# Work Question (Pulling a chain on the Moon)

• Arman777
In summary: I get it nowIn summary, the conversation discussed a situation at a lunar base where a uniform chain hangs over the edge of a horizontal platform. A machine exerted 1.0J of work to pull the rest of the chain onto the platform. The chain had a mass of 2.0kg and a length of 3.0m. Using the equations ##ρ_{chain}=M/L## and ##W_g=-\Delta U_g##, the conversation attempted to find the initial length of chain hanging over the edge. It was determined that the answer is 1.36m.
Arman777
Gold Member

## Homework Statement

At a lunar base,a uniform chain hangs over the edge of a horizontal platform.A machine does ##1.0J## of work in pulling the rest of the chain onto the platform.The chain has a mass of ##2.0 kg## and a length of ##3.0m##.What length was initally hanging over the edge ? On the moon,the gravitational acceleraiton is ##\frac 1 6## of ##9,8 \frac {m} {s^2}##

## Homework Equations

##ρ_{chain}=M/L##
##W_g=-ΔU_g##

## The Attempt at a Solution

Let's suppose ##L_1## length is needed to pull.
So chain density will be ##ρ=\frac {2kg} {3m}=0.66\frac {kg} {m}##
The mass of ##L_1## is## L_1ρ=M_1##

The gravitational work on ##M_1## is ##W_g## which its ##-M_1gL_1=1J##
##(L_!)^2ρg=6J##

which ##L_1=1.05m## but the answer is ##1.4m##

Theres a "-" sign which makes me uncomfortable.Also I am making wrong at some point

Hint: If the length hanging over the edge is L, does every bit of the mass have to pulled up a height L?

Also, the work is done by some force pulling the chain up, so its sign would be +. (It's work against gravity, not by gravity.)

Here what I did

##(M_1dm)g(L1-dl)=W-dw## dw is zero and ##(M_1-dm)=(L_1-dl)p##
Is this true ?
Doc Al said:
Also, the work is done by some force pulling the chain up, so its sign would be +. (It's work against gravity, not by gravity.)

yeah that's right but the work done by gravity on the chain is negative .The work done by machine to the chain is positive.Thats why I confused

Arman777 said:
Here what I did

##(M_1dm)g(L1-dl)=W-dw## dw is zero and ##(M_1-dm)=(L_1-dl)p##
Is this true ?
Not quite sure what you're doing here. Are you trying to set up an integral?

Hint: If some extended object changed height, what point on the object would you track to compute its change in gravitational PE?

Arman777 said:
yeah that's right but the work done by gravity on the chain is negative .The work done by machine to the chain is positive.Thats why I confused
The work done by the machine equals the change in gravitational PE; both are positive.

Doc Al said:
The work done by the machine equals the change in gravitational PE; both are positive.
yep I see now
Doc Al said:
Not quite sure what you're doing here. Are you trying to set up an integral?
I was...If the lengh decreases a bit what would be happen
Doc Al said:
Hint: If some extended object changed height, what point on the object would you track to compute its change in gravitational PE?
Ok let me try again

I found 1.36m by usig this equation
##(L_1)^2ρg=12J##
Is it true ?

Arman777 said:
I found 1.36m by usig this equation
##(L_1)^2ρg=12J##
Is it true ?
Don't just toss out an equation; show how you got the equation.

Consider a mass element dm of the hanging section of the chain. If the length hanging is L, what is the average distance that each mass element must be lifted to get to the platform?

##M_1a_gH=1J##

##H=\frac {L_1} {2}##; The center of mass of ##M_1## moves this much.
##M_1=L_1ρ##
so,
##(L_1)ρ\frac {g} {6}\frac {L_1} {2}=1J##

Arman777 said:
##M_1a_gH=1J##

##H=\frac {L_1} {2}##; The center of mass of ##M_1## moves this much.
##M_1=L_1ρ##
so,
##(L_1)ρ\frac {g} {6}\frac {L_1} {2}=1J##
Perfect!

Doc Al said:
Consider a mass element dm of the hanging section of the chain. If the length hanging is L, what is the average distance that each mass element must be lifted to get to the platform?

Average distance will be ##\frac {L_1} {2}##

##∫\frac {L_1} {2}dmg=1J## from 0 to M and so This is true I am sure but If I wanted to convert it to ##dl## ,

##dm=dlρ## so ;
##∫\frac {L_1} {2}gρdl## from 0 to ##L_1p## ?

Last edited:
Once you use the center of mass there's no need to integrate. But if you do, you'll get the same answer.

Doc Al said:
Once you use the center of mass there's no need to integrate. But if you do, you'll get the same answer.
Oh ok thank you

## 1. How does pulling a chain on the Moon differ from pulling a chain on Earth?

On the Moon, the force of gravity is approximately one-sixth of that on Earth. This means that it would take less effort to pull a chain on the Moon compared to on Earth.

## 2. Would pulling a chain on the Moon be easier or harder than pulling a chain on Earth?

Pulling a chain on the Moon would be easier because of the lower force of gravity. However, the lack of atmosphere and lower friction on the Moon's surface may make the chain more difficult to grip and manipulate.

## 3. How would the weight of the chain affect pulling it on the Moon?

The weight of the chain would still play a role in pulling it on the Moon, but it would be less significant due to the lower force of gravity. This means that a heavier chain may be easier to pull on the Moon compared to on Earth.

## 4. Would the length of the chain affect pulling it on the Moon?

The length of the chain would not have a significant impact on pulling it on the Moon. As long as the chain is within reach and can be gripped, the lower force of gravity would make it easier to pull regardless of its length.

## 5. Is there any other factor that would affect pulling a chain on the Moon?

The lack of atmosphere on the Moon would also play a role in pulling a chain. Without air resistance, the chain would experience less drag, making it easier to pull. Additionally, the lower friction on the Moon's surface may also affect the ease of pulling the chain.

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