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Question Number 46003 by MrW3 last updated on 19/Oct/18

Commented by MrW3 last updated on 20/Oct/18

$F i n d t h e a n g l e θ w h e n t h e r o p e s t a r t s$ $t o s l i d e . (μ = c o e f f i c i e n t o f f r i c t i o n)$
	Answered by MrW3 last updated on 20/Oct/18

	Commented by MrW3 last updated on 20/Oct/18
	$T=tension force in rope ρ=mass of rope of unit length μ=coefficient of friction length of free hanging rope AB is Rθ, hence tension in rope at point B is T_B =ρgRθ. dG=ρgRdϕ dN=dG cos ϕ+T dϕ=(ρgR cos ϕ+T)dϕ df=μdN=μ(ρgR cos ϕ+T)dϕ T+dG sin ϕ=T+dT+df ρgR sin ϕ dϕ=dT+μ(ρgR cos ϕ+T)dϕ ⇒(dT/dϕ)+μT=ρgR(sin ϕ−μ cos ϕ) ⇒T=((ρgR[2μ cos ϕ+(1−μ^2 )sin ϕ])/(1+μ^2 ))+Ce^(−μϕ) (see Q for more details) at point B: ϕ=0 and T=T_B =ρgRθ ⇒((ρgR2μ)/(1+μ^2 ))+C=ρgRθ ⇒C=ρgR(θ−((2μ)/(1+μ^2 )))=((ρgR)/(1+μ^2 ))[(1+μ^2 )θ−2μ] ⇒T=((ρgR)/(1+μ^2 )){2μ cos ϕ+(1−μ^2 )sin ϕ+[(1+μ^2 )θ−2μ]e^(−μϕ) } at point C: ϕ=π−θ, the rope starts to slide if T_C =0, i.e. ⇒(1−μ^2 )sin θ−2μcos θ+[(1+μ^2 )θ−2μ]e^(−μ(π−θ)) =0 this equation can be solved only numerically for θ. for example if μ=0.4, we get θ=42.7°. the relationship θ(μ) can be displayed in a diagram like this:$
	$T = t e n s i o n f o r c e i n r o p e$ $ρ = m a s s o f r o p e o f u n i t l e n g t h$ $μ = c o e f f i c i e n t o f f r i c t i o n$ $l e n g t h o f f r e e h a n g i n g r o p e A B i s$ $R θ, h e n c e t e n s i o n i n r o p e a t p o i n t B$ $i s T_{B} = ρ g R θ .$ $d G = ρ g R d φ$ $d N = d G \cos φ + T d φ = (ρ g R \cos φ + T) d φ$ $d f = μ d N = μ (ρ g R \cos φ + T) d φ$ $T + d G \sin φ = T + d T + d f$ $ρ g R \sin φ d φ = d T + μ (ρ g R \cos φ + T) d φ$ $\Rightarrow \frac{d T}{d φ} + μ T = ρ g R (\sin φ - μ \cos φ)$ $\Rightarrow T = \frac{ρ g R [2 μ \cos φ + (1 - μ^{2}) \sin φ]}{1 + μ^{2}} + {C e}^{- μ φ}$ $(s e e Q f o r m o r e d e t a i l s)$ $a t p o i n t B : φ = 0 a n d T = T_{B} = ρ g R θ$ $\Rightarrow \frac{ρ g R 2 μ}{1 + μ^{2}} + C = ρ g R θ$ $\Rightarrow C = ρ g R (θ - \frac{2 μ}{1 + μ^{2}}) = \frac{ρ g R}{1 + μ^{2}} [(1 + μ^{2}) θ - 2 μ]$ $\Rightarrow T = \frac{ρ g R}{1 + μ^{2}} {2 μ \cos φ + (1 - μ^{2}) \sin φ + [(1 + μ^{2}) θ - 2 μ] e^{- μ φ}}$ $a t p o i n t C : φ = π - θ,$ $t h e r o p e s t a r t s t o s l i d e i f T_{C} = 0, i . e .$ $\Rightarrow (1 - μ^{2}) \sin θ - 2 μ \cos θ + [(1 + μ^{2}) θ - 2 μ] e^{- μ (π - θ)} = 0$ $t h i s e q u a t i o n c a n b e s o l v e d o n l y$ $n u m e r i c a l l y f o r θ . f o r e x a m p l e i f$ $μ = 0.4, w e g e t θ = 42.7 ° .$ $t h e r e l a t i o n s h i p θ (μ) c a n b e$ $d i s p l a y e d i n a d i a g r a m l i k e t h i s :$
	Commented by MrW3 last updated on 20/Oct/18

	Commented by ajfour last updated on 20/Oct/18

	$E x c e l l e n t S i r .$
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