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
$${can}\:{you}\:{please}\:{try}\:{Mr}.{W} \\ $$
Answered by mr W last updated on 08/Dec/22
Commented by BHOOPENDRA last updated on 08/Dec/22
$${Nice}\:{sir} \\ $$
Commented by mr W last updated on 08/Dec/22
Commented by mr W last updated on 08/Dec/22
![α: direction of force φ: direction of motion, opposite to direction of friction force F_(friction) =μmg cos θ (F sin α)^2 +(F cos α−mg sin θ)^2 =(μmg cos θ)^2 F^2 −2mg cos α sin θ F−m^2 g^2 (sin^2 θ+μ^2 cos^2 θ)=0 ⇒F=mg [cos α sin θ+(√((1+cos^2 α) sin^2 θ+μ^2 cos^2 θ))] μmg cos θ sin φ=F sin α ⇒sin φ=(( sin α[cos α sin θ+(√((1+cos^2 α) sin^2 θ+μ^2 cos^2 θ))])/(μ cos θ)) ⇒φ=sin^(−1) (( sin α[cos α tan θ+(√((1+cos^2 α) tan^2 θ+μ^2 ))])/μ)](Q182354.png)
$$\alpha:\:{direction}\:{of}\:{force} \\ $$$$\phi:\:{direction}\:{of}\:{motion},\:{opposite}\:{to} \\ $$$$\:\:\:\:\:{direction}\:{of}\:{friction}\:{force} \\ $$$${F}_{{friction}} =\mu{mg}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$$\left({F}\:\mathrm{sin}\:\alpha\right)^{\mathrm{2}} +\left({F}\:\mathrm{cos}\:\alpha−{mg}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} =\left(\mu{mg}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} \\ $$$${F}^{\mathrm{2}} \:−\mathrm{2}{mg}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta\:{F}−{m}^{\mathrm{2}} {g}^{\mathrm{2}} \left(\mathrm{sin}^{\mathrm{2}} \:\theta+\mu^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow{F}={mg}\:\left[\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta+\sqrt{\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mu^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\right] \\ $$$$ \\ $$$$\mu{mg}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\phi={F}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\mathrm{sin}\:\phi=\frac{\:\mathrm{sin}\:\alpha\left[\mathrm{cos}\:\alpha\:\mathrm{sin}\:\theta+\sqrt{\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha\right)\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mu^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta}\right]}{\mu\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\phi=\mathrm{sin}^{−\mathrm{1}} \frac{\:\mathrm{sin}\:\alpha\left[\mathrm{cos}\:\alpha\:\mathrm{tan}\:\theta+\sqrt{\left(\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:\alpha\right)\:\mathrm{tan}^{\mathrm{2}} \:\theta+\mu^{\mathrm{2}} }\right]}{\mu} \\ $$