Question Number 174351 by ajfour last updated on 30/Jul/22
Answered by mr W last updated on 01/Aug/22
Commented by mr W last updated on 01/Aug/22
$$\mathrm{tan}\:\phi=\mu \\ $$$${T}={m}_{\mathrm{1}} {g} \\ $$$$\mathrm{2}{T}\:\mathrm{cos}\:\phi={mg} \\ $$$$\mathrm{2}{m}_{\mathrm{1}} {g}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}={mg} \\ $$$$\Rightarrow\frac{{m}_{\mathrm{1}} }{{m}}\:=\frac{\sqrt{\mathrm{1}+\mu^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\frac{{L}}{\mathrm{2}}\mathrm{cos}\:\theta={h}\:\mathrm{tan}\:\phi=\mu{h} \\ $$$$\Rightarrow\mathrm{cos}\:\theta=\frac{{L}}{\mathrm{2}\mu{h}} \\ $$
Commented by ajfour last updated on 01/Aug/22
![given m_1 =m T=mg N+Tcos φ=mg N=mg(1−cos φ) f=μN=μmg(1−cos φ)=Tsin φ ⇒ μ(1−cos φ)=sin φ μ=cot (φ/2) mg(((Lcos θ)/2))=(Tcos φ)(Lcos θ) +(Tsin φ)(Lsin θ) ⇒ tan θ=(1/2)−cos φ h=Lcos φ+Lsin θ (h/L)=cos φ+sin θ =cos (2cot^(−1) μ) +(1/( (√(1+(4/([1−2cos (2cot^(−1) μ)]^2 ))))))](Q174461.png)
$${given}\:\:\:{m}_{\mathrm{1}} ={m} \\ $$$${T}={mg} \\ $$$${N}+{T}\mathrm{cos}\:\phi={mg} \\ $$$${N}={mg}\left(\mathrm{1}−\mathrm{cos}\:\phi\right) \\ $$$${f}=\mu{N}=\mu{mg}\left(\mathrm{1}−\mathrm{cos}\:\phi\right)={T}\mathrm{sin}\:\phi \\ $$$$\Rightarrow\:\mu\left(\mathrm{1}−\mathrm{cos}\:\phi\right)=\mathrm{sin}\:\phi \\ $$$$\mu=\mathrm{cot}\:\frac{\phi}{\mathrm{2}}\:\: \\ $$$${mg}\left(\frac{{L}\mathrm{cos}\:\theta}{\mathrm{2}}\right)=\left({T}\mathrm{cos}\:\phi\right)\left({L}\mathrm{cos}\:\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({T}\mathrm{sin}\:\phi\right)\left({L}\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{cos}\:\phi \\ $$$${h}={L}\mathrm{cos}\:\phi+{L}\mathrm{sin}\:\theta \\ $$$$\frac{{h}}{{L}}=\mathrm{cos}\:\phi+\mathrm{sin}\:\theta \\ $$$$\:\:\:\:=\mathrm{cos}\:\left(\mathrm{2cot}^{−\mathrm{1}} \mu\right) \\ $$$$+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{\mathrm{4}}{\left[\mathrm{1}−\mathrm{2cos}\:\left(\mathrm{2cot}^{−\mathrm{1}} \mu\right)\right]^{\mathrm{2}} }}} \\ $$$$ \\ $$
Commented by ajfour last updated on 01/Aug/22
$${Sir},\:{for}\:{your}\:{solution}: \\ $$$${line}\:{of}\:{mg}\:{passes}\:{through}\:{the} \\ $$$${pulley}\:{needs}\:{justification}!?\:{sir}. \\ $$
Commented by Tawa11 last updated on 01/Aug/22
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Commented by mr W last updated on 02/Aug/22
$${the}\:{three}\:{forces}\:{acting}\:{on}\:{the}\:{rod}: \\ $$$${mg},\:{R},\:{T} \\ $$$${they}\:{must}\:{intersect}\:{at}\:{on}\:{point}. \\ $$$$ \\ $$$${m}_{\mathrm{1}} \:{must}\:{be}\:{less}\:{than}\:{m},\:{otherwise} \\ $$$${the}\:{rod}\:{can}\:{only}\:{have}\:{one}\:{single} \\ $$$${position}:\:{vertical}\:{position}. \\ $$
Commented by ajfour last updated on 02/Aug/22
$${say}\:\:{for}\:{example}\left({extreme}\:{case}\right) \\ $$$$\:\mu=\mathrm{cot}\:\frac{\phi}{\mathrm{2}}=\sqrt{\mathrm{3}}\:\:\Rightarrow\:\phi=\mathrm{60}° \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\frac{{h}}{{L}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\Rightarrow\:\:{h}=\frac{{L}}{\mathrm{2}} \\ $$$${N}+{T}\mathrm{cos}\:\phi={mg} \\ $$$${N}=\frac{{mg}}{\mathrm{2}} \\ $$$${f}={T}\mathrm{sin}\:\phi=\frac{{mg}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mu{N} \\ $$$$ \\ $$
Commented by ajfour last updated on 02/Aug/22
Commented by mr W last updated on 02/Aug/22
$${yes}\:{sir}. \\ $$
Commented by ajfour last updated on 02/Aug/22
$${in}\:{your}\:{solution} \\ $$$${sir},\:{how}\:\:\:\:\:\mathrm{2}{T}\mathrm{cos}\:\phi={mg} \\ $$$${there}\:{is}\:{even}\:{torque}\:{due}\:{to} \\ $$$${other}\:{component}\:{of}\:{tension}\:{as} \\ $$$${well}. \\ $$
Commented by mr W last updated on 03/Aug/22
$${that}\:{might}\:{be}\:{incorrect}. \\ $$
Commented by mr W last updated on 03/Aug/22
Commented by mr W last updated on 03/Aug/22
$${T}={mg} \\ $$$$\varphi=\pi−\mathrm{2}\phi \\ $$