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Question Number 182280 by BHOOPENDRA last updated on 06/Dec/22

Commented by BHOOPENDRA last updated on 06/Dec/22

Commented by BHOOPENDRA last updated on 07/Dec/22

$$canyoupleasetryMr.W$$

Answered by mr W last updated on 08/Dec/22

Commented by BHOOPENDRA last updated on 08/Dec/22

$$Nicesir$$

Commented by mr W last updated on 08/Dec/22

Commented by mr W last updated on 08/Dec/22

$$\alpha :directionofforce$$$$\varphi :directionofmotion,oppositeto$$$$directionoffrictionforce$$$$F_{friction}=\mu mg\cos \theta$$$$$$$${\left(F\sin \alpha \right)}^2+{(F\cos \alpha -mg\sin \theta )}^2={\left(\mu mg\cos \theta \right)}^2$$$$F^2-2mg\cos \alpha \sin \theta F-m^2{g}^2({\sin }^2\theta +{\mu }^2{\cos }^2\theta )=0$$$$\Rightarrow F=mg[\cos \alpha \sin \theta +\sqrt{(1+{\cos }^2\alpha ){\sin }^2\theta +{\mu }^2{\cos }^2\theta }]$$$$$$$$\mu mg\cos \theta \sin \varphi =F\sin \alpha$$$$\Rightarrow \sin \varphi =\frac{\sin \alpha [\cos \alpha \sin \theta +\sqrt{(1+{\cos }^2\alpha ){\sin }^2\theta +{\mu }^2{\cos }^2\theta }]}{\mu \cos \theta }$$$$\Rightarrow \varphi ={\sin }^{-1}\frac{\sin \alpha [\cos \alpha \tan \theta +\sqrt{(1+{\cos }^2\alpha ){\tan }^2\theta +{\mu }^2}]}{\mu }$$

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