Work Question (Pulling a chain on the Moon)

[Arman777]

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

 

[Doc Al]

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.)

 

[Arman777]

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 ?

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

 

[Doc Al]

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?

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.

 

[Arman777]

The work done by the machine equals the change in gravitational PE; both are positive.

yep I see now

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

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

 

[Arman777]

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

 

[Doc Al]

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?

 

[Arman777]

$M_1a_gH=1J$

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

 

[Doc Al]

$M_1a_gH=1J$

$H=\frac {L_1} {2}$; 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!

 

[Arman777]

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 $L_1p$ ?

 

[Doc Al]

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

 

[Arman777]

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