Solving Mass-on-Scale Problem with Falling Chain

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AI Thread Summary
A chain of mass M and length L is released from a vertical position, falling onto a scale. The velocity of the chain when a length x has fallen is given by V = sqrt(2gx). The forces acting on the scale include the weight of the fallen portion of the chain and the momentum change as it impacts the scale. The discussion focuses on expressing the mass flow rate (dm/dt) in terms of the chain's velocity and length, ultimately leading to the conclusion that the scale reads 3Mgx/L when the chain has fallen a length x. The solution is confirmed as correct by participants in the thread.
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Homework Statement


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

Homework Equations


dp = IMPULSE=F*dt
p=mv

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 can't 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.
 
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Dweirdo said:

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:)
 
Doc Al said:
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.
 
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.
 
Wooho!
thanks Al, i got 3Mgx/L
and it seems like the right answer.

Thanks,

Weirdo
 
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