WBJEE · Physics · Laws of Motion
A block of mass m rests on a horizontal table with a co-efficient of static friction \(\mu .\) What minimum force must be applied on the block to drag it on the table?
- A \(\frac{\mu}{\sqrt{1+\mu^{2}}} \mathrm{mg}\)
- B \(\frac{\mu-1}{\mu+1} \mathrm{mg}\)
- C \(\frac{\mu}{\sqrt{1-\mu^{2}}} \mathrm{mg}\)
- D \(\mu \mathrm{mg}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mu}{\sqrt{1+\mu^{2}}} \mathrm{mg}\)
Step-by-step Solution
Detailed explanation
Hint: \(F \sin \theta+N=m g \quad N=m g-F \sin \theta\) \(F \cos \theta=\mu(m g-F \sin \theta) \quad F=\frac{\mu m g}{\cos \theta+\mu \sin \theta}\), for \(F_{\min }, \frac{d}{d \theta}[\cos \theta+\mu \sin \theta]=0\) \(\therefore \tan \theta=\mu\)…
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