AP EAMCET · PHYSICS · Laws of Motion
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 & Solution
Correct Answer
(C) \(\frac{3 w}{4}\)
Step-by-step Solution
Detailed explanation
Man is sliding down the telegraphic pole with acceleration \(g / 4\). So, \(\begin{array}{rlrl} & m g-F & =\frac{m g}{4} \\ \Rightarrow & & F & =m g-\frac{m g}{4} \\ \Rightarrow & F & =\frac{3 m g}{4} \\ \Rightarrow & & F & =\frac{3 w}{4}\end{array}\)
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