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)

• A

  $\frac{\mathrm{W}}{4}$

• B

  $\frac{3\mathrm{W}}{4}$

• C

  $\frac{\mathrm{W}}{2}$

• D

  W