Question Number 27224 by mrW1 last updated on 03/Jan/18
Commented by mrW1 last updated on 03/Jan/18
$${The}\:{friction}\:{coefficient}\:{betweem}\:{the} \\ $$$${blocks}\:{is}\:\mu. \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{tension}\:{in}\:{the}\:{thread}. \\ $$$$\left.\mathrm{2}\right)\:{what}\:{is}\:{the}\:{inclination}\:{angle}\:{such} \\ $$$${that}\:{the}\:{tension}\:{in}\:{thread}\:{is}\:{maximum}? \\ $$
Commented by mrW1 last updated on 03/Jan/18
$${See}\:{also}\:{Q}\mathrm{27202} \\ $$
Commented by ajfour last updated on 03/Jan/18
Commented by ajfour last updated on 03/Jan/18
![Granted block slides: T=Nsin θ−μmgcos^2 θ =(mgcos θ)sin θ−μmgcos^2 θ (dT/dθ)=mg(cos 2θ+μsin 2θ) =0 for tan 2θ=−(1/μ) ; for this value (dT/dθ^2 )=mg(−2sin 2θ+2μcos 2θ) =mg(−(2/( (√(1+μ^2 ))))−((2μ^2 )/( (√(1+μ^2 ))))) =−2mg(√(1+μ^2 )) < 0 hence maximum tension for θ=(π/2)−(1/2)tan^(−1) ((1/μ)) =α (say). T_(max) =((mg)/2)(sin 2α−2μcos^2 α) =((mg)/2)[sin 2α−μ(1+cos 2α)] =((mg)/2)[(1/( (√(1+μ^2 ))))−μ(1−(μ/( (√(1+μ^2 )))))] =((mg)/2)[(√(1+μ^2 ))−μ] .](https://www.tinkutara.com/question/Q27229.png)
$${Granted}\:{block}\:{slides}: \\ $$$${T}={N}\mathrm{sin}\:\theta−\mu{mg}\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$=\left({mg}\mathrm{cos}\:\theta\right)\mathrm{sin}\:\theta−\mu{mg}\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$\frac{{dT}}{{d}\theta}={mg}\left(\mathrm{cos}\:\mathrm{2}\theta+\mu\mathrm{sin}\:\mathrm{2}\theta\right) \\ $$$$=\mathrm{0}\:\:{for}\:\:\mathrm{tan}\:\mathrm{2}\theta=−\frac{\mathrm{1}}{\mu}\:;\:{for}\:{this}\:{value} \\ $$$$\frac{{dT}}{{d}\theta^{\mathrm{2}} }={mg}\left(−\mathrm{2sin}\:\mathrm{2}\theta+\mathrm{2}\mu\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$\:\:\:\:\:\:={mg}\left(−\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}−\frac{\mathrm{2}\mu^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\right) \\ $$$$\:\:\:\:\:=−\mathrm{2}{mg}\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\:\:<\:\mathrm{0}\: \\ $$$$\:\:\:\:\:{hence}\:{maximum}\:{tension}\:{for} \\ $$$$\:\:\:\theta=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mu}\right)\:=\alpha\:\left({say}\right). \\ $$$${T}_{{max}} =\frac{{mg}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{2}\alpha−\mathrm{2}\mu\mathrm{cos}\:^{\mathrm{2}} \alpha\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{{mg}}{\mathrm{2}}\left[\mathrm{sin}\:\mathrm{2}\alpha−\mu\left(\mathrm{1}+\mathrm{cos}\:\mathrm{2}\alpha\right)\right] \\ $$$$\:\:\:\:\:\:\:\:=\frac{{mg}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}−\mu\left(\mathrm{1}−\frac{\mu}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:=\frac{{mg}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }−\mu\right]\:. \\ $$
Answered by mrW1 last updated on 03/Jan/18
![Q1: N=mg cos θ f=μN=μmg cos θ case 1: mg sin θ < f=μmg cos θ or tan θ<μ or θ<α=tan^(−1) μ block m rests on block M, ⇒T=0 case 2: θ≥α=tan^(−1) μ T=N sin θ−f cos θ T=mg cos θ sin θ−μmg cos^2 θ =((mg)/2)(sin 2θ−μ 2cos^2 θ) =((mg)/2)(sin 2θ−μ cos 2θ−μ) =((mg)/2)[(√(1+μ^2 )) ((1/( (√(1+μ^2 )))) sin 2θ−(μ/( (√(1+μ^2 )))) cos 2θ)−μ =((mg)/2)[(√(1+μ^2 )) (cos α sin 2θ−sin α cos 2θ)−μ] T=((mg)/2)[(√(1+μ^2 )) sin (2θ−α)−μ] with α=tan^(−1) μ Q2: T is maximum if 2θ−α=(π/2) or θ_m =(π/4)+(α/2)=(π/4)+(1/2)tan^(−1) μ=(π/2)−(1/2) tan^(−1) (1/μ) Example: μ=0.3, θ_m =53.3°](https://www.tinkutara.com/question/Q27233.png)
$${Q}\mathrm{1}: \\ $$$${N}={mg}\:\mathrm{cos}\:\theta \\ $$$${f}=\mu{N}=\mu{mg}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:{mg}\:\mathrm{sin}\:\theta\:<\:{f}=\mu{mg}\:\mathrm{cos}\:\theta \\ $$$${or}\:\mathrm{tan}\:\theta<\mu\:{or}\:\theta<\alpha=\mathrm{tan}^{−\mathrm{1}} \mu \\ $$$${block}\:{m}\:{rests}\:{on}\:{block}\:{M}, \\ $$$$\Rightarrow{T}=\mathrm{0} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:\theta\geqslant\alpha=\mathrm{tan}^{−\mathrm{1}} \mu \\ $$$${T}={N}\:\mathrm{sin}\:\theta−{f}\:\mathrm{cos}\:\theta \\ $$$${T}={mg}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta−\mu{mg}\:\mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$$=\frac{{mg}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{2}\theta−\mu\:\mathrm{2cos}^{\mathrm{2}} \:\theta\right) \\ $$$$=\frac{{mg}}{\mathrm{2}}\left(\mathrm{sin}\:\mathrm{2}\theta−\mu\:\mathrm{cos}\:\mathrm{2}\theta−\mu\right) \\ $$$$=\frac{{mg}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\:\mathrm{sin}\:\mathrm{2}\theta−\frac{\mu}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}\:\mathrm{cos}\:\mathrm{2}\theta\right)−\mu\right. \\ $$$$=\frac{{mg}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\:\left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\mathrm{2}\theta−\mathrm{sin}\:\alpha\:\mathrm{cos}\:\mathrm{2}\theta\right)−\mu\right] \\ $$$${T}=\frac{{mg}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }\:\mathrm{sin}\:\left(\mathrm{2}\theta−\alpha\right)−\mu\right] \\ $$$${with}\:\alpha=\mathrm{tan}^{−\mathrm{1}} \mu \\ $$$$ \\ $$$${Q}\mathrm{2}: \\ $$$${T}\:{is}\:{maximum}\:{if}\:\mathrm{2}\theta−\alpha=\frac{\pi}{\mathrm{2}} \\ $$$${or}\:\theta_{{m}} =\frac{\pi}{\mathrm{4}}+\frac{\alpha}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \mu=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mu} \\ $$$${Example}:\:\mu=\mathrm{0}.\mathrm{3},\:\theta_{{m}} =\mathrm{53}.\mathrm{3}° \\ $$
Commented by Tinkutara last updated on 04/Jan/18
Thank you!
Commented by mrW1 last updated on 03/Jan/18
https://play.google.com/store/apps/details?id=com.iudesk.android.photo.editor
Commented by mrW1 last updated on 03/Jan/18
$${I}\:{use}\:\boldsymbol{{Photo}}\:\boldsymbol{{Editor}}. \\ $$