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Question-170601




Question Number 170601 by mr W last updated on 27/May/22
Commented by mr W last updated on 29/May/22
an uniform thin rope with length L  is hanging at one end at a height of h   above a inclined plane as shown.   if the friction coefficient between   rope and the plane is μ, find the  minimum and maximum length of   that part of the rope which can lie on   the plane.  (L>h)
$${an}\:{uniform}\:{thin}\:{rope}\:{with}\:{length}\:{L} \\ $$$${is}\:{hanging}\:{at}\:{one}\:{end}\:{at}\:{a}\:{height}\:{of}\:{h}\: \\ $$$${above}\:{a}\:{inclined}\:{plane}\:{as}\:{shown}.\: \\ $$$${if}\:{the}\:{friction}\:{coefficient}\:{between}\: \\ $$$${rope}\:{and}\:{the}\:{plane}\:{is}\:\mu,\:{find}\:{the} \\ $$$${minimum}\:{and}\:{maximum}\:{length}\:{of}\: \\ $$$${that}\:{part}\:{of}\:{the}\:{rope}\:{which}\:{can}\:{lie}\:{on}\: \\ $$$${the}\:{plane}. \\ $$$$\left({L}>{h}\right) \\ $$
Answered by aleks041103 last updated on 28/May/22
Commented by aleks041103 last updated on 28/May/22
red curve → catenary  y=a cosh((x/a))  length function:  ∫(√(1+y′^2 ))dx=∫(√(1+sinh^2 ((x/a))))dx=  =∫cosh((x/a))dx=a sinh((x/a))  Let length on slope be l, then there is  a rope of length L−l hanging or  L−l=a(sinh((x_b /a))−sinh((x_a /a)))  also  y′(x_b )=tanθ=k  ⇒sinh((x_b /a))=k⇒x_b =a arcsinh(k)  and L−l=a(k−sinh((x_a /a)))  ⇒slope has eqn:  s(x)=k(x−a arcsinh(k)) + a(√(1+k^2 ))  ⇒ for x_a  we have  acosh((x_a /a))−s(x_a )=h  ⇒cosh((x_a /a))−k(x_a /a)=(h/a)+(√(1+k^2 ))−k arcsinh(k)  Also, the hanging part has two forces acting  T_1  at A at angle α to x and T_2  at B at angle  θ to x.  tanα=sinh((x_a /a))  ⇒sinα=((tanα)/( (√(1+tan^2 α))))=((sinh(x_a /a))/(cosh(x_a /a)))=tanh((x_a /a))  cosα=sech((x_a /a))  ⇒Equilibrium requires:  T_1 cosα=T_2 cosθ  −T_1 sinα+T_2 sinθ=((L−l)/L)mg  Also T_2 ≤(l/L)μmg cosθ  ⇒tgα≤(((l/L)μmg sinθcosθ+((l−L)/L)mg)/((l/L)μmgcos^2 θ))=  =tgθ−((L−l)/(μl))sec^2 θ=−((L−l)/(μl))(1+k^2 )+k≥sinh(x_a /a)  ⇒L−l≤a(((L−l)/(μl))(1+k^2 ))  ⇒l≤((a(1+k^2 ))/μ)⇒a≥((μl)/(1+k^2 ))  ...  A lot of inequalities left to do
$${red}\:{curve}\:\rightarrow\:{catenary} \\ $$$${y}={a}\:{cosh}\left(\frac{{x}}{{a}}\right) \\ $$$${length}\:{function}: \\ $$$$\int\sqrt{\mathrm{1}+{y}'^{\mathrm{2}} }{dx}=\int\sqrt{\mathrm{1}+{sinh}^{\mathrm{2}} \left(\frac{{x}}{{a}}\right)}{dx}= \\ $$$$=\int{cosh}\left(\frac{{x}}{{a}}\right){dx}={a}\:{sinh}\left(\frac{{x}}{{a}}\right) \\ $$$${Let}\:{length}\:{on}\:{slope}\:{be}\:{l},\:{then}\:{there}\:{is} \\ $$$${a}\:{rope}\:{of}\:{length}\:{L}−{l}\:{hanging}\:{or} \\ $$$${L}−{l}={a}\left({sinh}\left(\frac{{x}_{{b}} }{{a}}\right)−{sinh}\left(\frac{{x}_{{a}} }{{a}}\right)\right) \\ $$$${also} \\ $$$${y}'\left({x}_{{b}} \right)={tan}\theta={k} \\ $$$$\Rightarrow{sinh}\left(\frac{{x}_{{b}} }{{a}}\right)={k}\Rightarrow{x}_{{b}} ={a}\:{arcsinh}\left({k}\right) \\ $$$${and}\:{L}−{l}={a}\left({k}−{sinh}\left(\frac{{x}_{{a}} }{{a}}\right)\right) \\ $$$$\Rightarrow{slope}\:{has}\:{eqn}: \\ $$$${s}\left({x}\right)={k}\left({x}−{a}\:{arcsinh}\left({k}\right)\right)\:+\:{a}\sqrt{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{for}\:{x}_{{a}} \:{we}\:{have} \\ $$$${acosh}\left(\frac{{x}_{{a}} }{{a}}\right)−{s}\left({x}_{{a}} \right)={h} \\ $$$$\Rightarrow{cosh}\left(\frac{{x}_{{a}} }{{a}}\right)−{k}\frac{{x}_{{a}} }{{a}}=\frac{{h}}{{a}}+\sqrt{\mathrm{1}+{k}^{\mathrm{2}} }−{k}\:{arcsinh}\left({k}\right) \\ $$$${Also},\:{the}\:{hanging}\:{part}\:{has}\:{two}\:{forces}\:{acting} \\ $$$${T}_{\mathrm{1}} \:{at}\:{A}\:{at}\:{angle}\:\alpha\:{to}\:{x}\:{and}\:{T}_{\mathrm{2}} \:{at}\:{B}\:{at}\:{angle} \\ $$$$\theta\:{to}\:{x}. \\ $$$${tan}\alpha={sinh}\left(\frac{{x}_{{a}} }{{a}}\right) \\ $$$$\Rightarrow{sin}\alpha=\frac{{tan}\alpha}{\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}}=\frac{{sinh}\left({x}_{{a}} /{a}\right)}{{cosh}\left({x}_{{a}} /{a}\right)}={tanh}\left(\frac{{x}_{{a}} }{{a}}\right) \\ $$$${cos}\alpha={sech}\left(\frac{{x}_{{a}} }{{a}}\right) \\ $$$$\Rightarrow{Equilibrium}\:{requires}: \\ $$$${T}_{\mathrm{1}} {cos}\alpha={T}_{\mathrm{2}} {cos}\theta \\ $$$$−{T}_{\mathrm{1}} {sin}\alpha+{T}_{\mathrm{2}} {sin}\theta=\frac{{L}−{l}}{{L}}{mg} \\ $$$${Also}\:{T}_{\mathrm{2}} \leqslant\frac{{l}}{{L}}\mu{mg}\:{cos}\theta \\ $$$$\Rightarrow{tg}\alpha\leqslant\frac{\frac{{l}}{{L}}\mu{mg}\:{sin}\theta{cos}\theta+\frac{{l}−{L}}{{L}}{mg}}{\frac{{l}}{{L}}\mu{mgcos}^{\mathrm{2}} \theta}= \\ $$$$={tg}\theta−\frac{{L}−{l}}{\mu{l}}{sec}^{\mathrm{2}} \theta=−\frac{{L}−{l}}{\mu{l}}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)+{k}\geqslant{sinh}\frac{{x}_{{a}} }{{a}} \\ $$$$\Rightarrow{L}−{l}\leqslant{a}\left(\frac{{L}−{l}}{\mu{l}}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)\right) \\ $$$$\Rightarrow{l}\leqslant\frac{{a}\left(\mathrm{1}+{k}^{\mathrm{2}} \right)}{\mu}\Rightarrow{a}\geqslant\frac{\mu{l}}{\mathrm{1}+{k}^{\mathrm{2}} } \\ $$$$… \\ $$$${A}\:{lot}\:{of}\:{inequalities}\:{left}\:{to}\:{do} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 29/May/22
thanks for the solution sir!
$${thanks}\:{for}\:{the}\:{solution}\:{sir}! \\ $$
Answered by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
let′s have a look at an object on an  inclined plane.  such that the object is in equilibrium  on such a slope, i.e. such that it  doesn′t slide down, a friction force f  must act between the object and the  slope as shown. the friction coefficient  between them defines actually the   maximal possible friction force   between them.  f=μN with 0≤μ≤μ_(nax) .  though we speak of friction  coefficient μ, but actually we mean  the maximal possible friction  coefficient μ_(max) . the actual friction  coefficient μ must fulfull 0≤μ≤μ_(max) .  in following with μ i′ll denote the  actual friction coefficient between  rope and slope.  f=μN  N=Mg cos θ  such that the object doesn′t slide  down,   f−Mg sin θ=0  μMg cos θ−Mg sin θ=0  μ cos θ−sin θ=0  μ=((sin θ)/(cos θ))=tan θ≤μ_(max)   if the friction coefficient μ_(nax)  is  smaller than tan θ, an object can not  be in equilibrium on the slope.  since in our case a part of the rope  can lie on the slope, we must have   μ_(max) ≥tan θ.
$${let}'{s}\:{have}\:{a}\:{look}\:{at}\:{an}\:{object}\:{on}\:{an} \\ $$$${inclined}\:{plane}. \\ $$$${such}\:{that}\:{the}\:{object}\:{is}\:{in}\:{equilibrium} \\ $$$${on}\:{such}\:{a}\:{slope},\:{i}.{e}.\:{such}\:{that}\:{it} \\ $$$${doesn}'{t}\:{slide}\:{down},\:{a}\:{friction}\:{force}\:{f} \\ $$$${must}\:{act}\:{between}\:{the}\:{object}\:{and}\:{the} \\ $$$${slope}\:{as}\:{shown}.\:{the}\:{friction}\:{coefficient} \\ $$$${between}\:{them}\:{defines}\:{actually}\:{the}\: \\ $$$${maximal}\:{possible}\:{friction}\:{force}\: \\ $$$${between}\:{them}. \\ $$$${f}=\mu{N}\:{with}\:\mathrm{0}\leqslant\mu\leqslant\mu_{{nax}} . \\ $$$${though}\:{we}\:{speak}\:{of}\:{friction} \\ $$$${coefficient}\:\mu,\:{but}\:{actually}\:{we}\:{mean} \\ $$$${the}\:{maximal}\:{possible}\:{friction} \\ $$$${coefficient}\:\mu_{{max}} .\:{the}\:{actual}\:{friction} \\ $$$${coefficient}\:\mu\:{must}\:{fulfull}\:\mathrm{0}\leqslant\mu\leqslant\mu_{{max}} . \\ $$$${in}\:{following}\:{with}\:\mu\:{i}'{ll}\:{denote}\:{the} \\ $$$${actual}\:{friction}\:{coefficient}\:{between} \\ $$$${rope}\:{and}\:{slope}. \\ $$$${f}=\mu{N} \\ $$$${N}={Mg}\:\mathrm{cos}\:\theta \\ $$$${such}\:{that}\:{the}\:{object}\:{doesn}'{t}\:{slide} \\ $$$${down},\: \\ $$$${f}−{Mg}\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\mu{Mg}\:\mathrm{cos}\:\theta−{Mg}\:\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta=\mathrm{0} \\ $$$$\mu=\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta}=\mathrm{tan}\:\theta\leqslant\mu_{{max}} \\ $$$${if}\:{the}\:{friction}\:{coefficient}\:\mu_{{nax}} \:{is} \\ $$$${smaller}\:{than}\:\mathrm{tan}\:\theta,\:{an}\:{object}\:{can}\:{not} \\ $$$${be}\:{in}\:{equilibrium}\:{on}\:{the}\:{slope}. \\ $$$${since}\:{in}\:{our}\:{case}\:{a}\:{part}\:{of}\:{the}\:{rope} \\ $$$${can}\:{lie}\:{on}\:{the}\:{slope},\:{we}\:{must}\:{have}\: \\ $$$$\mu_{{max}} \geqslant\mathrm{tan}\:\theta. \\ $$
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
case 1: μ=tan θ  we see the rope can take the shape  as shown in diagram. the length of  the part which lies on the slope is  b=L−h.
$${case}\:\mathrm{1}:\:\mu=\mathrm{tan}\:\theta \\ $$$${we}\:{see}\:{the}\:{rope}\:{can}\:{take}\:{the}\:{shape} \\ $$$${as}\:{shown}\:{in}\:{diagram}.\:{the}\:{length}\:{of} \\ $$$${the}\:{part}\:{which}\:{lies}\:{on}\:{the}\:{slope}\:{is} \\ $$$${b}={L}−{h}. \\ $$
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
case 2: μ>tan θ  the friction force f is large enough  so that it can even allow an  addition tension force between the  two parts of the rope.
$${case}\:\mathrm{2}:\:\mu>\mathrm{tan}\:\theta \\ $$$${the}\:{friction}\:{force}\:{f}\:{is}\:{large}\:{enough} \\ $$$${so}\:{that}\:{it}\:{can}\:{even}\:{allow}\:{an} \\ $$$${addition}\:{tension}\:{force}\:{between}\:{the} \\ $$$${two}\:{parts}\:{of}\:{the}\:{rope}.\: \\ $$
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
say the length of the rope part which  lies on the inclined plane is b.  ρ=(m/L)  the friction force between rope and  the slope is f.  f=μρgb cos θ with tan θ<μ≤μ_(max)   tension in the rope at point B:  T_1 =f−ρgb sin θ=(μ cos θ−sin θ)ρgb  let μ′=μ cos θ−sin θ  T_1 =μ′ρgb  T_0 =T_1  cos θ=μ′ρgb cos θ  define a=(T_0 /(ρg))=μ′b cos θ  the hanging rope from point A to B   is part of a catenary. w.r.t. the  coordinate system as shown,  y=a cosh (x/a)  y′=sinh (x/a)  s=a sinh (x/a)    tan θ=sinh (x_B /a) ⇒(x_B /a)=sinh^(−1)  (tan θ)  s_1 =CB^(⌢) =a sinh (x_B /a)=a tan θ  AC^(⌢) =L−b−s_1 =a sinh (x_A /a)  L−b−a tan θ=a sinh (x_A /a)  ⇒(x_A /a)=sinh^(−1)  (((L−b)/a)−tan θ)  y_A −y_B =h−(x_A +x_B ) tan θ  a cosh (x_A /a)−a cosh (x_B /a)=h−(x_A +x_B ) tan θ  cosh (x_A /a)−cosh (x_B /a)=(h/a)−((x_A /a)+(x_B /a)) tan θ  cosh [sinh^(−1)  (((L−b)/a)−tan θ)]−cosh [sinh^(−1)  (tan θ)]=(h/a)−[sinh^(−1)  (((L−b)/a)−tan θ)+sinh^(−1)  (tan θ)] tan θ  let (b/L)=(1/λ)  cosh [sinh^(−1)  (((λ−1)/(μ′ cos θ))−tan θ)]−cosh [sinh^(−1)  (tan θ)]=((λh)/(μ′L cos θ))−[sinh^(−1)  (((λ−1)/(μ′ cos θ))−tan θ)+sinh^(−1)  (tan θ)] tan θ  ⇒[sinh^(−1)  (((𝛌−1)/(𝛍′ cos 𝛉))−tan 𝛉)+sinh^(−1)  (tan 𝛉)]sin 𝛉=1+((𝛌h)/(𝛍′L))−(√(1+(((𝛌−1)/(𝛍′)))^2 −2(((𝛌−1)/(𝛍′))) sin 𝛉))  this equation shows the relationship  between λ and the actual friction  coefficient μ (tan θ<μ≤μ_(max) ).  generally this relationship looks like  as shown in the diagram below.    example:  h=3m, L=5m, θ=30°, μ_(max) =1.5  ⇒(b_(min) /L)=0.33202 at μ=1.5  ⇒(b_(max) /L)=0.43165 at μ=0.7
$${say}\:{the}\:{length}\:{of}\:{the}\:{rope}\:{part}\:{which} \\ $$$${lies}\:{on}\:{the}\:{inclined}\:{plane}\:{is}\:{b}. \\ $$$$\rho=\frac{{m}}{{L}} \\ $$$${the}\:{friction}\:{force}\:{between}\:{rope}\:{and} \\ $$$${the}\:{slope}\:{is}\:{f}. \\ $$$${f}=\mu\rho{gb}\:\mathrm{cos}\:\theta\:{with}\:\mathrm{tan}\:\theta<\mu\leqslant\mu_{{max}} \\ $$$${tension}\:{in}\:{the}\:{rope}\:{at}\:{point}\:{B}: \\ $$$${T}_{\mathrm{1}} ={f}−\rho{gb}\:\mathrm{sin}\:\theta=\left(\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)\rho{gb} \\ $$$${let}\:\mu'=\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta \\ $$$${T}_{\mathrm{1}} =\mu'\rho{gb} \\ $$$${T}_{\mathrm{0}} ={T}_{\mathrm{1}} \:\mathrm{cos}\:\theta=\mu'\rho{gb}\:\mathrm{cos}\:\theta \\ $$$${define}\:{a}=\frac{{T}_{\mathrm{0}} }{\rho{g}}=\mu'{b}\:\mathrm{cos}\:\theta \\ $$$${the}\:{hanging}\:{rope}\:{from}\:{point}\:{A}\:{to}\:{B}\: \\ $$$${is}\:{part}\:{of}\:{a}\:{catenary}.\:{w}.{r}.{t}.\:{the} \\ $$$${coordinate}\:{system}\:{as}\:{shown}, \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${y}'=\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$${s}={a}\:\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$$ \\ $$$$\mathrm{tan}\:\theta=\mathrm{sinh}\:\frac{{x}_{{B}} }{{a}}\:\Rightarrow\frac{{x}_{{B}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right) \\ $$$${s}_{\mathrm{1}} =\overset{\frown} {{CB}}={a}\:\mathrm{sinh}\:\frac{{x}_{{B}} }{{a}}={a}\:\mathrm{tan}\:\theta \\ $$$$\overset{\frown} {{AC}}={L}−{b}−{s}_{\mathrm{1}} ={a}\:\mathrm{sinh}\:\frac{{x}_{{A}} }{{a}} \\ $$$${L}−{b}−{a}\:\mathrm{tan}\:\theta={a}\:\mathrm{sinh}\:\frac{{x}_{{A}} }{{a}} \\ $$$$\Rightarrow\frac{{x}_{{A}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{{L}−{b}}{{a}}−\mathrm{tan}\:\theta\right) \\ $$$${y}_{{A}} −{y}_{{B}} ={h}−\left({x}_{{A}} +{x}_{{B}} \right)\:\mathrm{tan}\:\theta \\ $$$${a}\:\mathrm{cosh}\:\frac{{x}_{{A}} }{{a}}−{a}\:\mathrm{cosh}\:\frac{{x}_{{B}} }{{a}}={h}−\left({x}_{{A}} +{x}_{{B}} \right)\:\mathrm{tan}\:\theta \\ $$$$\mathrm{cosh}\:\frac{{x}_{{A}} }{{a}}−\mathrm{cosh}\:\frac{{x}_{{B}} }{{a}}=\frac{{h}}{{a}}−\left(\frac{{x}_{{A}} }{{a}}+\frac{{x}_{{B}} }{{a}}\right)\:\mathrm{tan}\:\theta \\ $$$$\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{{L}−{b}}{{a}}−\mathrm{tan}\:\theta\right)\right]−\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)\right]=\frac{{h}}{{a}}−\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{{L}−{b}}{{a}}−\mathrm{tan}\:\theta\right)+\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)\right]\:\mathrm{tan}\:\theta \\ $$$${let}\:\frac{{b}}{{L}}=\frac{\mathrm{1}}{\lambda} \\ $$$$\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{\lambda−\mathrm{1}}{\mu'\:\mathrm{cos}\:\theta}−\mathrm{tan}\:\theta\right)\right]−\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)\right]=\frac{\lambda{h}}{\mu'{L}\:\mathrm{cos}\:\theta}−\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{\lambda−\mathrm{1}}{\mu'\:\mathrm{cos}\:\theta}−\mathrm{tan}\:\theta\right)+\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)\right]\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\left[\boldsymbol{\mathrm{sinh}}^{−\mathrm{1}} \:\left(\frac{\boldsymbol{\lambda}−\mathrm{1}}{\boldsymbol{\mu}'\:\boldsymbol{\mathrm{cos}}\:\boldsymbol{\theta}}−\boldsymbol{\mathrm{tan}}\:\boldsymbol{\theta}\right)+\boldsymbol{\mathrm{sinh}}^{−\mathrm{1}} \:\left(\boldsymbol{\mathrm{tan}}\:\boldsymbol{\theta}\right)\right]\boldsymbol{\mathrm{sin}}\:\boldsymbol{\theta}=\mathrm{1}+\frac{\boldsymbol{\lambda{h}}}{\boldsymbol{\mu}'\boldsymbol{{L}}}−\sqrt{\mathrm{1}+\left(\frac{\boldsymbol{\lambda}−\mathrm{1}}{\boldsymbol{\mu}'}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\boldsymbol{\lambda}−\mathrm{1}}{\boldsymbol{\mu}'}\right)\:\boldsymbol{\mathrm{sin}}\:\boldsymbol{\theta}} \\ $$$${this}\:{equation}\:{shows}\:{the}\:{relationship} \\ $$$${between}\:\lambda\:{and}\:{the}\:{actual}\:{friction} \\ $$$${coefficient}\:\mu\:\left(\mathrm{tan}\:\theta<\mu\leqslant\mu_{{max}} \right). \\ $$$${generally}\:{this}\:{relationship}\:{looks}\:{like} \\ $$$${as}\:{shown}\:{in}\:{the}\:{diagram}\:{below}. \\ $$$$ \\ $$$${example}: \\ $$$${h}=\mathrm{3}{m},\:{L}=\mathrm{5}{m},\:\theta=\mathrm{30}°,\:\mu_{{max}} =\mathrm{1}.\mathrm{5} \\ $$$$\Rightarrow\frac{{b}_{{min}} }{{L}}=\mathrm{0}.\mathrm{33202}\:{at}\:\mu=\mathrm{1}.\mathrm{5} \\ $$$$\Rightarrow\frac{{b}_{{max}} }{{L}}=\mathrm{0}.\mathrm{43165}\:{at}\:\mu=\mathrm{0}.\mathrm{7} \\ $$
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
point 1:   this is the case 1 with  μ=tan θ and (b/L)=((L−h)/L)=1−(h/L)=0.4  point 3:   this is the situation when the friction  reaches its maximum and the part of  rope which lies on the rope is the  minimum.  point 2:  this is the situation when the part of  rope on the slope has its maximum  length. we can not put more rope on  the slope and still keep it in equilibrium.
$${point}\:\mathrm{1}:\: \\ $$$${this}\:{is}\:{the}\:{case}\:\mathrm{1}\:{with} \\ $$$$\mu=\mathrm{tan}\:\theta\:{and}\:\frac{{b}}{{L}}=\frac{{L}−{h}}{{L}}=\mathrm{1}−\frac{{h}}{{L}}=\mathrm{0}.\mathrm{4} \\ $$$${point}\:\mathrm{3}:\: \\ $$$${this}\:{is}\:{the}\:{situation}\:{when}\:{the}\:{friction} \\ $$$${reaches}\:{its}\:{maximum}\:{and}\:{the}\:{part}\:{of} \\ $$$${rope}\:{which}\:{lies}\:{on}\:{the}\:{rope}\:{is}\:{the} \\ $$$${minimum}. \\ $$$${point}\:\mathrm{2}: \\ $$$${this}\:{is}\:{the}\:{situation}\:{when}\:{the}\:{part}\:{of} \\ $$$${rope}\:{on}\:{the}\:{slope}\:{has}\:{its}\:{maximum} \\ $$$${length}.\:{we}\:{can}\:{not}\:{put}\:{more}\:{rope}\:{on} \\ $$$${the}\:{slope}\:{and}\:{still}\:{keep}\:{it}\:{in}\:{equilibrium}. \\ $$
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
case 3: μ<tan θ  the friction force is too small so that  the hanging part of the rope must   hold the the part lying on the slope.
$${case}\:\mathrm{3}:\:\mu<\mathrm{tan}\:\theta \\ $$$${the}\:{friction}\:{force}\:{is}\:{too}\:{small}\:{so}\:{that} \\ $$$${the}\:{hanging}\:{part}\:{of}\:{the}\:{rope}\:{must}\: \\ $$$${hold}\:{the}\:{the}\:{part}\:{lying}\:{on}\:{the}\:{slope}. \\ $$
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
Commented by mr W last updated on 29/May/22
say the length of the rope part which  lies on the inclined plane is b.  the friction force between rope and  the slope is f.  f=μρgb cos θ with 0≤μ<tan θ  tension in the rope at point B:  T_1 =ρgb sin θ−f=(sin θ−μ cos θ)ρgb  let μ′=μ cos θ−sin θ  T_1 =−μ′ρgb  T_0 =T_1  cos θ=−μ′ρgb cos θ  define a=(T_0 /(ρg))=−μ′b cos θ  the hanging rope from point A to B   is part of a catenary. w.r.t. the  coordinate system as shown,  y=a cosh (x/a)  y′=sinh (x/a)  s=a sinh (x/a)    tan θ=sinh (x_B /a) ⇒(x_B /a)=sinh^(−1)  (tan θ)  s_1 =CB^(⌢) =a sinh (x_B /a)=a tan θ  AC^(⌢) =L−b+s_1 =a sinh (x_A /a)  L−b+a tan θ=a sinh (x_A /a)  ⇒(x_A /a)=sinh^(−1)  (((L−b)/a)+tan θ)  y_A −y_B =h+(x_A −x_B ) tan θ  a cosh (x_A /a)−a cosh (x_B /a)=h+(x_A −x_B ) tan θ  cosh (x_A /a)−cosh (x_B /a)=(h/a)+((x_A /a)−(x_B /a)) tan θ  cosh [sinh^(−1)  (((L−b)/a)+tan θ)]−cosh [sinh^(−1)  (tan θ)]=(h/a)+[sinh^(−1)  (((L−b)/a)+tan θ)−sinh^(−1)  (tan θ)] tan θ  let (b/L)=(1/λ)  cosh [sinh^(−1)  (−((λ−1)/(μ′ cos θ))+tan θ)]−cosh [sinh^(−1)  (tan θ)]=−((λh)/(μ′L cos θ))+[sinh^(−1)  (−((λ−1)/(μ′ cos θ))+tan θ)−sinh^(−1)  (tan θ)] tan θ  ⇒[sinh^(−1)  (((𝛌−1)/(𝛍′ cos 𝛉))−tan 𝛉)+sinh^(−1)  (tan 𝛉)]sin 𝛉=1− ((𝛌h)/(𝛍′))−(√(1+(((𝛌−1)/(𝛍′)))^2 −2(((𝛌−1)/(𝛍′)))sin 𝛉))  this equation shows the relationship  between λ and the actual friction  coefficient μ (0≤μ<tan θ).  generally this relationship looks like  as shown in the diagram below.    example:  h=3m, L=5m, θ=30°, μ_(max) =1.5  ⇒(b_(min) /L)=0.22196 at μ=0 (point 4)  ⇒(b_(max) /L)=0.43165 at μ=0.7 (point 2)
$${say}\:{the}\:{length}\:{of}\:{the}\:{rope}\:{part}\:{which} \\ $$$${lies}\:{on}\:{the}\:{inclined}\:{plane}\:{is}\:{b}. \\ $$$${the}\:{friction}\:{force}\:{between}\:{rope}\:{and} \\ $$$${the}\:{slope}\:{is}\:{f}. \\ $$$${f}=\mu\rho{gb}\:\mathrm{cos}\:\theta\:{with}\:\mathrm{0}\leqslant\mu<\mathrm{tan}\:\theta \\ $$$${tension}\:{in}\:{the}\:{rope}\:{at}\:{point}\:{B}: \\ $$$${T}_{\mathrm{1}} =\rho{gb}\:\mathrm{sin}\:\theta−{f}=\left(\mathrm{sin}\:\theta−\mu\:\mathrm{cos}\:\theta\right)\rho{gb} \\ $$$${let}\:\mu'=\mu\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta \\ $$$${T}_{\mathrm{1}} =−\mu'\rho{gb} \\ $$$${T}_{\mathrm{0}} ={T}_{\mathrm{1}} \:\mathrm{cos}\:\theta=−\mu'\rho{gb}\:\mathrm{cos}\:\theta \\ $$$${define}\:{a}=\frac{{T}_{\mathrm{0}} }{\rho{g}}=−\mu'{b}\:\mathrm{cos}\:\theta \\ $$$${the}\:{hanging}\:{rope}\:{from}\:{point}\:{A}\:{to}\:{B}\: \\ $$$${is}\:{part}\:{of}\:{a}\:{catenary}.\:{w}.{r}.{t}.\:{the} \\ $$$${coordinate}\:{system}\:{as}\:{shown}, \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${y}'=\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$${s}={a}\:\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$$ \\ $$$$\mathrm{tan}\:\theta=\mathrm{sinh}\:\frac{{x}_{{B}} }{{a}}\:\Rightarrow\frac{{x}_{{B}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right) \\ $$$${s}_{\mathrm{1}} =\overset{\frown} {{CB}}={a}\:\mathrm{sinh}\:\frac{{x}_{{B}} }{{a}}={a}\:\mathrm{tan}\:\theta \\ $$$$\overset{\frown} {{AC}}={L}−{b}+{s}_{\mathrm{1}} ={a}\:\mathrm{sinh}\:\frac{{x}_{{A}} }{{a}} \\ $$$${L}−{b}+{a}\:\mathrm{tan}\:\theta={a}\:\mathrm{sinh}\:\frac{{x}_{{A}} }{{a}} \\ $$$$\Rightarrow\frac{{x}_{{A}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{{L}−{b}}{{a}}+\mathrm{tan}\:\theta\right) \\ $$$${y}_{{A}} −{y}_{{B}} ={h}+\left({x}_{{A}} −{x}_{{B}} \right)\:\mathrm{tan}\:\theta \\ $$$${a}\:\mathrm{cosh}\:\frac{{x}_{{A}} }{{a}}−{a}\:\mathrm{cosh}\:\frac{{x}_{{B}} }{{a}}={h}+\left({x}_{{A}} −{x}_{{B}} \right)\:\mathrm{tan}\:\theta \\ $$$$\mathrm{cosh}\:\frac{{x}_{{A}} }{{a}}−\mathrm{cosh}\:\frac{{x}_{{B}} }{{a}}=\frac{{h}}{{a}}+\left(\frac{{x}_{{A}} }{{a}}−\frac{{x}_{{B}} }{{a}}\right)\:\mathrm{tan}\:\theta \\ $$$$\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{{L}−{b}}{{a}}+\mathrm{tan}\:\theta\right)\right]−\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)\right]=\frac{{h}}{{a}}+\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\frac{{L}−{b}}{{a}}+\mathrm{tan}\:\theta\right)−\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)\right]\:\mathrm{tan}\:\theta \\ $$$${let}\:\frac{{b}}{{L}}=\frac{\mathrm{1}}{\lambda} \\ $$$$\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(−\frac{\lambda−\mathrm{1}}{\mu'\:\mathrm{cos}\:\theta}+\mathrm{tan}\:\theta\right)\right]−\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)\right]=−\frac{\lambda{h}}{\mu'{L}\:\mathrm{cos}\:\theta}+\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(−\frac{\lambda−\mathrm{1}}{\mu'\:\mathrm{cos}\:\theta}+\mathrm{tan}\:\theta\right)−\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta\right)\right]\:\mathrm{tan}\:\theta \\ $$$$\Rightarrow\left[\boldsymbol{\mathrm{sinh}}^{−\mathrm{1}} \:\left(\frac{\boldsymbol{\lambda}−\mathrm{1}}{\boldsymbol{\mu}'\:\boldsymbol{\mathrm{cos}}\:\boldsymbol{\theta}}−\boldsymbol{\mathrm{tan}}\:\boldsymbol{\theta}\right)+\boldsymbol{\mathrm{sinh}}^{−\mathrm{1}} \:\left(\boldsymbol{\mathrm{tan}}\:\boldsymbol{\theta}\right)\right]\boldsymbol{\mathrm{sin}}\:\boldsymbol{\theta}=\mathrm{1}−\:\frac{\boldsymbol{\lambda{h}}}{\boldsymbol{\mu}'}−\sqrt{\mathrm{1}+\left(\frac{\boldsymbol{\lambda}−\mathrm{1}}{\boldsymbol{\mu}'}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\boldsymbol{\lambda}−\mathrm{1}}{\boldsymbol{\mu}'}\right)\boldsymbol{\mathrm{sin}}\:\boldsymbol{\theta}} \\ $$$${this}\:{equation}\:{shows}\:{the}\:{relationship} \\ $$$${between}\:\lambda\:{and}\:{the}\:{actual}\:{friction} \\ $$$${coefficient}\:\mu\:\left(\mathrm{0}\leqslant\mu<\mathrm{tan}\:\theta\right). \\ $$$${generally}\:{this}\:{relationship}\:{looks}\:{like} \\ $$$${as}\:{shown}\:{in}\:{the}\:{diagram}\:{below}. \\ $$$$ \\ $$$${example}: \\ $$$${h}=\mathrm{3}{m},\:{L}=\mathrm{5}{m},\:\theta=\mathrm{30}°,\:\mu_{{max}} =\mathrm{1}.\mathrm{5} \\ $$$$\Rightarrow\frac{{b}_{{min}} }{{L}}=\mathrm{0}.\mathrm{22196}\:{at}\:\mu=\mathrm{0}\:\left({point}\:\mathrm{4}\right) \\ $$$$\Rightarrow\frac{{b}_{{max}} }{{L}}=\mathrm{0}.\mathrm{43165}\:{at}\:\mu=\mathrm{0}.\mathrm{7}\:\left({point}\:\mathrm{2}\right) \\ $$
Commented by mr W last updated on 30/May/22
Commented by Tawa11 last updated on 06/Oct/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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