up using W = U = mg(h₀ - h). I got the value as 7.2 J. But the answer was 3.6 N. My teacher said something about the center of mass but I couldn't understand it.

 

[olivermsun]

You aren't pulling the entire 60 cm section upward by 60 cm. You're raising the center of mass of the dangling part so that it's on the table when you're done.

 

[Yashbhatt]

But won't we still have to apply force after the center of mass is on the table(which in this case would be at the midpoint of the chain) to pull the rest of the chain upwards is on the table?

P.S. The question says "...pulling the entire chain upwards."

 

[Nathanael]

Only the very end of the chain moves up by 60cm. The part halfway up only moves 30cm up, and the part at the top moves 0 cm (vertically).
(etc. etc. etc. for every point on the chain)

So the average height that the chain moves up is only 30cm

The way to solve this is to look at the center of mass of the 60cm part of the chain. How high up does this center of mass move up? What is the change in energy?

But won't we still have to apply force after the center of mass is on the table(which in this case would be at the midpoint of the chain) to pull the rest of the chain upwards is on the table?

If you marked off the center of mass (before pulling it up) and then pulled the chain so that the mark was on the table, the new center of mass would not be on the table. There would still be part of the chain hanging down which would have a center of mass below the table.
Forgive me for not being very concise but hopefully it helps.

 

[Yashbhatt]

Only the very end of the chain moves up by 60cm. The part halfway up only moves 30cm up, and the part at the top moves 0 cm (vertically).
(etc. etc. etc. for every point on the chain)

So the average height that the chain moves up is only 30cm

The way to solve this is to look at the center of mass of the 60cm part of the chain. How high up does this center of mass move up? What is the change in energy?


If you marked off the center of mass (before pulling it up) and then pulled the chain so that the mark was on the table, the new center of mass would not be on the table. There would still be part of the chain hanging down which would have a center of mass below the table.



Forgive me for not being very concise but hopefully it helps.

But that's like sum of infinite series. You keep on finding the new center of mass and pulling it up.

 

[Satvik Pandey]

We can assume that total mass of chain hanging down the table is located at the CM of the part of chain below the table.You just need to find the work done in lifting that point(CM).

 

[Yashbhatt]

We can assume that total mass of chain hanging down the table is located at the CM of the part of chain below the table.You just need to find the work done in lifting that point(CM).

But even after the CM is on the table, you have to pull the remaining part.

 

[Nathanael]

But that's like sum of infinite series. You keep on finding the new center of mass and pulling it up.

There's always a complicated way to solve a problem.

But very often, as in this case, there is a simple way.

No energy is lost to friction, so conservation of energy will be useful to find the work done.

 

[Yashbhatt]

There's always a complicated way to solve a problem.

But very often, as in this case, there is a simple way.

No energy is lost to friction, so conservation of energy will be useful to find the work done.

Okay. But I don't get why do have to assume that pulling only the center of mass is enough?

 

[olivermsun]

You don't assume that. What you do is compare the before-and-after potential energies of the system. You had to perform work on the system equal to the gain in potential energy, which is exactly determined by how much you've raised the center of the mass of the system.

 

[ehild]

See the picture. If the hanging part has n units (links) and each link has mass m and length ΔL you have to exert F=mg force to balance the weight of the hanging chain and do W(n)=(nm)g ΔL work to lift the chain by one link. The hanging part becomes one link shorter. To lift the next link, you exert (n-1)mg force and do W(n-1) = mg(n-1)ΔL work, and so on.

Initially, the hanging part consisted of N links. So your work to lift all the links of the chain is

W= W(N)+W(N-1).W(N-2)+... +W(2)+W(1)= mgΔL[N+(N-1)+(M-2)+...2+1].

It is an arithmetic series. Yo know that 1+2+...(N-1)+N= (1+N)N /2≈ N²2 if N >>1.

But m=M/N and ΔL = L/N, so the total work is W=(M/N)g(L/N) N²/2 = MGL/2, as if you pulled up a point mass M by length L/2.


ehild

 

[Yashbhatt]

See the picture. If the hanging part has n units (links) and each link has mass m and length ΔL you have to exert F=mg force to balance the weight of the hanging chain and do W(n)=(nm)g ΔL work to lift the chain by one link. The hanging part becomes one link shorter. To lift the next link, you exert (n-1)mg force and do W(n-1) = mg(n-1)ΔL work, and so on.

Initially, the hanging part consisted of N links. So your work to lift all the links of the chain is

W= W(N)+W(N-1).W(N-2)+... +W(2)+W(1)= mgΔL[N+(N-1)+(M-2)+...2+1].

It is an arithmetic series. Yo know that 1+2+...(N-1)+N= (1+N)N /2≈ N²2 if N >>1.

But m=M/N and ΔL = L/N, so the total work is W=(M/N)g(L/N) N²/2 = MGL/2, as if you pulled up a point mass M by length L/2.


ehild

Great explanation. Just a question. You mentioned that the sum is N² /2 if N >> 1. But what it's only a few links?

 

[olivermsun]

Great explanation. Just a question. You mentioned that the sum is N² /2 if N >> 1. But what it's only a few links?

If N = 3, then the sum 1 + 2 + ... + N = 1 + 2 + 3.

 

[ehild]

You have to pull the last link only from half of its length. So W=mgΔL[N+N-1+N-2+...3+2]+mgΔL/2.

(You can grab the last link at the middle)

ehild

 

[Yashbhatt]

If N = 3, then the sum 1 + 2 + ... + N = 1 + 2 + 3.

But then one cannot say that it is equivalent to N² / 2.

 

[olivermsun]

But then one cannot say that it is equivalent to N² / 2.

It is approximately true for N » 1 (N "large"), which is what ehild said.

 

[Yashbhatt]

It is approximately true for N»1 (N "large), which is what ehild said.

And what I say is that what if you have 3 or 4 links hanging there. In that case, it won't be N² / 2 but N(N + 1)/2. So, we won't get the L/2 thing.

 

[olivermsun]

You have a chain of mass M composed of 3 links, each with length ΔL, hence the total length of the chain L = 3ΔL.

Link #1's center of gravity is in the middle of the link, at height -ΔL/2. Therefore it needs to be pulled up by ΔL/2 and its potential energy increases by (M/3)gΔL/2.

Link #2 needs to be pulled up the length of the previous link, ΔL, plus another ΔL/2. Its potential energy increases by (M/3)g * 3ΔL/2.

Link #3 needs to be pulled up 2ΔL for the previous links and another ΔL/2. Its potential energy increases by (M/3)g * 5ΔL/2.

Total ΔPE = (M/3)gΔL/2 + 3(M/3)gΔL/2 + 5(M/3)gΔL/2 = Mg(1 + 3 + 5)/3 *ΔL/2 = Mg * 3ΔL/2 = Mg L/2.

 

[ehild]

But then one cannot say that it is equivalent to N² / 2.

You are right. In fact, you have to pull up the upmost link only by ΔL/2, while the others move up by ΔL.
So you grab the upmost link. You need to move it up only by ΔL/2 then its CM will be in level of the platform, and you lay it down, and move horizontally till the top of the next link reaches the platform. Your work is 2mgΔL+mgΔL/2. Lifting the next link needs mgΔL+mgΔL/2 work, and at last, your work is mgΔL/2. The total work is mgΔL[(2+1+0)+3˙1/2]=4.5 mgΔL.
In general, it is $W=mgΔL(\frac{N(N-1)}{2}+N\frac{1}{2})=\frac{N^2}{2}$

ehild

 

[Murtuza Tipu]

For such type of sum
general formula can be found out
work done =MgL/2n^2

where M is total mass of the whole chain
g is gravitational constant
L is total length of chain
n is the length of hanging chain in terms of L
Suppose L is 100 and length of hanging part is 25 then length of hanging part in terms of L is L/4 hence in this example n =4.
This formula can be derived easily.

 

[Murtuza Tipu]

And what I say is that what if you have 3 or 4 links hanging there. In that case, it won't be N² / 2 but N(N + 1)/2. So, we won't get the L/2 thing.

Just take the ratio
it is possible that ratio my be an non terminating rational number
just square it and use it in the formula

And if only 3 or 4 links are hanging then
n= no. of total links /no. of links which are hanging
so just take the ratio

thing which you said that it would be n (n+1)/2 is not true and suppose you think that it 8 true give a logical explanation or proof.

Here n is a ratio hence every time we will get n is terms of total length or total links multiplied by an irrational number.

 

[olivermsun]

thing which you said that it would be n (n+1)/2 is not true and suppose you think that it 8 true give a logical explanation or proof.

Here n is a ratio hence every time we will get n is terms of total length or total links multiplied by an irrational number.

$?$

 

[Yashbhatt]

Just take the ratio
it is possible that ratio my be an non terminating rational number
just square it and use it in the formula

And if only 3 or 4 links are hanging then
n= no. of total links /no. of links which are hanging
so just take the ratio

thing which you said that it would be n (n+1)/2 is not true and suppose you think that it 8 true give a logical explanation or proof.

Here n is a ratio hence every time we will get n is terms of total length or total links multiplied by an irrational number.

I said that in that case n >> 1 is not true. So, one cannot approximate it to n^2/2.