Calculating the Force on a Falling Chain: Solving a Momentum Problem

[cordyceps]

Homework Statement

A chain of length L and total mass M is released from rest with its lower end just touching the top of a table. Find the force exerted by the table on the chain after the chain has fallen through a distance x. (Assume each link comes to rest the instant it reaches the table.)

Homework Equations

p= mv
ek= (1/2)mv^2

The Attempt at a Solution

First, I tried to find the force by taking the derivative of kinetic energy:
ek= (1/2)mv^2
m= M(L-x)/L, v^2= 2xg
ek= Mxg(L-x)/L
d/dt(ek)= Mgv-2Mgxv/L

F= [d/dt(ek)]/v= Mg-2Mgx/L
But this turns out to be the wrong answer. Any advice? Thanks.

 

[Carid]

Assuming that the mass is distributed equally along the chain, what proportion of the mass M is on the table when a length x out of a total length L is on the table?

 

[cordyceps]

mass= M(x/L)
But I don't think its as simple as the normal force of the mass on the table: Mg(x/L).

 

[Carid]

Well the answer does seem a bit simple and obvious. But I think it is right.
Let's ask ourselves what force, if any, does the table exert on the part of the chain which is falling?

 

[cordyceps]

In addition to the normal force, I think there is also a force somehow related to the motion of the part of the chain that is in the air. Mg(x/L) is not the right answer according to the book.

 

[Carid]

OK a force is characterised by a change in momentum.

At the beginning the momentum of the chain is zero.

After the chain has fallen X the part of the chain in motion has a certain momentum.

Maybe they want you to express the instantaneous change in momentum.

 

[cordyceps]

Thanks for the help.
So d/dt[ (M(L-X)/L) * sqrt(sgX) ] = ... Mg - 3Mgx/L
-> Force= 3Mgx/L - Mg
This is fairly close to the answer = 3Mgx/L. Why is there the Mg?