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Question Number 177214 by mr W last updated on 02/Oct/22

Commented by mr W last updated on 02/Oct/22

$$thefrictioncoefficientbetweenthe$$$$thinrodandthegroundaswellasthe$$$$wallis\mu .thelengthoftherodisL.$$$$findtheconditionwhichaandb$$$$mustfulfillsuchthattherodcan$$$$restinequilibriumandisaboutto$$$$slipasshown.$$

Commented by mahdipoor last updated on 02/Oct/22

$$AB=(-a,b,c)\sqrt{c^2+a^2+b^2}=L^2$$$$F_A=(p,q,s)s=N_B⩾0$$$$F_B=(u,v,w)u=N_A⩾0$$$$W=(0,0,-mg)$$$$\mathrm{\Sigma }F=ma=0=F_B+F_A+W$$$$\Rightarrow F_A=(-u,-v,mg-w)$$$$\mathrm{\Sigma }{M}_A=0=AA\times F_A+AB\times F_B+\frac{1}{2}AB\times W$$$$=(bw-\frac{bmg}{2}-vc,cu+aw-\frac{amg}{2},-av-bu)$$$$\Rightarrow \{\begin{array}{l}{F}_A=(-u,\frac{b}{a}u,\frac{mg}{2}+\frac{c}{a}u)\\ F_B=(u,-\frac{b}{a}u,\frac{mg}{2}-\frac{c}{a}u)\end{array}$$$$\{\begin{array}{l}\mu \mid N_A\mid ⩾\mid T_A\mid \\ \mu \mid N_B\mid ⩾\mid T_B\mid \end{array}\Rightarrow$$$$A:\{\begin{array}{l}0⩾(\frac{b^2+c^2}{a^2}-{\mu }^2)u^2-\left(\frac{cmg}{a}\right)u+{\left(\frac{mg}{2}\right)}^2\\ 0⩽(\frac{c^2}{a^2}-\frac{b^2+a^2}{{\mu }^2{a}^2})u^2+\left(\frac{cmg}{a}\right)u+{\left(\frac{mg}{2}\right)}^2\end{array}$$$$for\mathrm{\forall }a,bif\mathrm{\exists }u⩾0forA,rodcanrest.$$$$andifatleastoneoftheinequalitiesin$$$$Aequalzero,rodisaboutslip$$

Commented by mr W last updated on 02/Oct/22

$$thanksforyoursolutionsir!$$$$ineedsometimetofollowit.$$

Answered by mr W last updated on 03/Oct/22

Commented by mr W last updated on 03/Oct/22

$$\mathit{b}\mathit{a}\mathit{c}\mathit{k}\mathit{g}\mathit{r}\mathit{o}\mathit{u}\mathit{n}\mathit{d}\mathit{k}\mathit{n}\mathit{o}\mathit{w}\mathit{l}\mathit{e}\mathit{d}\mathit{g}\mathit{e}1$$$$anobject{doesn}^{\prime }tslip,whentheangle$$$$\theta betweentheresultantcontactforce$$$$andthenormalofthecontactsurface$$$$islessthan{\theta }^{\ast }whichis{\tan }^{-1}\mu .$$$$theobjectisabouttoslip,$$$$when\theta ={\theta }^{\ast }={\tan }^{-1}\mu .$$

Commented by mr W last updated on 02/Oct/22

Commented by mr W last updated on 03/Oct/22

$$let\xi =\frac{a}{L},\eta =\frac{b}{L}$$$$c=\sqrt{a^2+b^2}=L\sqrt{{\xi }^2+{\eta }^2}$$$$h=\sqrt{L^2-a^2-b^2}=L\sqrt{1-{\xi }^2-{\eta }^2}$$$$\sin \varphi =\frac{a}{c}=\frac{\xi }{\sqrt{{\xi }^2+{\eta }^2}}$$$$$$$$\tan {\theta }_1⩽\mu$$$$\tan {\theta }_2⩽\mu$$$$\Rightarrow \cos {\theta }_2⩾\frac{1}{\sqrt{1+{\mu }^2}}$$$$$$$$\tan \beta =\frac{c}{h}=\frac{\sqrt{{\xi }^2+{\eta }^2}}{\sqrt{1-{\xi }^2-{\eta }^2}}$$$$\sin \phi \times \sin \varphi =\cos {\theta }_2$$$$\Rightarrow \sin \phi =\frac{\cos {\theta }_2}{\sin \varphi }⩾\frac{1}{\xi }\sqrt{\frac{{\xi }^2+{\eta }^2}{1+{\mu }^2}}$$$$\Rightarrow \tan \phi ⩾\sqrt{\frac{{\xi }^2+{\eta }^2}{{\mu }^2{\xi }^2-{\eta }^2}}$$$$$$$$\frac{CD}{CB}=\frac{\sin (\beta +\phi )}{\sin \phi }=\frac{\sin \beta }{\tan \phi }+\cos \beta$$$$\frac{CD}{AC}=\frac{\sin (\beta -{\theta }_1)}{\sin {\theta }_1}=\frac{\sin \beta }{\tan {\theta }_1}-\cos \beta$$$$\frac{\sin \beta }{\tan {\theta }_1}-\cos \beta =\frac{\sin \beta }{\tan \phi }+\cos \beta$$$$\frac{1}{\tan {\theta }_1}-\frac{1}{\tan \phi }=\frac{2}{\tan \beta }$$$$\frac{1}{\mu }-\frac{\sqrt{{\mu }^2{\xi }^2-{\eta }^2}}{\sqrt{{\xi }^2+{\eta }^2}}⩽\frac{2\sqrt{1-{\xi }^2-{\eta }^2}}{\sqrt{{\xi }^2+{\eta }^2}}$$$$\overline{)\begin{array}{c}2\sqrt{1-{\xi }^2-{\eta }^2}+\sqrt{{\mu }^2{\xi }^2-{\eta }^2}-\frac{\sqrt{{\xi }^2+{\eta }^2}}{\mu }⩾0\end{array}}$$

Commented by mr W last updated on 03/Oct/22

$$\mathit{b}\mathit{a}\mathit{c}\mathit{k}\mathit{g}\mathit{r}\mathit{o}\mathit{u}\mathit{n}\mathit{d}\mathit{k}\mathit{n}\mathit{o}\mathit{w}\mathit{l}\mathit{e}\mathit{d}\mathit{g}\mathit{e}2$$$$whenanobjectisinequilibriumunder$$$$theactionofthreeforces,thenthese$$$$threeforcesmustintersectatthe$$$$samepointandtheirsum\left(vector\right)is$$$$zero.$$

Commented by mr W last updated on 03/Oct/22

Commented by mr W last updated on 02/Oct/22

Commented by mr W last updated on 02/Oct/22

Commented by mr W last updated on 02/Oct/22

$$=END=$$

Commented by mr W last updated on 03/Oct/22

$$followingdiagramshowstherange$$$$foraandbsuchthattherodisin$$$$equilibrium.$$$$example\mu =0.5.$$

Commented by mr W last updated on 02/Oct/22

Commented by mahdipoor last updated on 02/Oct/22

$$verynicemethodforanswer!$$$$but,$$$$inlemma1onlycorrectnessof\mu N⩽T$$$$and$$$$inlemma2only\mathrm{\Sigma }Momentiszero$$$$areguaranteed.$$$$howabout\mathrm{\Sigma }F=0?$$$$if\mathrm{\Sigma }F=0thenlemma2istrue.$$$$ithinkbalanceofforcesisnotinsolution$$$$and{that}^{\prime }swhy\left(mg\right)isineffective$$$$inequilibriumconditions.$$

Commented by mr W last updated on 03/Oct/22

$$1)\mu N⩽Tmeans\theta ⩽{\tan }^{-1}\mu$$$$2)hereijustwanttosay:$$$$ifequilibrium,then3forces$$$$intersectatsamepoint.$$$$itisnotsaid:if3forcesintersect$$$$atsamepoint,thenequilbrium.$$$$certainlyiknow\sum _{1,2,3}{F}_{i}shouldbe$$$$zeroforequilibrium.butforsolving$$$$thisproblem,ionlyneedtoknow$$$$‘‘ifequilibrium,then3forces$$$$intersectatsame{point}^{″}.that$$$$\mathrm{\Sigma }F=0isalsotrue,isclear.buti$$$${don}^{\prime }tneedthisforsolvingthe$$$$problem.$$$$3)forsolvingtheproblem,only$$$$thedirectionsoftheforcesare$$$$interesting,nottheirsizes.$$$$thereforetheweightoftherodmg$$$${doesn}^{\prime }taffecttheresult.infact$$$$itismoreaquestionofgeometry.$$

Commented by mahdipoor last updated on 02/Oct/22

$$thankyou,youareright!\mathrm{♡}$$

Commented by mahdipoor last updated on 03/Oct/22

$$andonemorequestion:$$$$inspecialcase:let{a}^2+b^2=L^2(\Rightarrow h=0$$$$and{\xi }^2+{\eta }^2=1)$$$$itsmeans,rodishorizontal$$$$itsseemsthatinthiscase,0⩽\xi ⩽1$$$$and\mu isineffective$$$$fromyouranswer:$$$$0+\sqrt{{\mu }^2{\xi }^2-{\eta }^2}-\frac{1}{\mu }⩾0\Rightarrow \xi ⩾\frac{1}{\mu }$$$$whatiswrong?$$

Commented by Tawa11 last updated on 03/Oct/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by mr W last updated on 03/Oct/22

$$\mathit{t}\mathit{o}\mathit{m}\mathit{a}\mathit{h}\mathit{d}\mathit{i}\mathit{p}\mathit{o}\mathit{o}\mathit{r}\mathit{s}\mathit{i}\mathit{r}:$$$$thanksforyourfeedback!$$$$iappreciatesuchsubstantialfeedback$$$$verymuch.$$$$yourquestionisinteresting.ialso$$$$noticedthisissueasisawthegraph$$$$forthecase\mu =2,whichi{didn}^{\prime }tpost.$$$$iconsideredtostudyitandtryto$$$$explainitlater,whenihavemoretime.$$$$sinceyouhaveaskedthesame$$$$questionaboutthisissue,ihavespent$$$$sometimetotakeacloserlookatit.$$$$ithinkitisnotafaultinthe$$$$solution.itcanbeexplainedas$$$$following.$$

Commented by mr W last updated on 03/Oct/22

Commented by mr W last updated on 03/Oct/22

$$incaseof\mu ⩾1,{\theta }^{\ast }={\tan }^{-1}\mu ⩾45°.$$$$inthiscasetherodcanlieinavery$$$$flatposition,closetohorizontalor$$$$eveninhorizontalposition,andstill$$$$bekeptinequilbrium.wecanseein$$$$thediagramabovethatwithinthe$$$$angles{\theta }^{\ast }contactforces{R}_{1}and{R}_2$$$$canbefoundwhichmeetthemgat$$$$thesamepoint,thereforeaequilibrium$$$$stateispossible.certainlyaabsolutely$$$$horizontalequilibriumpositionis$$$$justtheory.butmathematicallyitis$$$$possible.$$

Commented by mr W last updated on 03/Oct/22

Commented by mr W last updated on 03/Oct/22

Commented by mr W last updated on 03/Oct/22

$$for\mu <1,{\theta }^{\ast }<45°,wecanseethatan$$$$equilibriumstateinsuchaflat$$$$positionisnotpossible,becausewe$$$$cannotfindwithintheangles{\theta }^{\ast }such$$$$contactforces{R}_{1}and{R}_{2}whichmeet$$$$themgatthesamepoint.inthiscase$$$$c<L,i.e.{\xi }^2+{\eta }^2<1.$$

Commented by mr W last updated on 03/Oct/22

Commented by mahdipoor last updated on 04/Oct/22

$$exactly$$$$inthisanswer,itisimportantthatthe$$$$threeforcesareconcurrent$$$$inhorizontalpositon,theforcesareparallel$$$$andarenotconcurrent.$$$$thatiswhytheanswerforthiscaseisfals$$

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