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Question Number 15485 by tawa tawa last updated on 11/Jun/17

Answered by mrW1 last updated on 11/Jun/17

$$\mathrm{Fcos}\alpha ⩾\mu (\mathrm{mg}-\mathrm{Fsin}\alpha )$$$$\mathrm{F}(\cos \alpha +\mu \sin \alpha )⩾\mu \mathrm{mg}$$$$\mathrm{F}⩾\frac{\mu }{\cos \alpha +\mu \sin \alpha }\mathrm{mg}$$$$\mathrm{a})\alpha =0$$$$\mathrm{F}⩾0.45\times 8\times 10=36\mathrm{N}$$$$$$$$\mathrm{b})\alpha =60°$$$$\mathrm{F}⩾\frac{0.45}{\cos 60°+0.45\times \sin 60°}\times 8\times 10=0.51\times 80=40.5\mathrm{N}$$$$$$$$\mathrm{c})\alpha =-60°$$$$\mathrm{F}⩾\frac{0.45}{\cos 60°-0.45\times \sin 60°}\times 8\times 10=4.08\times 80=326.4\mathrm{N}$$$$$$$$\mathrm{Find}\mathrm{the}\mathrm{best}\mathrm{angle}\alpha \mathrm{such}\mathrm{that}\mathrm{F}=\min .$$$$\mathrm{when}\cos \alpha +\mu \sin \alpha \mathrm{is}\mathrm{maximum}$$$$\cos \alpha +\mu \sin \alpha =\sqrt{1+{\mu }^2}(\frac{1}{\sqrt{1+{\mu }^2}}\cos \alpha +\frac{\mu }{\sqrt{1+{\mu }^2}}\sin \alpha )$$$$=\sqrt{1+{\mu }^2}(\cos \theta \cos \alpha +\sin \theta \sin \alpha )$$$$=\sqrt{1+{\mu }^2}\cos (\alpha -\theta )$$$$$$$$\mathrm{F}\mathrm{is}\mathrm{mininal}\mathrm{when}$$$$\alpha =\theta ={\cos }^{-1}\frac{1}{\sqrt{1+{\mu }^2}}=24.2°$$$${\mathrm{F}}_{\min }=\frac{\mu }{\sqrt{1+{\mu }^2}}\times \mathrm{mg}=\frac{0.45}{\sqrt{1+0.{45}^2}}\times 8\times 10=32.8\mathrm{N}$$

Commented by tawa tawa last updated on 11/Jun/17

$$\mathrm{Wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.\mathrm{i}\mathrm{appreciate}\mathrm{your}\mathrm{effort}.$$

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