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Question Number 52468 by mr W last updated on 08/Jan/19

Commented by mr W last updated on 08/Jan/19

$$Nofriction.$$$$Find\theta =?$$

Commented by ajfour last updated on 08/Jan/19

Commented by ajfour last updated on 08/Jan/19

$$letmutualreactionforce(notshown$$$$infigure)beT.$$$$F\cos \alpha -T\sin \theta =mg..\left(i\right)$$$$N\cos \beta +T\sin \theta =Mg...\left(ii\right)$$$$F\sin \alpha =T\cos \theta =N\sin \beta ...\left(iii\right)$$$$hencefrom\left(i\right)\mathbf{÷}\left(ii\right)\mathrm{\&}using\left(iii\right)$$$$\frac{\frac{\cos \theta \cos \alpha }{\sin \alpha }-\sin \theta }{\frac{\cos \theta \cos \beta }{\sin \beta }+\sin \theta }=\lambda (=\frac{m}{M})$$$$\Rightarrow \frac{1}{\tan \alpha }-\frac{\lambda }{\tan \beta }=(\lambda +1)\tan \theta$$$$\Rightarrow \theta ={\tan }^{-1}\left[\frac{\tan \beta -\lambda \tan \alpha }{(\lambda +1)\tan \alpha \tan \beta }\right].$$

Commented by mr W last updated on 08/Jan/19

$$thankyousir!$$$$answeriscorrect.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 08/Jan/19

Commented by mr W last updated on 08/Jan/19

$$thankyousir!$$

Answered by mr W last updated on 08/Jan/19

Commented by mr W last updated on 08/Jan/19

$$totalweightofbothballs=(M+m)g$$$$itspositionisinthelineCD$$$$AD=\frac{M(R+r)}{M+m}$$$$BD=\frac{m(R+r)}{M+m}$$$$in\mathrm{\Delta }ABC:$$$$\mathrm{\angle }A=\frac{\pi }{2}-\alpha -\theta =\frac{\pi }{2}-(\alpha +\theta )$$$$\mathrm{\angle }B=\frac{\pi }{2}-\beta +\theta =\frac{\pi }{2}-(\beta -\theta )$$$$CD=\frac{AD\times \sin \mathrm{\angle }A}{\sin \alpha }=\frac{M(R+r)}{M+m}\times \frac{\cos (\alpha +\theta )}{\sin \alpha }$$$$CD=\frac{BD\times \sin \mathrm{\angle }B}{\sin \beta }=\frac{m(R+r)}{M+m}\times \frac{\cos (\beta -\theta )}{\sin \beta }$$$$\Rightarrow \frac{M(R+r)}{M+m}\times \frac{\cos (\alpha +\theta )}{\sin \alpha }=\frac{m(R+r)}{M+m}\times \frac{\cos (\beta -\theta )}{\sin \beta }$$$$\Rightarrow \frac{\cos (\alpha +\theta )}{\sin \alpha }=\frac{m}{M}\times \frac{\cos (\beta -\theta )}{\sin \beta }$$$$with\lambda =\frac{m}{M}$$$$\Rightarrow \frac{\cos \theta }{\tan \alpha }-\sin \theta =\lambda (\frac{\cos \theta }{\tan \beta }+\sin \theta )$$$$\Rightarrow \tan \theta =\frac{1}{1+\lambda }(\frac{1}{\tan \alpha }-\frac{\lambda }{\tan \beta })$$$$\Rightarrow \theta ={\tan }^{-1}\left[\frac{1}{1+\lambda }(\frac{1}{\tan \alpha }-\frac{\lambda }{\tan \beta })\right]$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$thankyousir...inmycalculationihaveput$$$$cos(\theta +\beta )...butshouldbecos(\beta -\theta )$$$$puttingthatcos(\beta -\theta )...icouldhavereached$$$$destination..$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

Commented by mr W last updated on 09/Jan/19

$$thanksforyoureffortssir!$$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Jan/19

$$thankyousir...$$

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