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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

Fcos α≥μ(mg−Fsin α) F(cos α+μsin α)≥μmg F≥(μ/(cos α+μsin α))mg a) α=0 F≥0.45×8×10=36 N b) α=60° F≥((0.45)/(cos 60°+0.45×sin 60°))×8×10=0.51×80=40.5 N c) α=−60° F≥((0.45)/(cos 60°−0.45×sin 60°))×8×10=4.08×80=326.4 N Find the best angle α such that F=min. when cos α+μsin α is maximum cos α+μsin α =(√(1+μ^2 ))((1/(√(1+μ^2 )))cos α+(μ/(√(1+μ^2 )))sin α) =(√(1+μ^2 ))(cos θcos α+sin θsin α) =(√(1+μ^2 )) cos (α−θ) F is mininal when α=θ=cos^(−1) (1/(√(1+μ^2 )))=24.2° F_(min) =(μ/(√(1+μ^2 )))×mg=((0.45)/(√(1+0.45^2 )))×8×10=32.8 N

$$\mathrm{Fcos}\:\alpha\geqslant\mu\left(\mathrm{mg}−\mathrm{Fsin}\:\alpha\right) \\ $$$$\mathrm{F}\left(\mathrm{cos}\:\alpha+\mu\mathrm{sin}\:\alpha\right)\geqslant\mu\mathrm{mg} \\ $$$$\mathrm{F}\geqslant\frac{\mu}{\mathrm{cos}\:\alpha+\mu\mathrm{sin}\:\alpha}\mathrm{mg} \\ $$$$\left.\mathrm{a}\right)\:\alpha=\mathrm{0} \\ $$$$\mathrm{F}\geqslant\mathrm{0}.\mathrm{45}×\mathrm{8}×\mathrm{10}=\mathrm{36}\:\mathrm{N} \\ $$$$ \\ $$$$\left.\mathrm{b}\right)\:\alpha=\mathrm{60}° \\ $$$$\mathrm{F}\geqslant\frac{\mathrm{0}.\mathrm{45}}{\mathrm{cos}\:\mathrm{60}°+\mathrm{0}.\mathrm{45}×\mathrm{sin}\:\mathrm{60}°}×\mathrm{8}×\mathrm{10}=\mathrm{0}.\mathrm{51}×\mathrm{80}=\mathrm{40}.\mathrm{5}\:\mathrm{N} \\ $$$$ \\ $$$$\left.\mathrm{c}\right)\:\alpha=−\mathrm{60}° \\ $$$$\mathrm{F}\geqslant\frac{\mathrm{0}.\mathrm{45}}{\mathrm{cos}\:\mathrm{60}°−\mathrm{0}.\mathrm{45}×\mathrm{sin}\:\mathrm{60}°}×\mathrm{8}×\mathrm{10}=\mathrm{4}.\mathrm{08}×\mathrm{80}=\mathrm{326}.\mathrm{4}\:\mathrm{N} \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{best}\:\mathrm{angle}\:\alpha\:\mathrm{such}\:\mathrm{that}\:\mathrm{F}=\mathrm{min}. \\ $$$$\mathrm{when}\:\mathrm{cos}\:\alpha+\mu\mathrm{sin}\:\alpha\:\mathrm{is}\:\mathrm{maximum} \\ $$$$\mathrm{cos}\:\alpha+\mu\mathrm{sin}\:\alpha\:=\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\left(\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\mathrm{cos}\:\alpha+\frac{\mu}{\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\mathrm{sin}\:\alpha\right) \\ $$$$\:=\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\left(\mathrm{cos}\:\theta\mathrm{cos}\:\alpha+\mathrm{sin}\:\theta\mathrm{sin}\:\alpha\right) \\ $$$$\:=\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\:\mathrm{cos}\:\left(\alpha−\theta\right) \\ $$$$ \\ $$$$\mathrm{F}\:\mathrm{is}\:\mathrm{mininal}\:\mathrm{when} \\ $$$$\alpha=\theta=\mathrm{cos}^{−\mathrm{1}} \:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}=\mathrm{24}.\mathrm{2}° \\ $$$$\mathrm{F}_{\mathrm{min}} =\frac{\mu}{\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}×\mathrm{mg}=\frac{\mathrm{0}.\mathrm{45}}{\sqrt{\mathrm{1}+\mathrm{0}.\mathrm{45}^{\mathrm{2}} }}×\mathrm{8}×\mathrm{10}=\mathrm{32}.\mathrm{8}\:\mathrm{N} \\ $$

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

Wow, God bless you sir. i appreciate your effort.

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

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