Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

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?

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

Commented by ajfour last updated on 24/Apr/20

solid, Sir.

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

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!

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

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!

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com