Weight registered of a falling chain Let's say I have a  metallic chain. Suppose I hold one of its tip of in my hand let the chain hang freely downward such that the other tip touches a weighing scale. At this point, with  me holding the chain the weight, 
registered on the machine will be zero. 
But once I let go of the chain, how will the reading on the machine change? Will 
it show the same reading all the way down (ie total weight of the chain)
or will it only show the weight of the part of the chain that it is in contact with.
I don't think this is one of those 'lift' problems since there, the weighing scale itself is NOT moving. 
Intuitively I think the weight registered will increase, but what is confusing me 
is that if I stand on a weighing scale, my head is not touching the scale,
in the same way different parts of the chain are not touching the scale while falling. 
 A: Using the 'momentum theorem' $F\Delta t=\Delta p$—（1）， and the weighing scale (at time $t$) results from two contributions:
One is the weight of the part of the chain that lie on the machine, which is equal to $\rho \frac{gt^2}{2}g$—（2）, where $\rho $ is the line density of the chain, $\frac{gt^2}{2}$ is the length of the lying part of the chain with $g$ the acceleration of gravity.
The other contribution comes from the force $\Delta F$ that acts on the falling chain to stop it. Specifically, imagine that at time $t$, the contacting point A on the chain will change to B after a very 'small' time interval $\Delta t$, and during this progress, the 'small' part of the chain (with length $v_t\Delta t$, where $v_t=gt$ is the velocity of the point A at time $t$) that connects A and B was stopped by the force $\Delta F$ which comes from the machine. Thus, apply Eq.(1) to this 'small' part, we have $(\Delta F-\Delta m\cdot g)\Delta t=\Delta m\cdot v_t$, where $\Delta m=\rho \cdot v_t\Delta t$ is the mass of the 'small' part, then we arrive at $\Delta F=\rho v_t^2=\rho g^2t^2(\Delta t\rightarrow0)$—（3）.
Finally, the weighing scale (at time $t$) $=Eq.(2)+Eq.(3)=\frac{3}{2}\rho g^2t^2$, which is 3 times of Eq.(2)—the weight of the lying part of the chain on the machine.
Remark: Referring to what confusing you in your last paragraph, one key difference between "you stand on the scale" and "the different part of the chain standing on the scale" is that the former is static while the latter is moving downward with acceleration of gravity $g$. Thus, back to your question, at any instant time $t$ the reading of the scale will not depend on the length(hence the mass) of the falling part of the chain.
A: "Intuitive" answer:
The weight registered on the scale would be the sum of two components. One component is due to the accumulating links which have come to rest on the scale (or each other), and this would increase approximately stepwise as each successive link comes in contact with the scale. The other component would be due to the currently arriving link decelerating to a stop; once it has come to a stop, there will no longer be any deceleration force, and it becomes part of the accumulating set of links in the first component. An approximation of this deceleration component would be a sharp spike in force (weight) as each link contacts the scale (or accumulating set of links already at rest) and is brought to a stop. Since there is no force communicated up the chain, the entire portion of it not yet at rest on the scale would fall with a 1g acceleration. This means that the first links to settle on the scale will exert minimal deceleration force on the scale as they come to rest, whereas the final links to land will exert a larger deceleration force, as they will have fallen through a longer distance, and had more time to accelerate to a greater velocity before encountering the scale (or other stopped links).
So, a graph of the scale reading might be approximated by a "staircase" of equal step heights superimposed with an upward spike at the outer edge of each step; the spikes would be very small at the bottom, increasing in amplitude in a parabolic progression from step to step; this whole view would be in turn distorted such that the horizontal "run" (time dimension) of each step would decrease in a parabolic progression. 
