How Much Work Is Needed to Pull a Hanging Chain Onto a Table?

  • Thread starter NAkid
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In summary, the conversation discusses the method for finding the work required to pull a hanging chain back onto a frictionless table. The correct method is to use energy methods or to integrate the force, which is not constant due to the changing mass distribution. The correct answer can be found by integrating (9.8*density*x*dx) from 0.325 to 0, where density is the linear density of the chain.
  • #1
NAkid
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[SOLVED] this should be really easy but..

Homework Statement


A chain is held on a frictionless table with one-fourth of its length hanging over the edge, as shown in the figure. If the chain has length L = 1.30m and mass m = 2.6kg, how much work is required to pull the hanging part back onto the table?


Homework Equations





The Attempt at a Solution


ΔX=1.3*.25=.325
mass=2.6
g=9.8
F=m*a
W=FΔX
W=(2.6)(9.8)(.325)=8.281J


not the right answer
 
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  • #2
should i not assume that the mass of the chain is evenly distributed over the length of the chain?
 
  • #3
NAkid said:

The Attempt at a Solution


ΔX=1.3*.25=.325
mass=2.6
g=9.8
F=m*a
W=FΔX
W=(2.6)(9.8)(.325)=8.281J


not the right answer
Since the force is not constant, to use this method you'd need to integrate. (The integration is easy though.) Why not use energy methods instead? How does the gravitational PE change?
 
  • #4
NAkid said:
should i not assume that the mass of the chain is evenly distributed over the length of the chain?

Yes, you should. But you can't use F*dx to get the work, since F isn't constant. As more of the chain is pulled onto the table F decreases. So you want to write F as a function of x, the amount of chain left hanging, and then integrate F(x)*dx.
 
  • #5
so, integrate (2.6/1.3)dx from 0 to .325?
 
  • #6
wait no that would be if the mass wasn't distributed evenly, never mind.
 
  • #7
NAkid said:
wait no that would be if the mass wasn't distributed evenly, never mind.

The overhanging mass is the density of the chain times x. What other factor do you need to get a force?
 
  • #8
using energy methods, i have W=mgy1-mgy2
you're lifting from y1=-.325 to y2=0
is the mass proportional to the chain or is it constant?
 
Last edited:
  • #9
The overhanging mass is the density of the chain times x. What other factor do you need to get a force?

you need acceleration..
 
  • #10
NAkid said:
using energy methods, i have W=mgy1-mgy2
you're lifting from y1=-.325 to y2=0
is the mass proportionate to the chain or is it constant?
Hint: Consider the part initially hanging. Where's the center of mass of that piece? How does it change when the chain is pulled up? (Alternatively, what's the average distance that the hanging mass elements must be raised?)

The mass of any piece of chain is proportional to its length.
 
  • #11
NAkid said:
The overhanging mass is the density of the chain times x. What other factor do you need to get a force?

you need acceleration..

You need g. Integrate g*density*x*dx from 0.325 to 0. Using energy is also instructive. See if you get the same thing.
 
  • #12
hm, integral of (9.8xdx) from .325 to 0 = .517, which is not what the answer i should get
 
  • #13
NAkid said:
hm, integral of (9.8xdx) from .325 to 0 = .517, which is not what the answer i should get

Put in the density. (I omitted that when I first posted).
 
  • #14
ah ok, i thought there was something missing, density = mass/volume --> 2.6/...?
 
  • #15
Linear density. (total mass)/(total length).
 
  • #16
never mind ---- got it.
 

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