Browse Questions

# 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 pole is equal to in terms of man's weight w

$\begin {array} {1 1} (a)\;\frac{w}{4} & \quad (b)\;\frac{w}{2} \\ (c)\;\frac{3w}{4} & \quad (d)\;w \end {array}$

$(c) \frac{3w}{4}$