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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 ×(h/(sin θ))=mg×((l cos θ)/(2 )) N_1 =mg×((l sin θ cos θ)/(2h)) N=mg−N_1 cos θ=mg(1−((l sin θ cos^2 θ)/(2h))) f=N_1 sin θ=mg ((l sin^2 θ cos θ)/(2h)) f=μN mg ((l sin^2 θ cos θ)/(2h))=μ mg(1−((l sin θ cos^2 θ)/(2h))) ((l sin^2 θ cos θ)/(2h))=μ(1−((l sin θ cos^2 θ)/(2h))) μ=((l sin^2 θ cos θ)/(2h))×((2h)/(2h−l sin θ cos^2 θ)) μ=((l sin^2 θ cos θ)/(2h−l sin θ cos^2 θ))

$${N}_{\mathrm{1}} ×\frac{{h}}{\mathrm{sin}\:\theta}={mg}×\frac{{l}\:\mathrm{cos}\:\theta}{\mathrm{2}\:} \\ $$$${N}_{\mathrm{1}} ={mg}×\frac{{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}} \\ $$$${N}={mg}−{N}_{\mathrm{1}} \:\mathrm{cos}\:\theta={mg}\left(\mathrm{1}−\frac{{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}{h}}\right) \\ $$$${f}={N}_{\mathrm{1}} \:\mathrm{sin}\:\theta={mg}\:\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}} \\ $$$${f}=\mu{N} \\ $$$${mg}\:\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}}=\mu\:{mg}\left(\mathrm{1}−\frac{{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}{h}}\right) \\ $$$$\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}}=\mu\left(\mathrm{1}−\frac{{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}{h}}\right) \\ $$$$\mu=\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}}×\frac{\mathrm{2}{h}}{\mathrm{2}{h}−{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\mu=\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}−{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$

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 (1) is correct.

$${Option}\:\left(\mathrm{1}\right)\:{is}\:{correct}. \\ $$

Commented by Tinkutara last updated on 23/Mar/18

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

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