IIT JEE , AIEEE Forum - Physics , Mechanics - How much is the force created due to the motion of the falling chain ?

elessar_iitkgp

The answer given in the book seems to be incorrect. My approach is a bit different from urs.
The left portion of the string is a variable mass system in which mass is entering.
By Merchersky's equation,
m(dv/dt) = F_ext + u_rel (dm/dt) = F_ext + u_rel (dm/dx)(dx/dt)
As the left portion of the chain is in equilibrium,
0 = {T-(m/L)[(L/2)+(x/2)]}j +(-

2gxj)(m/L)(

2gx)
T = (mg/L)[(L/2) + (5x/2)]
= (W/2)[1+(5x/L)]

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