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 105594 by ajfour last updated on 30/Jul/20

Commented by ajfour last updated on 30/Jul/20

Find maximum x, for y>0. The two cylinders have radii r.

$${Find}\:{maximum}\:\boldsymbol{{x}},\:{for}\:{y}>\mathrm{0}. \\ $$$${The}\:{two}\:{cylinders}\:{have}\:{radii}\:\boldsymbol{{r}}. \\ $$

Commented by ajfour last updated on 30/Jul/20

Commented by ajfour last updated on 30/Jul/20

s=f=g lets call mg=w scos θ+Nsin θ+mg=R=(g/μ) g+ssin θ=Ncos θ F=g R+f = 2mg Nsin θ+scos θ+f=mg −−−−−−−−−−−−−−− ⇒ gcos θ+((g(1+sin θ)sin θ+w)/(cos θ))=(g/μ) (g/μ)+g=2w ⇒ (w/g)=(1/2)(1+(1/μ)) ((g(1+sin θ)sin θ)/(cos θ))+gcos θ+g=w ⇒ (1/2)(1+(1/μ))=((1/μ)−cos θ)cos θ −sin θ(1+sin θ) = 1+cos θ+((sin θ)/(cos θ))(1+sin θ) ⇒ (1/μ)(cos θ−(1/2))=sin θ+(3/2) ⇒ (1/2)(1+((3+2sin θ)/(2cos θ−1)))=1+((1+sin θ)/(cos θ)) ⇒ 2cos θ−1=cos θ ⇒ cos θ=1 ⇒ y=0.

$${s}={f}={g} \\ $$$${lets}\:{call}\:{mg}={w} \\ $$$${s}\mathrm{cos}\:\theta+{N}\mathrm{sin}\:\theta+{mg}={R}=\frac{{g}}{\mu} \\ $$$${g}+{s}\mathrm{sin}\:\theta={N}\mathrm{cos}\:\theta \\ $$$${F}={g} \\ $$$${R}+{f}\:=\:\mathrm{2}{mg} \\ $$$${N}\mathrm{sin}\:\theta+{s}\mathrm{cos}\:\theta+{f}={mg} \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\Rightarrow\:\:\:{g}\mathrm{cos}\:\theta+\frac{{g}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)\mathrm{sin}\:\theta+{w}}{\mathrm{cos}\:\theta}=\frac{{g}}{\mu} \\ $$$$\frac{{g}}{\mu}+{g}=\mathrm{2}{w}\:\:\:\:\Rightarrow\:\:\:\frac{{w}}{{g}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mu}\right) \\ $$$$\frac{{g}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}+{g}\mathrm{cos}\:\theta+{g}={w} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mu}\right)=\left(\frac{\mathrm{1}}{\mu}−\mathrm{cos}\:\theta\right)\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}+\mathrm{cos}\:\theta+\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}\left(\mathrm{1}+\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mu}\left(\mathrm{cos}\:\theta−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{sin}\:\theta+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{3}+\mathrm{2sin}\:\theta}{\mathrm{2cos}\:\theta−\mathrm{1}}\right)=\mathrm{1}+\frac{\mathrm{1}+\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\:\:\mathrm{2cos}\:\theta−\mathrm{1}=\mathrm{cos}\:\theta \\ $$$$\Rightarrow\:\:\:\:\:\mathrm{cos}\:\theta=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:{y}=\mathrm{0}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com