Question

a man slides down a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional between man and pole is equal to in terms of man's weight W.

Answer 1

A man slides down a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional between man and pole is equal to in terms of man's weight $W$.

Solution

According to Newton's second law

$$ma = mg - F_{fr} = m \left(\frac{1}{4} g\right) = \frac{1}{4} mg \gg F_{fr} = mg - \frac{1}{4} mg = \frac{3}{4} mg = \frac{3}{4} W$$