66Suppose at any instant the top of the chain is at height h. At this instant the velocity of the chain is v = √2g(L-h).
In the next δt interval of time the length vδt of the chain loses it momentum to the pan. So the momentum imparted to the pan in the next δt interval is
mL (vδt) v
So the force which must have acted on the falling chain is momentum/time i.e.
mL v2
By Newton's third law, the same force must act on the pan as well. This force is besides the weight of the chain already lying on the pan which is mL (L-h) g.
Hence the total force (which is the reading shown)
= mL (L-h) g + mL v2 =mL (L-h) g + mL 2g(L-h) = 3mg(1 &ndash hL)
This formula holds as long as the fall continues after which the reading drops suddenly to mg.
