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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 $w$
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between man and pole is equal to in terms of man's weight $w$
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Verified Answer
The correct answer is:
$\frac{3 w}{4}$
Man is sliding down the telegraphic pole with acceleration $g / 4$. So,
$\begin{array}{rlrl} & m g-F & =\frac{m g}{4} \\ \Rightarrow & & F & =m g-\frac{m g}{4} \\ \Rightarrow & F & =\frac{3 m g}{4} \\ \Rightarrow & & F & =\frac{3 w}{4}\end{array}$
$\begin{array}{rlrl} & m g-F & =\frac{m g}{4} \\ \Rightarrow & & F & =m g-\frac{m g}{4} \\ \Rightarrow & F & =\frac{3 m g}{4} \\ \Rightarrow & & F & =\frac{3 w}{4}\end{array}$
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