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
$${N}_{\mathrm{1}} ×\frac{{h}}{\mathrm{sin}\:\theta}={mg}×\frac{{l}\:\mathrm{cos}\:\theta}{\mathrm{2}\:} \\ $$$${N}_{\mathrm{1}} ={mg}×\frac{{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}} \\ $$$${N}={mg}−{N}_{\mathrm{1}} \:\mathrm{cos}\:\theta={mg}\left(\mathrm{1}−\frac{{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}{h}}\right) \\ $$$${f}={N}_{\mathrm{1}} \:\mathrm{sin}\:\theta={mg}\:\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}} \\ $$$${f}=\mu{N} \\ $$$${mg}\:\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}}=\mu\:{mg}\left(\mathrm{1}−\frac{{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}{h}}\right) \\ $$$$\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}}=\mu\left(\mathrm{1}−\frac{{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{2}{h}}\right) \\ $$$$\mu=\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}}×\frac{\mathrm{2}{h}}{\mathrm{2}{h}−{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$$\mu=\frac{{l}\:\mathrm{sin}^{\mathrm{2}} \:\theta\:\mathrm{cos}\:\theta}{\mathrm{2}{h}−{l}\:\mathrm{sin}\:\theta\:\mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$
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
$${Option}\:\left(\mathrm{1}\right)\:{is}\:{correct}. \\ $$
Commented by Tinkutara last updated on 23/Mar/18
Thank you very much Sir! I got the answer.