Question Number 108133 by ajfour last updated on 14/Aug/20
Commented by ajfour last updated on 14/Aug/20
$${Rod}\:{AB}\:{inside}\:{a}\:{smooth}\:{sphere} \\ $$$${of}\:{radius}\:{R},\:{Find}\:{Normal} \\ $$$${reactions}\:{at}\:{A}\:{and}\:{B}. \\ $$
Answered by mr W last updated on 15/Aug/20
Commented by mr W last updated on 15/Aug/20
$${let}\:\lambda=\frac{{m}}{{M}} \\ $$$${AC}=\frac{{M}×\frac{{a}+{b}}{\mathrm{2}}+{m}×{a}}{{M}+{m}}=\frac{\left(\mathrm{1}+\mathrm{2}\lambda\right){a}+{b}}{\mathrm{2}\left(\mathrm{1}+\lambda\right)} \\ $$$${CB}=\frac{{a}+\left(\mathrm{1}+\mathrm{2}\lambda\right){b}}{\mathrm{2}\left(\mathrm{1}+\lambda\right)} \\ $$$$\mathrm{cos}\:\gamma=\frac{{a}+{b}}{\mathrm{2}{R}} \\ $$$$\Rightarrow\gamma=\mathrm{cos}^{−\mathrm{1}} \left\{\frac{{a}+{b}}{\mathrm{2}{R}}\right\} \\ $$$$\frac{\mathrm{sin}\:\left(\alpha+\gamma\right)}{\mathrm{sin}\:\alpha}=\frac{{AO}}{{AC}}=\frac{\mathrm{2}\left(\mathrm{1}+\lambda\right){R}}{\left(\mathrm{1}+\mathrm{2}\lambda\right){a}+{b}} \\ $$$$\mathrm{cos}\:\gamma+\frac{\mathrm{sin}\:\gamma}{\mathrm{tan}\:\alpha}=\frac{\mathrm{2}\left(\mathrm{1}+\lambda\right){R}}{\left(\mathrm{1}+\mathrm{2}\lambda\right){a}+{b}} \\ $$$$\Rightarrow\alpha=\mathrm{tan}^{−\mathrm{1}} \left\{\frac{\sqrt{\mathrm{1}−\left(\frac{{a}+{b}}{\mathrm{2}{R}}\right)^{\mathrm{2}} }}{\frac{\mathrm{2}\left(\mathrm{1}+\lambda\right){R}}{\left(\mathrm{1}+\mathrm{2}\lambda\right){a}+{b}}−\frac{{a}+{b}}{\mathrm{2}{R}}}\right\} \\ $$$$\Rightarrow\beta=\mathrm{tan}^{−\mathrm{1}} \left\{\frac{\sqrt{\mathrm{1}−\left(\frac{{a}+{b}}{\mathrm{2}{R}}\right)^{\mathrm{2}} }}{\frac{\mathrm{2}\left(\mathrm{1}+\lambda\right){R}}{{a}+\left(\mathrm{1}+\mathrm{2}\lambda\right){b}}−\frac{{a}+{b}}{\mathrm{2}{R}}}\right\} \\ $$$$\frac{{N}_{\mathrm{1}} }{\mathrm{sin}\:\beta}=\frac{\left({M}+{m}\right){g}}{\mathrm{sin}\:\mathrm{2}\gamma} \\ $$$$\Rightarrow{N}_{\mathrm{1}} =\frac{\mathrm{sin}\:\beta\:\left({M}+{m}\right){g}}{\mathrm{sin}\:\mathrm{2}\gamma} \\ $$$$\Rightarrow{N}_{\mathrm{2}} =\frac{\mathrm{sin}\:\alpha\:\left({M}+{m}\right){g}}{\mathrm{sin}\:\mathrm{2}\gamma} \\ $$
Commented by ajfour last updated on 15/Aug/20
$${So}\:{quick},\:{and}\:{superb}\:{solution}, \\ $$$${thanks}\:{a}\:{lot}\:{Sir}. \\ $$
Commented by otchereabdullai@gmail.com last updated on 15/Aug/20
$$\mathrm{The}\:\mathrm{world}\:\mathrm{mathematician}/\:\mathrm{physicist} \\ $$$$\mathrm{I}\:\mathrm{always}\:\mathrm{become}\:\mathrm{very}\:\mathrm{happy}\:\mathrm{when}\:\mathrm{i}\:\mathrm{see} \\ $$$$\mathrm{your}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{a}\:\mathrm{question} \\ $$
Answered by ajfour last updated on 15/Aug/20
Commented by ajfour last updated on 15/Aug/20
$${N}\mathrm{cos}\:\left(\alpha+\theta\right)={F}\mathrm{cos}\:\left(\alpha−\theta\right)\:\:\:\:…\left({i}\right) \\ $$$${p}=\sqrt{{R}^{\mathrm{2}} −\left(\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{cos}\:\alpha=\frac{{a}+{b}}{\mathrm{2}{R}} \\ $$$${N}\mathrm{sin}\:\left(\alpha+\theta\right)+{F}\mathrm{sin}\:\left(\alpha−\theta\right)=\left({M}+{m}\right){g} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:……\left({ii}\right) \\ $$$${Torque}\:{balance}\:{about}\:{centre}\:{of}\:{sphere} \\ $$$${mg}\left(\frac{{b}−{a}}{\mathrm{2}}−{p}\mathrm{tan}\:\theta\right)\mathrm{cos}\:\theta={Mg}\left({p}\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\:\:\:\mathrm{tan}\:\theta=\frac{{m}\left({b}−{a}\right)}{\mathrm{2}{p}\left({M}+{m}\right)} \\ $$$${Now}\:\:{using}\:\left({i}\right)\:{in}\:\:\left({ii}\right) \\ $$$$\:\:\:{N}\left\{\mathrm{sin}\:\left(\alpha+\theta\right)+\mathrm{sin}\:\left(\alpha−\theta\right)\frac{\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{cos}\:\left(\alpha−\theta\right)}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({M}+{m}\right){g} \\ $$$$\Rightarrow\:\:{N}=\frac{\left({M}+{m}\right){g}\mathrm{cos}\:\left(\alpha−\theta\right)}{\mathrm{sin}\:\mathrm{2}\alpha}\:\:\:\:\& \\ $$$$\:\:\:\:\:\:\:\:{F}=\frac{\left({M}+{m}\right){g}\mathrm{cos}\:\left(\alpha+\theta\right)}{\mathrm{sin}\:\mathrm{2}\alpha}\:\:\:\blacksquare \\ $$
Commented by mr W last updated on 15/Aug/20
$${perfect}\:{solution}! \\ $$