A piece of uniform string hangs vertically so that its free end just touches horizontal surface of a table . The upper end of the string is now released . Show that at any instant during the falling of string , the total force on the surface is three times the weight of that part of string lying on the surface .

A piece of uniform string hangs vertically so that its free end just t

1.	A piece of uniform string hangs vertically so that its free end just touches horizontal surface of a table . The upper end of the string is now released . Show that at any instant during the falling of string , the total force on the surface is three times the weight of that part of string lying on the surface .
Answer» Solution :LET m be the mass per unit length of string and X the length of string at any instant on the table . If differential length dx of string further falls in time dt, then velocityv of this element will be given by ` v^(2)=u^(2)+2gx` or `v^(2)=0+2gx`....(1) or `v^(2)=2gx`....(2) Net force on the surface of the table `F=(dp)/(dt)`+weight of string lying on the surface `=(m dx .v )/(dt)+mxg=mv(dx)/(dt) +mxg =mv^(2)+mgx` Using (1) , `F=m(2gx)+mgx=3(mx)G` But `mx=M^(1)= `weight of string lying on the table `:. F =3M^(1)g`