Question-90509

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Question Number 90509 by ajfour last updated on 24/Apr/20

Commented by ajfour last updated on 24/Apr/20

If the brown hemisphere is about to slide, find θ.

$${If}\:{the}\:{brown}\:{hemisphere}\:{is}\:{about} \\ $$$${to}\:{slide},\:{find}\:\theta. \\ $$

Commented by mr W last updated on 24/Apr/20

$${solid}\:{or}\:{hollow}? \\ $$

Commented by ajfour last updated on 24/Apr/20

$${solid},\:{Sir}. \\ $$

Answered by mr W last updated on 24/Apr/20

Commented by mr W last updated on 25/Apr/20

R=radius ϕ=30° φ=tan^(−1) μ OM=λR with λ=(3/8) for solid, λ=(1/2) for hollow ((OS)/(sin φ))=((OA)/(sin (60°+φ)))=((2R)/( (√3)cos φ+sin φ)) ⇒OS=((2μR)/( (√3)+μ)) OS cos 30°=OM cos θ (((√3)μR)/( (√3)+μ))=λR cos θ cos θ=(((√3)μ)/(λ((√3)+μ))) ⇒θ=cos^(−1) (((√3)μ)/(λ((√3)+μ)))

$${R}={radius} \\ $$$$\varphi=\mathrm{30}° \\ $$$$\phi=\mathrm{tan}^{−\mathrm{1}} \mu \\ $$$${OM}=\lambda{R} \\ $$$${with}\:\lambda=\frac{\mathrm{3}}{\mathrm{8}}\:{for}\:{solid},\:\lambda=\frac{\mathrm{1}}{\mathrm{2}}\:{for}\:{hollow} \\ $$$$\frac{{OS}}{\mathrm{sin}\:\phi}=\frac{{OA}}{\mathrm{sin}\:\left(\mathrm{60}°+\phi\right)}=\frac{\mathrm{2}{R}}{\:\sqrt{\mathrm{3}}\mathrm{cos}\:\phi+\mathrm{sin}\:\phi} \\ $$$$\Rightarrow{OS}=\frac{\mathrm{2}\mu{R}}{\:\sqrt{\mathrm{3}}+\mu} \\ $$$${OS}\:\mathrm{cos}\:\mathrm{30}°={OM}\:\mathrm{cos}\:\theta\: \\ $$$$\frac{\sqrt{\mathrm{3}}\mu{R}}{\:\sqrt{\mathrm{3}}+\mu}=\lambda{R}\:\mathrm{cos}\:\theta\: \\ $$$$\mathrm{cos}\:\theta=\frac{\sqrt{\mathrm{3}}\mu}{\lambda\left(\sqrt{\mathrm{3}}+\mu\right)} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{3}}\mu}{\lambda\left(\sqrt{\mathrm{3}}+\mu\right)} \\ $$

Commented by ajfour last updated on 25/Apr/20

$${Nice}\:{method},\:{right}\:{answer},\:{Sir}! \\ $$

Answered by ajfour last updated on 25/Apr/20

Commented by ajfour last updated on 25/Apr/20

Ncos 30°=μG ⇒ G=(((√3)N)/(2μ)) Nsin 30°+G=mg ⇒ G=mg−(N/2) ⇒ N(((√3)/(2μ))+(1/2))=mg s=((3R)/8) (for solid) , s=(R/2) (hollow) Torque balance about ground point: mgscos θ=NRcos 30° cos θ=(((√3)R)/2)((N/(mgs)))=(((√3)R)/s)((μ/( (√3)+μ))) ⇒ θ=cos^(−1) [(((√3)μR)/(s((√3)+μ)))].

$${N}\mathrm{cos}\:\mathrm{30}°=\mu{G} \\ $$$$\Rightarrow\:\:{G}=\frac{\sqrt{\mathrm{3}}{N}}{\mathrm{2}\mu} \\ $$$${N}\mathrm{sin}\:\mathrm{30}°+{G}={mg} \\ $$$$\Rightarrow\:\:{G}={mg}−\frac{{N}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{N}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\mu}+\frac{\mathrm{1}}{\mathrm{2}}\right)={mg} \\ $$$$\:\:{s}=\frac{\mathrm{3}{R}}{\mathrm{8}}\:\:\:\left({for}\:{solid}\right)\:\:,\:{s}=\frac{{R}}{\mathrm{2}}\:\:\left({hollow}\right) \\ $$$${Torque}\:{balance}\:{about}\:{ground}\:{point}: \\ $$$$\:\:\:\:{mgs}\mathrm{cos}\:\theta={NR}\mathrm{cos}\:\mathrm{30}° \\ $$$$\mathrm{cos}\:\theta=\frac{\sqrt{\mathrm{3}}{R}}{\mathrm{2}}\left(\frac{{N}}{{mgs}}\right)=\frac{\sqrt{\mathrm{3}}{R}}{{s}}\left(\frac{\mu}{\:\sqrt{\mathrm{3}}+\mu}\right) \\ $$$$\:\:\Rightarrow\:\:\theta=\mathrm{cos}^{−\mathrm{1}} \left[\frac{\sqrt{\mathrm{3}}\mu{R}}{{s}\left(\sqrt{\mathrm{3}}+\mu\right)}\right]. \\ $$$$\:\:\:\:\:\:\: \\ $$$$ \\ $$

Commented by mr W last updated on 25/Apr/20

$${clear}\:{and}\:{comprehensible}! \\ $$

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