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Question Number 48285 by ajfour last updated on 21/Nov/18

Commented by ajfour last updated on 21/Nov/18

$$Findmaximum\theta (<\pi /2)when$$$$equilibriumprevailstoo.$$

Commented by ajfour last updated on 22/Nov/18

Answered by mr W last updated on 23/Nov/18

$$\alpha =inclinationofrodLw.r.t.ground$$$$\beta =inclinationofrodl$$$$\lambda =l/L,\varrho =m/M$$$$L\sin \alpha =l\sin \beta$$$$\Rightarrow \sin \beta =\frac{\sin \alpha }{\lambda }$$$$\cos \beta =\sqrt{1-\frac{{\sin }^2\alpha }{{\lambda }^2}}=\frac{\sqrt{{\lambda }^2-{\sin }^2\alpha }}{\lambda }$$$$N_1(L\cos \alpha +l\cos \beta )=Mg(\frac{L\cos \alpha }{2}+l\cos \beta )+mg\frac{l\cos \beta }{2}$$$$\Rightarrow N_1=\frac{MgL\cos \alpha +(2M+m)gl\cos \beta }{2(L\cos \alpha +l\cos \beta )}$$$$\Rightarrow N_2=\frac{(M+2m)gL\cos \alpha +mgl\cos \beta }{2(L\cos \alpha +l\cos \beta )}$$$$$$$$N_{1}L\cos \alpha -Mg\frac{L\cos \alpha }{2}-f_{1}L\sin \alpha =0$$$$f_1=\frac{2N_1\cos \alpha -Mg\cos \alpha }{2\sin \alpha }=\mu N_1\Rightarrow noslippingatpointA$$$$2N_1(\cos \alpha -\mu \sin \alpha )=Mg\cos \alpha$$$$\frac{MgL\cos \alpha +(2M+m)gl\cos \beta }{(L\cos \alpha +l\cos \beta )}(\cos \alpha -\mu \sin \alpha )=Mg\cos \alpha$$$$\frac{1+\lambda (2+\varrho )\frac{\cos \beta }{\cos \alpha }}{(1+\lambda \frac{\cos \beta }{\cos \vartheta })}(1-\mu \tan \alpha )=1$$$$\Rightarrow (1-\mu \tan \alpha )[1+(2+\varrho )\frac{\sqrt{{\lambda }^2-{\sin }^2\alpha }}{\cos \alpha }]=1+\frac{\sqrt{{\lambda }^2-{\sin }^2\alpha }}{\cos \alpha }...\left(i\right)$$$$\Rightarrow \alpha ={\alpha }_1=...$$$$similarly$$$$N_{2}l\cos \beta -mg\frac{l\cos \beta }{2}-f_{2}l\sin \beta =0$$$$f_2=\frac{2N_2\cos \beta -mg\cos \beta }{2\sin \beta }=\mu N_2\Rightarrow noslippingatpointB$$$$2N_2(\cos \beta -\mu \sin \beta )=mg\cos \beta$$$$\frac{(M+2m)gL\cos \alpha +mgl\cos \beta }{(L\cos \alpha +l\cos \beta )}(\cos \beta -\mu \sin \beta )=mg\cos \beta$$$$\frac{(1+2\varrho )\frac{\cos \alpha }{\cos \beta }+\varrho \lambda }{(\frac{\cos \alpha }{\cos \beta }+\lambda )}(1-\mu \tan \beta )=\varrho$$$$\Rightarrow \frac{(1+2\varrho )\cos \alpha +\varrho \sqrt{{\lambda }^2-{\sin }^2\alpha }}{\cos \alpha +\sqrt{{\lambda }^2-{\sin }^2\alpha }}(1-\frac{\mu \sin \alpha }{\sqrt{{\lambda }^2-{\sin }^2\alpha }})=\varrho ...\left(ii\right)$$$$\Rightarrow \alpha ={\alpha }_2=...$$$$$$$$\Rightarrow \alpha =max({\alpha }_1,{\alpha }_2)$$$$\Rightarrow \beta ={\sin }^{-1}\left(\sin \alpha /\lambda \right)$$$$\Rightarrow \theta =\pi -\alpha -\beta$$$$$$$$Example:\lambda =l/L=0.75,\varrho =m/M=1.5,\mu =0.5$$$${\alpha }_1=41.5624°from\left(i\right)$$$${\alpha }_2=32.2635°from\left(ii\right)$$$$\Rightarrow \alpha =41.5624°$$$$\Rightarrow \beta ={\sin }^{-1}\left(\sin \alpha /\lambda \right)=62.20°$$$$\Rightarrow {\theta }_{max}=180-41.56-62.20=76.24°$$

Commented by ajfour last updated on 23/Nov/18

$$Thankssirforsolutionand$$$$objectiontomysolutionaswell.$$

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