Dunno....don't remember the answer..would you care to elaborate?
A chain of mass M and length l is suspended vertically with its
lowest end touching a scale. The chain is released and falls onto the
scale.
What is the reading of the scale when a length of chain, x, has fallen?
(Neglect the size of individual links.)
-
UP 0 DOWN 0 1 4
4 Answers
Anirudh Kumar
·2009-12-17 01:29:37
Weight of part of chain already on the scale = Mxg/L .....1
now velocity of the small part ( dm) which falls instantly on the scale = √2gx
thus impulse on the scale = dm √2gx
=> F.dt=dm √2gx (f=impulsive force)
=> F= (M/L)*dx/dt*(√2gx)
=>F= v2M/L
=>f=2gxM/L .............2
thus net force = 3Mgx/L
thus reading = 3MxL