Question Number 20412 by Tinkutara last updated on 26/Aug/17
$$\mathrm{A}\:\mathrm{block}\:\mathrm{is}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{rough}\:\mathrm{horizontal} \\ $$$$\mathrm{surface}.\:\mathrm{The}\:\mathrm{minimum}\:\mathrm{force}\:\mathrm{required} \\ $$$$\mathrm{to}\:\mathrm{slide}\:\mathrm{the}\:\mathrm{block}\:\mathrm{is} \\ $$
Commented by Tinkutara last updated on 26/Aug/17
Commented by ajfour last updated on 26/Aug/17
![Pcos θ=f_(max) =μN N+Psin θ=mg So Pcos θ=μ(mg−Psin θ) ⇒ P=((μmg)/(cos θ+μsin θ)) =(μ/( (√(1+μ^2 ))))×((mg)/(sin [θ+tan^(−1) ((1/μ))])) P is least when θ=(π/2)−cot^(−1) μ or when θ=tan^(−1) μ P_(minimum) =((μmg)/( (√(1+μ^2 )))) .](https://www.tinkutara.com/question/Q20414.png)
$${P}\mathrm{cos}\:\theta={f}_{{max}} =\mu{N} \\ $$$${N}+{P}\mathrm{sin}\:\theta={mg} \\ $$$${So}\:\:\:{P}\mathrm{cos}\:\theta=\mu\left({mg}−{P}\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\:\:\:{P}=\frac{\mu{mg}}{\mathrm{cos}\:\theta+\mu\mathrm{sin}\:\theta} \\ $$$$\:\:\:=\frac{\mu}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}×\frac{{mg}}{\mathrm{sin}\:\left[\theta+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mu}\right)\right]} \\ $$$${P}\:{is}\:{least}\:{when}\:\theta=\frac{\pi}{\mathrm{2}}−\mathrm{cot}^{−\mathrm{1}} \mu \\ $$$${or}\:{when}\:\theta=\mathrm{tan}^{−\mathrm{1}} \mu \\ $$$${P}_{{minimum}} =\frac{\mu{mg}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\:. \\ $$
Commented by Tinkutara last updated on 26/Aug/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$