Question Number 46003 by MrW3 last updated on 19/Oct/18
Commented by MrW3 last updated on 20/Oct/18
$${Find}\:{the}\:{angle}\:\theta\:{when}\:{the}\:{rope}\:{starts} \\ $$$${to}\:{slide}.\:\left(\mu={coefficient}\:{of}\:{friction}\right) \\ $$
Answered by MrW3 last updated on 20/Oct/18
Commented by MrW3 last updated on 20/Oct/18
![T=tension force in rope ρ=mass of rope of unit length μ=coefficient of friction length of free hanging rope AB is Rθ, hence tension in rope at point B is T_B =ρgRθ. dG=ρgRdϕ dN=dG cos ϕ+T dϕ=(ρgR cos ϕ+T)dϕ df=μdN=μ(ρgR cos ϕ+T)dϕ T+dG sin ϕ=T+dT+df ρgR sin ϕ dϕ=dT+μ(ρgR cos ϕ+T)dϕ ⇒(dT/dϕ)+μT=ρgR(sin ϕ−μ cos ϕ) ⇒T=((ρgR[2μ cos ϕ+(1−μ^2 )sin ϕ])/(1+μ^2 ))+Ce^(−μϕ) (see Q for more details) at point B: ϕ=0 and T=T_B =ρgRθ ⇒((ρgR2μ)/(1+μ^2 ))+C=ρgRθ ⇒C=ρgR(θ−((2μ)/(1+μ^2 )))=((ρgR)/(1+μ^2 ))[(1+μ^2 )θ−2μ] ⇒T=((ρgR)/(1+μ^2 )){2μ cos ϕ+(1−μ^2 )sin ϕ+[(1+μ^2 )θ−2μ]e^(−μϕ) } at point C: ϕ=π−θ, the rope starts to slide if T_C =0, i.e. ⇒(1−μ^2 )sin θ−2μcos θ+[(1+μ^2 )θ−2μ]e^(−μ(π−θ)) =0 this equation can be solved only numerically for θ. for example if μ=0.4, we get θ=42.7°. the relationship θ(μ) can be displayed in a diagram like this:](https://www.tinkutara.com/question/Q46017.png)
$${T}={tension}\:{force}\:{in}\:{rope} \\ $$$$\rho={mass}\:{of}\:{rope}\:{of}\:{unit}\:{length} \\ $$$$\mu={coefficient}\:{of}\:{friction} \\ $$$$ \\ $$$${length}\:{of}\:{free}\:{hanging}\:{rope}\:{AB}\:{is} \\ $$$${R}\theta,\:{hence}\:{tension}\:{in}\:{rope}\:{at}\:{point}\:{B} \\ $$$${is}\:{T}_{{B}} =\rho{gR}\theta. \\ $$$$ \\ $$$${dG}=\rho{gRd}\varphi \\ $$$${dN}={dG}\:\mathrm{cos}\:\varphi+{T}\:{d}\varphi=\left(\rho{gR}\:\mathrm{cos}\:\varphi+{T}\right){d}\varphi \\ $$$${df}=\mu{dN}=\mu\left(\rho{gR}\:\mathrm{cos}\:\varphi+{T}\right){d}\varphi \\ $$$$ \\ $$$${T}+{dG}\:\mathrm{sin}\:\varphi={T}+{dT}+{df} \\ $$$$\rho{gR}\:\mathrm{sin}\:\varphi\:{d}\varphi={dT}+\mu\left(\rho{gR}\:\mathrm{cos}\:\varphi+{T}\right){d}\varphi \\ $$$$\Rightarrow\frac{{dT}}{{d}\varphi}+\mu{T}=\rho{gR}\left(\mathrm{sin}\:\varphi−\mu\:\mathrm{cos}\:\varphi\right) \\ $$$$\Rightarrow{T}=\frac{\rho{gR}\left[\mathrm{2}\mu\:\mathrm{cos}\:\varphi+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\mathrm{sin}\:\varphi\right]}{\mathrm{1}+\mu^{\mathrm{2}} }+{Ce}^{−\mu\varphi} \\ $$$$\left({see}\:{Q} \:{for}\:{more}\:{details}\right) \\ $$$$ \\ $$$${at}\:{point}\:{B}:\:\varphi=\mathrm{0}\:{and}\:{T}={T}_{{B}} =\rho{gR}\theta \\ $$$$\Rightarrow\frac{\rho{gR}\mathrm{2}\mu}{\mathrm{1}+\mu^{\mathrm{2}} }+{C}=\rho{gR}\theta \\ $$$$\Rightarrow{C}=\rho{gR}\left(\theta−\frac{\mathrm{2}\mu}{\mathrm{1}+\mu^{\mathrm{2}} }\right)=\frac{\rho{gR}}{\mathrm{1}+\mu^{\mathrm{2}} }\left[\left(\mathrm{1}+\mu^{\mathrm{2}} \right)\theta−\mathrm{2}\mu\right] \\ $$$$\Rightarrow{T}=\frac{\rho{gR}}{\mathrm{1}+\mu^{\mathrm{2}} }\left\{\mathrm{2}\mu\:\mathrm{cos}\:\varphi+\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\mathrm{sin}\:\varphi+\left[\left(\mathrm{1}+\mu^{\mathrm{2}} \right)\theta−\mathrm{2}\mu\right]{e}^{−\mu\varphi} \right\} \\ $$$$ \\ $$$${at}\:{point}\:{C}:\:\varphi=\pi−\theta, \\ $$$${the}\:{rope}\:{starts}\:{to}\:{slide}\:{if}\:{T}_{{C}} =\mathrm{0},\:{i}.{e}. \\ $$$$\Rightarrow\left(\mathrm{1}−\mu^{\mathrm{2}} \right)\mathrm{sin}\:\theta−\mathrm{2}\mu\mathrm{cos}\:\theta+\left[\left(\mathrm{1}+\mu^{\mathrm{2}} \right)\theta−\mathrm{2}\mu\right]{e}^{−\mu\left(\pi−\theta\right)} =\mathrm{0} \\ $$$${this}\:{equation}\:{can}\:{be}\:{solved}\:{only} \\ $$$${numerically}\:{for}\:\theta.\:{for}\:{example}\:{if} \\ $$$$\mu=\mathrm{0}.\mathrm{4},\:{we}\:{get}\:\theta=\mathrm{42}.\mathrm{7}°. \\ $$$$ \\ $$$${the}\:{relationship}\:\theta\left(\mu\right)\:{can}\:{be} \\ $$$${displayed}\:{in}\:{a}\:{diagram}\:{like}\:{this}: \\ $$
Commented by MrW3 last updated on 20/Oct/18
Commented by ajfour last updated on 20/Oct/18
$${Excellent}\:{Sir}. \\ $$