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Question Number 32218 by Tinkutara last updated on 21/Mar/18

Answered by mrW2 last updated on 21/Mar/18

Commented by mrW2 last updated on 22/Mar/18

$$N_1\times \frac{h}{\sin \theta }=mg\times \frac{l\cos \theta }{2}$$$$N_1=mg\times \frac{l\sin \theta \cos \theta }{2h}$$$$N=mg-N_1\cos \theta =mg(1-\frac{l\sin \theta {\cos }^2\theta }{2h})$$$$f=N_1\sin \theta =mg\frac{l{\sin }^2\theta \cos \theta }{2h}$$$$f=\mu N$$$$mg\frac{l{\sin }^2\theta \cos \theta }{2h}=\mu mg(1-\frac{l\sin \theta {\cos }^2\theta }{2h})$$$$\frac{l{\sin }^2\theta \cos \theta }{2h}=\mu (1-\frac{l\sin \theta {\cos }^2\theta }{2h})$$$$\mu =\frac{l{\sin }^2\theta \cos \theta }{2h}\times \frac{2h}{2h-l\sin \theta {\cos }^2\theta }$$$$\mu =\frac{l{\sin }^2\theta \cos \theta }{2h-l\sin \theta {\cos }^2\theta }$$

Commented by Tinkutara last updated on 22/Mar/18

Sir can you please recheck your answer or options are wrong?

Commented by mrW2 last updated on 22/Mar/18

$$Option\left(1\right)iscorrect.$$

Commented by Tinkutara last updated on 23/Mar/18

Thank you very much Sir! I got the answer. ��������

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