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

  $\frac{w}{4}$
• B

  $\frac{w}{2}$
• Correct answer
  C

  $\frac{3 w}{4}$
• D

  $w$

Solution

Man is sliding down the telegraphic pole with acceleration $g / 4$. So, $\begin{array}{rlrl} & m g-F & =\frac{m g}{4} \\ \Rightarrow & & F & =m g-\frac{m g}{4} \\ \Rightarrow & F & =\frac{3 m g}{4} \\ \Rightarrow & & F & =\frac{3 w}{4}\end{array}$

Hence, the correct answer is option: 'C'.