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

• A. $\frac{w}{4}$
• B. $\frac{w}{2}$
• C. $\frac{3 w}{4}$
• D. $w$

Answer 1

Answer: (C)

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