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Falling Chain

  • Thread starter Dweirdo
  • Start date
174
0
1. The problem statement, all variables and given/known data
A chain of mass M and length L is suspended vertically with the lower end touching a scale.
the chain is released and falls onto the scale.
what is the reading of the chain when a length x is fallen?
neglect the size of individual links

2. Relevant equations
dp = IMPULSE=F*dt
p=mv

3. The attempt at a solution
Well this is what I've done so far.
the velocity of the specific part of the chain when it hits the scale is V= sqrt(2gx)
F1= Mgx/L -weight of a X part of the chain.
now the second force is quite a problem.
F2=dp/dt=d(mv)/dt =V(dm/dt)......
what do I do from here??
i need to express dm/dt with the information i got, but cant find a way... =trying to translate it to words: the rate the mass hits the chain or...? I'm kinda stuck.
Any help appreciated!
Thank You.
 
Last edited:

Doc Al

Mentor
44,684
1,004
3. The attempt at a solution
Well this is what I've done so far.
the velocity of the specific part of the chain when it hits the scale is V= sqrt(2gx)
F1= Mgx/L -weight of a X part of the chain.
Good.
now the second force is quite a problem.
F2=dp/dt=d(mv)/dt =V(dm/dt)......
what do I do from here??
Express dm in terms of dx. (You're doing fine. :wink:)
 
174
0
Good.

Express dm in terms of dx. (You're doing fine. :wink:)
well i thought about v*M/L (for dm/dt)the only thing i found that works with the units mass/seconds)
but which v do i place here if it's right?

for dm alone it's xM/L?
Thanks Al.
 

Doc Al

Mentor
44,684
1,004
dm = M/L dx, so dm/dt = M/L dx/dt = M/L v, where v is the speed of the piece of chain (dm) hitting the scale, which you already found in post #1.
 
174
0
Wooho!
thanks Al, i got 3Mgx/L
and it seems like the right answer.

Thanks,

Weirdo
 

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