Question

A man slides down on a telegraphic pole with an acceleration equal to one fourth of acceleration due to gravity. the frictional force between man and the pole is equal to [in terms of man's weight w]

Answer 1

From FBD shown in the figure above,
$W - F= ma \\ = > \: \: W - F = m \frac{g}{4}$
$Putting \: \: \: \: \: (mg = W ) \\ \\ we \: get \\ W - F \: = \frac{W}{4 } \\ = > \: \: F = W - \frac{W}{4 } \: \\ = > \: \: F = \frac{3W}{4}$

Answer 2

$\frac{3w}{4}$