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.

Explanation

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 velocity v of this element will be given by
`v^(2) = u^(2) + 2gx`
` or v^(2) = 0 + 2 gx` ........ (1)
` or v^(2) = gx` ......... (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`