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Question Number 131767 by ajfour last updated on 08/Feb/21

Commented by ajfour last updated on 08/Feb/21

$$FindforceofNormalreaction$$$$betweenthecylindershanging$$$$onabeltwrap.$$

Answered by mr W last updated on 08/Feb/21

Commented by mr W last updated on 08/Feb/21

$$T=tensioninbelt$$$$N=normalcontactforce$$$$C=comofbothcylinders$$$$CDisperpendiculartohorizontal$$$$AB=R+r$$$$AC=\frac{M(R+r)}{M+m}$$$$CB=\frac{m(R+r)}{M+m}$$$$leta=DE=DF$$$$2({\tan }^{-1}\frac{r}{a}+{\tan }^{-1}\frac{R}{a})=45°\left(given\right)$$$${\tan }^{-1}\frac{r}{a}+{\tan }^{-1}\frac{R}{a}=\frac{45°}{2}$$$$\frac{\frac{r}{a}+\frac{R}{a}}{1-\frac{r}{a}\times \frac{R}{a}}=\tan \frac{45°}{2}=\sqrt{2}-1$$$$(\sqrt{2}-1)a^2-(R+r)a-(\sqrt{2}-1)Rr=0$$$$\Rightarrow a=\frac{R+r+\sqrt{R^2+r^2+2(7-4\sqrt{2})Rr}}{2(\sqrt{2}-1)}$$$$\alpha ={\tan }^{-1}\frac{a}{r}=\frac{\pi }{2}-{\tan }^{-1}\frac{r}{a}$$$$\beta ={\tan }^{-1}\frac{a}{R}=\frac{\pi }{2}-{\tan }^{-1}\frac{R}{a}$$$$\frac{\sin \gamma }{BC}=\frac{\sin (\gamma +\beta )}{DB}$$$$\frac{DB}{BC}=\frac{\sin (\gamma +\beta )}{\sin \gamma }=\cos \beta +\frac{\sin \beta }{\tan \gamma }$$$$\frac{(M+m)\sqrt{a^2+R^2}}{m(R+r)}=\frac{R}{\sqrt{a^2+R^2}}+\frac{a}{\tan \gamma \sqrt{a^2+R^2}}$$$$\Rightarrow \tan \gamma =\frac{a}{\frac{(M+m)(a^2+R^2)}{m(R+r)}-R}$$$$\Rightarrow \gamma ={\tan }^{-1}\frac{m(R+r)a}{(M+m)(a^2+R^2)-m(R+r)R}$$$$\varphi =\frac{\pi }{2}+{\sin }^{-1}\frac{R-r}{R+r}$$$$\delta =\pi -\varphi -2{\tan }^{-1}\frac{R}{a}=\frac{\pi }{2}-{\sin }^{-1}\frac{R-r}{R+r}-2{\tan }^{-1}\frac{R}{a}$$$$\phi =\frac{\pi }{2}-\beta -\gamma$$$$\Rightarrow \phi =\frac{\pi }{2}-{\tan }^{-1}\frac{a}{R}-{\tan }^{-1}\frac{m(R+r)a}{(M+m)(a^2+R^2)-m(R+r)R}$$$$\theta =\frac{\pi }{2}-\frac{\pi }{8}-\delta$$$$\Rightarrow \theta ={\sin }^{-1}\frac{R-r}{R+r}+2{\tan }^{-1}\frac{R}{a}-\frac{\pi }{8}$$$$$$$$2T\cos \frac{45°}{2}=(M+m)g$$$$\Rightarrow T=\frac{\sqrt{2(2-\sqrt{2})}(M+m)g}{2}$$$$N\cos \phi =T(\sin \frac{45°}{2}+\cos \theta )$$$$N=\sqrt{2(2-\sqrt{2})}(M+m)g\times \frac{\frac{\sqrt{2}-1}{\sqrt{2(2-\sqrt{2})}}+\cos ({\sin }^{-1}\frac{R-r}{R+r}+2{\tan }^{-1}\frac{R}{a}-\frac{\pi }{8})}{2\sin ({\tan }^{-1}\frac{a}{R}+{\tan }^{-1}\frac{m(R+r)a}{(M+m)(a^2+R^2)-m(R+r)R})}$$$$\Rightarrow \frac{N}{(M+m)g}=\frac{\sqrt{2}-1+\sqrt{2(2-\sqrt{2})}\cos ({\sin }^{-1}\frac{R-r}{R+r}+2{\tan }^{-1}\frac{R}{a}-\frac{\pi }{8})}{2\cos ({\tan }^{-1}\frac{R}{a}-{\tan }^{-1}\frac{m(R+r)a}{(M+m)(a^2+R^2)-m(R+r)R})}$$

Commented by ajfour last updated on 08/Feb/21

$$LooksafabulousdesignSir.$$$$I{couldn}^{\prime }tdareattempt,letme$$$$sometimefollowing...$$

Commented by otchereabdullai@gmail.com last updated on 10/Feb/21

$$\mathrm{You}\mathrm{are}\mathrm{gifted}\mathrm{prof}\mathrm{W}$$

Answered by ajfour last updated on 08/Feb/21

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