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Question-175949




Question Number 175949 by mr W last updated on 10/Sep/22
Commented by mr W last updated on 10/Sep/22
a mass m is fixed at the midpoint  of an uniform thin rope with unit   mass ρ. find the minimum length of  the rope such that it can be kept in   equilibrium as shown.   (no friction)  as example: m=ρd
$${a}\:{mass}\:{m}\:{is}\:{fixed}\:{at}\:{the}\:{midpoint} \\ $$$${of}\:{an}\:{uniform}\:{thin}\:{rope}\:{with}\:{unit}\: \\ $$$${mass}\:\rho.\:{find}\:{the}\:{minimum}\:{length}\:{of} \\ $$$${the}\:{rope}\:{such}\:{that}\:{it}\:{can}\:{be}\:{kept}\:{in}\: \\ $$$${equilibrium}\:{as}\:{shown}.\: \\ $$$$\left({no}\:{friction}\right) \\ $$$${as}\:{example}:\:{m}=\rho{d} \\ $$
Answered by mr W last updated on 14/Sep/22
Commented by mr W last updated on 11/Sep/22
Commented by mr W last updated on 13/Sep/22
let L=total length of the rope  say length of rope part BC is h,  then the length of part AB=(L/2)−h  AB is a segment of a catenary.  T_2 =hρg  T_1 sin θ_1 =((mg)/2)  T_0 =T_2 cos θ_2 =T_1 cos θ_1     with a=(T_0 /(ρg)) the eqn. of the catenary is:  y=a cosh (x/a)  tan θ=y′=sinh (x/a)  s=a sinh (x/a)    at point A:  tan θ_1 =sinh (d_1 /a)  s_1 =a sinh (d_1 /a)    at point B:  tan θ_2 =sinh ((d_1 +(d/2))/a)  s_2 =a sinh ((d_1 +(d/2))/a)    T_0 =T_1 cos θ_1   aρg=((mg)/(2 tan θ_1 ))  tan θ_1 =(m/(2ρa))  sinh (d_1 /a)=(m/(2ρa))  ⇒(d_1 /a)=sinh^(−1)  (m/(2ρa))    T_0 =T_2 cos θ_2   aρg=hρg cos θ_2   cos θ_2 =(a/h) ⇒tan θ_2 =((√(1−((a/h))^2 ))/(a/h))  tan θ_2 =sinh ((d_1 +(d/2))/a)=((√(1−((a/h))^2 ))/(a/h))  ⇒((d_1 +(d/2))/a)=sinh^(−1)  ((√(1−((a/h))^2 ))/(a/h))  ⇒(d/(2a))+sinh^(−1)  (m/(2ρa))=sinh^(−1)  ((√(1−((a/h))^2 ))/(a/h))  let k=(m/(ρd)), η=(a/h), ξ=(d/(2a))  ⇒ξ+sinh^(−1)  (kξ)=sinh^(−1)  ((√(1−η^2 ))/η)  ⇒((√(1−η^2 ))/η)=sinh [ξ+sinh^(−1)  (kξ)]  ⇒η=(1/( (√(1+sinh^2  [ξ+sinh^(−1)  (kξ)]))))    s_2 −s_1 =(L/2)−h  a(sinh ((d_1 +(d/2))/a)−sinh (d_1 /a))=(L/2)−h  a(((√(1−((a/h))^2 ))/(a/h))−(m/(2ρa)))=(L/2)−h  L+(m/ρ)=2h[1+(√(1−((a/h))^2 ))]  Φ=(L/d)+k=((1+(√(1−η^2 )))/(ηξ))  Φ=((sinh [ξ+sinh^(−1)  (kξ)]+(√(1+sinh^2  [ξ+sinh^(−1)  (kξ)])))/ξ)  minimum L ⇔  (dΦ/dξ)=0    λ=(h/d)=(1/(2ηξ))=((√(1+sinh^2  [ξ+sinh^(−1)  (kξ)]))/(2ξ))    examples:  m=0 ⇒k=0:  Φ_(min) =e ⇒L_(min) =ed    m=ρd ⇒k=1:  Φ_(min) =5.3268 ⇒L_(min) =4.3268d    m=2ρd ⇒k=2:  Φ_(min) =7.7568 ⇒L_(min) =5.7568d
$${let}\:{L}={total}\:{length}\:{of}\:{the}\:{rope} \\ $$$${say}\:{length}\:{of}\:{rope}\:{part}\:{BC}\:{is}\:{h}, \\ $$$${then}\:{the}\:{length}\:{of}\:{part}\:{AB}=\frac{{L}}{\mathrm{2}}−{h} \\ $$$${AB}\:{is}\:{a}\:{segment}\:{of}\:{a}\:{catenary}. \\ $$$${T}_{\mathrm{2}} ={h}\rho{g} \\ $$$${T}_{\mathrm{1}} \mathrm{sin}\:\theta_{\mathrm{1}} =\frac{{mg}}{\mathrm{2}} \\ $$$${T}_{\mathrm{0}} ={T}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{2}} ={T}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} \\ $$$$ \\ $$$${with}\:{a}=\frac{{T}_{\mathrm{0}} }{\rho{g}}\:{the}\:{eqn}.\:{of}\:{the}\:{catenary}\:{is}: \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$$\mathrm{tan}\:\theta={y}'=\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$${s}={a}\:\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$$ \\ $$$${at}\:{point}\:{A}: \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\mathrm{sinh}\:\frac{{d}_{\mathrm{1}} }{{a}} \\ $$$${s}_{\mathrm{1}} ={a}\:\mathrm{sinh}\:\frac{{d}_{\mathrm{1}} }{{a}} \\ $$$$ \\ $$$${at}\:{point}\:{B}: \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\mathrm{sinh}\:\frac{{d}_{\mathrm{1}} +\frac{{d}}{\mathrm{2}}}{{a}} \\ $$$${s}_{\mathrm{2}} ={a}\:\mathrm{sinh}\:\frac{{d}_{\mathrm{1}} +\frac{{d}}{\mathrm{2}}}{{a}} \\ $$$$ \\ $$$${T}_{\mathrm{0}} ={T}_{\mathrm{1}} \mathrm{cos}\:\theta_{\mathrm{1}} \\ $$$${a}\rho{g}=\frac{{mg}}{\mathrm{2}\:\mathrm{tan}\:\theta_{\mathrm{1}} } \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{{m}}{\mathrm{2}\rho{a}} \\ $$$$\mathrm{sinh}\:\frac{{d}_{\mathrm{1}} }{{a}}=\frac{{m}}{\mathrm{2}\rho{a}} \\ $$$$\Rightarrow\frac{{d}_{\mathrm{1}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\frac{{m}}{\mathrm{2}\rho{a}} \\ $$$$ \\ $$$${T}_{\mathrm{0}} ={T}_{\mathrm{2}} \mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$${a}\rho{g}={h}\rho{g}\:\mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$$\mathrm{cos}\:\theta_{\mathrm{2}} =\frac{{a}}{{h}}\:\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\sqrt{\mathrm{1}−\left(\frac{{a}}{{h}}\right)^{\mathrm{2}} }}{\frac{{a}}{{h}}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\mathrm{sinh}\:\frac{{d}_{\mathrm{1}} +\frac{{d}}{\mathrm{2}}}{{a}}=\frac{\sqrt{\mathrm{1}−\left(\frac{{a}}{{h}}\right)^{\mathrm{2}} }}{\frac{{a}}{{h}}} \\ $$$$\Rightarrow\frac{{d}_{\mathrm{1}} +\frac{{d}}{\mathrm{2}}}{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{1}−\left(\frac{{a}}{{h}}\right)^{\mathrm{2}} }}{\frac{{a}}{{h}}} \\ $$$$\Rightarrow\frac{{d}}{\mathrm{2}{a}}+\mathrm{sinh}^{−\mathrm{1}} \:\frac{{m}}{\mathrm{2}\rho{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{1}−\left(\frac{{a}}{{h}}\right)^{\mathrm{2}} }}{\frac{{a}}{{h}}} \\ $$$${let}\:{k}=\frac{{m}}{\rho{d}},\:\eta=\frac{{a}}{{h}},\:\xi=\frac{{d}}{\mathrm{2}{a}} \\ $$$$\Rightarrow\xi+\mathrm{sinh}^{−\mathrm{1}} \:\left({k}\xi\right)=\mathrm{sinh}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{1}−\eta^{\mathrm{2}} }}{\eta} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{1}−\eta^{\mathrm{2}} }}{\eta}=\mathrm{sinh}\:\left[\xi+\mathrm{sinh}^{−\mathrm{1}} \:\left({k}\xi\right)\right] \\ $$$$\Rightarrow\eta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{sinh}^{\mathrm{2}} \:\left[\xi+\mathrm{sinh}^{−\mathrm{1}} \:\left({k}\xi\right)\right]}} \\ $$$$ \\ $$$${s}_{\mathrm{2}} −{s}_{\mathrm{1}} =\frac{{L}}{\mathrm{2}}−{h} \\ $$$${a}\left(\mathrm{sinh}\:\frac{{d}_{\mathrm{1}} +\frac{{d}}{\mathrm{2}}}{{a}}−\mathrm{sinh}\:\frac{{d}_{\mathrm{1}} }{{a}}\right)=\frac{{L}}{\mathrm{2}}−{h} \\ $$$${a}\left(\frac{\sqrt{\mathrm{1}−\left(\frac{{a}}{{h}}\right)^{\mathrm{2}} }}{\frac{{a}}{{h}}}−\frac{{m}}{\mathrm{2}\rho{a}}\right)=\frac{{L}}{\mathrm{2}}−{h} \\ $$$${L}+\frac{{m}}{\rho}=\mathrm{2}{h}\left[\mathrm{1}+\sqrt{\mathrm{1}−\left(\frac{{a}}{{h}}\right)^{\mathrm{2}} }\right] \\ $$$$\Phi=\frac{{L}}{{d}}+{k}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−\eta^{\mathrm{2}} }}{\eta\xi} \\ $$$$\Phi=\frac{\mathrm{sinh}\:\left[\xi+\mathrm{sinh}^{−\mathrm{1}} \:\left({k}\xi\right)\right]+\sqrt{\mathrm{1}+\mathrm{sinh}^{\mathrm{2}} \:\left[\xi+\mathrm{sinh}^{−\mathrm{1}} \:\left({k}\xi\right)\right]}}{\xi} \\ $$$${minimum}\:{L}\:\Leftrightarrow\:\:\frac{{d}\Phi}{{d}\xi}=\mathrm{0} \\ $$$$ \\ $$$$\lambda=\frac{{h}}{{d}}=\frac{\mathrm{1}}{\mathrm{2}\eta\xi}=\frac{\sqrt{\mathrm{1}+\mathrm{sinh}^{\mathrm{2}} \:\left[\xi+\mathrm{sinh}^{−\mathrm{1}} \:\left({k}\xi\right)\right]}}{\mathrm{2}\xi} \\ $$$$ \\ $$$$\underline{{examples}:} \\ $$$${m}=\mathrm{0}\:\Rightarrow{k}=\mathrm{0}: \\ $$$$\Phi_{{min}} ={e}\:\Rightarrow{L}_{{min}} ={ed} \\ $$$$ \\ $$$${m}=\rho{d}\:\Rightarrow{k}=\mathrm{1}: \\ $$$$\Phi_{{min}} =\mathrm{5}.\mathrm{3268}\:\Rightarrow{L}_{{min}} =\mathrm{4}.\mathrm{3268}{d} \\ $$$$ \\ $$$${m}=\mathrm{2}\rho{d}\:\Rightarrow{k}=\mathrm{2}: \\ $$$$\Phi_{{min}} =\mathrm{7}.\mathrm{7568}\:\Rightarrow{L}_{{min}} =\mathrm{5}.\mathrm{7568}{d} \\ $$
Commented by mr W last updated on 11/Sep/22
Commented by mr W last updated on 11/Sep/22
Commented by Tawa11 last updated on 15/Sep/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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