Question Number 46344 by MrW3 last updated on 24/Oct/18
Commented by MrW3 last updated on 25/Oct/18
$${A}\:{rod}\:{of}\:{mass}\:{M}\:{and}\:{length}\:{l}\:{is}\:{supported} \\ $$$${at}\:{one}\:{end}\:{through}\:{a}\:{hinge}.\:{the}\:{other}\:{free} \\ $$$${end}\:{is}\:{hold}\:{horizontally}\:{by}\:{a}\:{rope} \\ $$$${as}\:{shown}.\:{the}\:{rope}\:{has}\:{a}\:{mass}\:\rho\:{per} \\ $$$${unit}\:{length}. \\ $$$${Find}\:{the}\:{max}.\:{tension}\:{in}\:{the}\:{rope}. \\ $$
Commented by ajfour last updated on 24/Oct/18
Commented by ajfour last updated on 24/Oct/18
![Tdθ = ρg((dy/(sin θ)))cos θ ....(i) Tcos θ = (T+dT)cos (θ+dθ) ...(ii) From eq.(ii) Tcos θ = T_0 cos 𝛃 = T_e cos 𝛂 = c ________________________ ⇒ cos α = (c/T_e ) , cos β = (c/T_0 ) ..(I) ________________________ using (ii) in (i) ∫_0 ^( h) ρgdy = c∫_α ^( β) sec θ tan θ dθ ⇒ c(sec β−sec α)= ρgh ________________________ ⇒ T_0 −T_e =𝛒gh ...... (II) ________________________ Torque about left end only on bridge (excluding rope): T_e Lsin 𝛂 = ((MgL)/2) ⇒ (√(T_e ^( 2) −c^2 )) = ((Mg)/2) ⇒ T_e = (√(c^2 +(((Mg)/2))^2 )) ....(A) And using (A) in (II) T_0 =𝛒gh+(√(c^2 +(((Mg)/2))^2 )) ...(B) Torque about left end of bridge, on bridge+rope : _____________________ T_0 hcos β = ∫_0 ^( l) (ρgdl)x+((MgL)/2) ..(iii) _____________________ but from (i) ρgdx = Tdθ = csec θdθ ⇒ ∫_0 ^( x) ρgdx = c∫_β ^( θ) sec θdθ ⇒ ρgx = cln (((sec θ+tan θ)/(sec β+tan β))) ..(iii) from (iii) ⇒ ch = ∫_β ^( α) (Tsec θdθ)x+((MgL)/2) using (iii) ch = c^2 ∫_β ^( α) (((sec^2 θdθ)ln (..))/(ρg))+((MgL)/2) ⇒ ch−((MgL)/2) = (c^2 /(ρg))[(tan θ)ln (((sec θ+tan θ)/(sec β+tan β)))∣_β ^α −∫_β ^( α) tan θsec θdθ ] ⇒ ((ρg)/c^2 )(ch−((MgL)/2))= (tan α)ln (((sec α+tan α)/(sec β+tan β))) + sec β−sec α but c(sec β−sec α)= ρgh tan α = (√(sec^2 α−1)) =(√((T_e ^( 2) /c^2 )−1)) = ((Mg)/(2c)) ; hence ⇒ −((ρg)/(2c^2 ))(MgL)=(((Mg)/(2c)))ln [((cos β)/(cos α))×((1+sin α)/(1+sin β))] ⇒ ln [((T_e +T_e sin α)/(T_0 +T_0 sin β))]= −((ρgL)/c) ⇒ ln [((T_0 +(√(T_0 ^( 2) −c^2 )))/(T_e +(√(T_e ^( 2) −c^2 )))) ]= ((ρgL)/c) ________________________ Answer : ln(((T_0 +(√(T_0 ^( 2) −c^2 )))/(T_e +((Mg)/2)))) = ((𝛒gL)/c) −−−−−−−−−−−−−−− where [see (A) & (B)] T_e = (√(c^2 +(((Mg)/2))^2 )) T_0 = 𝛒gh+(√(c^2 +(((Mg)/2))^2 )) ________________________.](https://www.tinkutara.com/question/Q46357.png)
$$\:\:{Td}\theta\:=\:\rho{g}\left(\frac{{dy}}{\mathrm{sin}\:\theta}\right)\mathrm{cos}\:\theta\:\:\:\:\:\:….\left({i}\right) \\ $$$$\:\:{T}\mathrm{cos}\:\theta\:=\:\left({T}+{dT}\right)\mathrm{cos}\:\left(\theta+{d}\theta\right)\:\:…\left({ii}\right) \\ $$$$\:\:{From}\:\:{eq}.\left({ii}\right) \\ $$$$\:\:\:{T}\mathrm{cos}\:\theta\:=\:\boldsymbol{{T}}_{\mathrm{0}} \mathrm{cos}\:\boldsymbol{\beta}\:=\:\boldsymbol{{T}}_{\boldsymbol{{e}}} \mathrm{cos}\:\boldsymbol{\alpha}\:=\:\boldsymbol{{c}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\mathrm{cos}\:\alpha\:=\:\frac{{c}}{{T}_{{e}} }\:\:,\:\:\mathrm{cos}\:\beta\:=\:\frac{{c}}{{T}_{\mathrm{0}} }\:\:\:..\left(\boldsymbol{{I}}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${using}\:\left({ii}\right)\:{in}\:\:\left({i}\right) \\ $$$$\:\:\int_{\mathrm{0}} ^{\:\:{h}} \rho{gdy}\:=\:{c}\int_{\alpha} ^{\:\:\beta} \mathrm{sec}\:\theta\:\mathrm{tan}\:\theta\:{d}\theta\: \\ $$$$\Rightarrow\:\:{c}\left(\mathrm{sec}\:\beta−\mathrm{sec}\:\alpha\right)=\:\rho{gh}\:\: \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\Rightarrow\:\:\:\boldsymbol{{T}}_{\mathrm{0}} −\boldsymbol{{T}}_{\boldsymbol{{e}}} =\boldsymbol{\rho{gh}}\:\:\:\:……\:\left(\boldsymbol{{II}}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:{Torque}\:{about}\:{left}\:{end}\:{only}\:{on} \\ $$$$\:\:\:{bridge}\:\left({excluding}\:{rope}\right): \\ $$$$\:\:\:\:\boldsymbol{{T}}_{\boldsymbol{{e}}} \boldsymbol{{L}}\mathrm{sin}\:\boldsymbol{\alpha}\:=\:\frac{\boldsymbol{{MgL}}}{\mathrm{2}}\:\:\:\: \\ $$$$\Rightarrow\:\sqrt{{T}_{{e}} ^{\:\:\mathrm{2}} −{c}^{\mathrm{2}} }\:=\:\frac{{Mg}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\boldsymbol{{T}}_{\boldsymbol{{e}}} =\:\sqrt{\boldsymbol{{c}}^{\mathrm{2}} +\left(\frac{\boldsymbol{{Mg}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:\:\:\:….\left({A}\right) \\ $$$${And}\:{using}\:\left({A}\right)\:{in}\:\left({II}\right)\: \\ $$$$\:\:\boldsymbol{{T}}_{\mathrm{0}} \:=\boldsymbol{\rho{gh}}+\sqrt{\boldsymbol{{c}}^{\mathrm{2}} +\left(\frac{\boldsymbol{{Mg}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:…\left({B}\right) \\ $$$${Torque}\:{about}\:{left}\:{end}\:{of}\:{bridge}, \\ $$$$\:\:\:\:{on}\:{bridge}+{rope}\:: \\ $$$$\:\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:{T}_{\mathrm{0}} {h}\mathrm{cos}\:\beta\:=\:\int_{\mathrm{0}} ^{\:\:{l}} \left(\rho{gdl}\right){x}+\frac{{MgL}}{\mathrm{2}}\:\:..\left({iii}\right) \\ $$$$\:\:\:\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\:{but}\:\:{from}\:\left({i}\right) \\ $$$$\:\:\:\:\:\:\rho{gdx}\:=\:{Td}\theta\:=\:{c}\mathrm{sec}\:\theta{d}\theta \\ $$$$\Rightarrow\:\:\int_{\mathrm{0}} ^{\:\:{x}} \rho{gdx}\:=\:{c}\int_{\beta} ^{\:\:\theta} \mathrm{sec}\:\theta{d}\theta \\ $$$$\Rightarrow\:\:\:\rho{gx}\:=\:{c}\mathrm{ln}\:\left(\frac{\mathrm{sec}\:\theta+\mathrm{tan}\:\theta}{\mathrm{sec}\:\beta+\mathrm{tan}\:\beta}\right)\:..\left({iii}\right) \\ $$$${from}\:\left({iii}\right) \\ $$$$\Rightarrow\:\:\boldsymbol{{ch}}\:=\:\int_{\beta} ^{\:\:\alpha} \left({T}\mathrm{sec}\:\theta{d}\theta\right){x}+\frac{{MgL}}{\mathrm{2}} \\ $$$${using}\:\:\left({iii}\right) \\ $$$$\:\:\:\boldsymbol{{ch}}\:=\:\boldsymbol{{c}}^{\mathrm{2}} \int_{\beta} ^{\:\:\alpha} \frac{\left(\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta\right)\mathrm{ln}\:\left(..\right)}{\rho{g}}+\frac{{MgL}}{\mathrm{2}} \\ $$$$\Rightarrow\:{ch}−\frac{{MgL}}{\mathrm{2}}\:=\:\frac{{c}^{\mathrm{2}} }{\rho{g}}\left[\left(\mathrm{tan}\:\theta\right)\mathrm{ln}\:\left(\frac{\mathrm{sec}\:\theta+\mathrm{tan}\:\theta}{\mathrm{sec}\:\beta+\mathrm{tan}\:\beta}\right)\mid_{\beta} ^{\alpha} \right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:−\int_{\beta} ^{\:\:\alpha} \mathrm{tan}\:\theta\mathrm{sec}\:\theta{d}\theta\:\right] \\ $$$$\Rightarrow\:\:\frac{\rho{g}}{{c}^{\mathrm{2}} }\left({ch}−\frac{{MgL}}{\mathrm{2}}\right)=\:\left(\mathrm{tan}\:\alpha\right)\mathrm{ln}\:\left(\frac{\mathrm{sec}\:\alpha+\mathrm{tan}\:\alpha}{\mathrm{sec}\:\beta+\mathrm{tan}\:\beta}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\mathrm{sec}\:\beta−\mathrm{sec}\:\alpha\:\:\: \\ $$$${but}\:\: \\ $$$$\:\:{c}\left(\mathrm{sec}\:\beta−\mathrm{sec}\:\alpha\right)=\:\rho{gh}\:\: \\ $$$$\:\:\mathrm{tan}\:\alpha\:=\:\sqrt{\mathrm{sec}\:^{\mathrm{2}} \alpha−\mathrm{1}}\:=\sqrt{\frac{{T}_{{e}} ^{\:\:\mathrm{2}} }{{c}^{\mathrm{2}} }−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{Mg}}{\mathrm{2}{c}}\:\:\:;\:\:{hence} \\ $$$$\Rightarrow\:−\frac{\rho{g}}{\mathrm{2}{c}^{\mathrm{2}} }\left({MgL}\right)=\left(\frac{{Mg}}{\mathrm{2}{c}}\right)\mathrm{ln}\:\left[\frac{\mathrm{cos}\:\beta}{\mathrm{cos}\:\alpha}×\frac{\mathrm{1}+\mathrm{sin}\:\alpha}{\mathrm{1}+\mathrm{sin}\:\beta}\right] \\ $$$$\Rightarrow\:\mathrm{ln}\:\left[\frac{{T}_{{e}} +{T}_{{e}} \mathrm{sin}\:\alpha}{{T}_{\mathrm{0}} +{T}_{\mathrm{0}} \mathrm{sin}\:\beta}\right]=\:−\frac{\rho{gL}}{{c}} \\ $$$$\Rightarrow\:\:\mathrm{ln}\:\left[\frac{{T}_{\mathrm{0}} +\sqrt{{T}_{\mathrm{0}} ^{\:\:\mathrm{2}} −{c}^{\mathrm{2}} }}{{T}_{{e}} +\sqrt{{T}_{{e}} ^{\:\:\mathrm{2}} −{c}^{\mathrm{2}} }}\:\right]=\:\frac{\rho{gL}}{{c}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${Answer}\:: \\ $$$$\:\mathrm{ln}\left(\frac{\boldsymbol{{T}}_{\mathrm{0}} +\sqrt{\boldsymbol{{T}}_{\mathrm{0}} ^{\:\:\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} }}{\boldsymbol{{T}}_{{e}} +\frac{\boldsymbol{{Mg}}}{\mathrm{2}}}\right)\:=\:\frac{\boldsymbol{\rho{gL}}}{\boldsymbol{{c}}} \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\:\:\:{where}\:\:\left[{see}\:\left({A}\right)\:\&\:\left({B}\right)\right] \\ $$$$\:\:\boldsymbol{{T}}_{\boldsymbol{{e}}} =\:\sqrt{\boldsymbol{{c}}^{\mathrm{2}} +\left(\frac{\boldsymbol{{Mg}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\:\:\:\: \\ $$$$\:\:\:\boldsymbol{{T}}_{\mathrm{0}} =\:\boldsymbol{\rho{gh}}+\sqrt{\boldsymbol{{c}}^{\mathrm{2}} +\left(\frac{\boldsymbol{{Mg}}}{\mathrm{2}}\right)^{\mathrm{2}} }\: \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. \\ $$
Commented by ajfour last updated on 24/Oct/18
![T_0 = ρgh+(√(c^2 +(((Mg)/2))^2 )) And for c ρgh+(√(c^2 +(((Mg)/2))^2 ))+ (√((ρgh+(√(c^2 +(((Mg)/2))^2 )) )^2 −c^2 )) = [(√(c^2 +(((Mg)/2))^2 ))+((Mg)/2)]e^((ρgL)/c) If further i let λ=(c/((Mg/2))) ; μ=((ρgh)/((Mg/2))) ; m=(L/h) ⇒ λln (((μ+(√(λ^2 +1))+(√((μ+(√(λ^2 +1)) )^2 −λ^2 )))/( (√(λ^2 +1)))) ) = μm T_0 = (μ+(√(λ^2 +1)))((Mg)/2) for L=4 , h=3, (ρ_B /ρ) = 50 (ρ_B /ρ) = ((Mg)/(ρgL)) = 2×(((Mg/2))/(ρgh))×(h/L)=(2/(μm)) ⇒ 50=((3×2)/(4μ)) ⇒ μ=(3/(100)) =0.03 with m=(L/h)=(4/3) λ=1.3664 , T_(max) =T_0 ≈ 1.723(((Mg)/2)) .](https://www.tinkutara.com/question/Q46372.png)
$${T}_{\mathrm{0}} =\:\rho{gh}+\sqrt{{c}^{\mathrm{2}} +\left(\frac{{Mg}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${And}\:{for}\:{c} \\ $$$$\rho{gh}+\sqrt{{c}^{\mathrm{2}} +\left(\frac{{Mg}}{\mathrm{2}}\right)^{\mathrm{2}} }+ \\ $$$$\sqrt{\left(\rho{gh}+\sqrt{{c}^{\mathrm{2}} +\left(\frac{{Mg}}{\mathrm{2}}\right)^{\mathrm{2}} }\:\right)^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$$\:\:=\:\left[\sqrt{{c}^{\mathrm{2}} +\left(\frac{{Mg}}{\mathrm{2}}\right)^{\mathrm{2}} }+\frac{{Mg}}{\mathrm{2}}\right]{e}^{\frac{\rho{gL}}{{c}}} \:\: \\ $$$${If}\:{further}\:{i}\:{let} \\ $$$$\:\lambda=\frac{{c}}{\left({Mg}/\mathrm{2}\right)}\:\:;\:\:\mu=\frac{\rho{gh}}{\left({Mg}/\mathrm{2}\right)}\:;\:{m}=\frac{{L}}{{h}} \\ $$$$\Rightarrow\:\lambda\mathrm{ln}\:\left(\frac{\mu+\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}+\sqrt{\left(\mu+\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}\:\right)^{\mathrm{2}} −\lambda^{\mathrm{2}} }}{\:\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mu{m} \\ $$$$\:\:{T}_{\mathrm{0}} \:=\:\left(\mu+\sqrt{\lambda^{\mathrm{2}} +\mathrm{1}}\right)\frac{{Mg}}{\mathrm{2}}\: \\ $$$${for}\:\:{L}=\mathrm{4}\:,\:\:\:{h}=\mathrm{3},\:\:\:\frac{\rho_{{B}} }{\rho}\:=\:\mathrm{50} \\ $$$$\:\frac{\rho_{{B}} }{\rho}\:=\:\frac{{Mg}}{\rho{gL}}\:=\:\mathrm{2}×\frac{\left({Mg}/\mathrm{2}\right)}{\rho{gh}}×\frac{{h}}{{L}}=\frac{\mathrm{2}}{\mu{m}} \\ $$$$\Rightarrow\:\mathrm{50}=\frac{\mathrm{3}×\mathrm{2}}{\mathrm{4}\mu}\:\Rightarrow\:\mu=\frac{\mathrm{3}}{\mathrm{100}}\:=\mathrm{0}.\mathrm{03} \\ $$$$\:\:\:\:{with}\:{m}=\frac{{L}}{{h}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\lambda=\mathrm{1}.\mathrm{3664}\:\:,\:\:{T}_{{max}} ={T}_{\mathrm{0}} \approx\:\mathrm{1}.\mathrm{723}\left(\frac{{Mg}}{\mathrm{2}}\right)\:. \\ $$
Commented by MrW3 last updated on 07/Nov/18
$${your}\:{answer}\:{is}\:{correct}\:{sir}! \\ $$$${thank}\:{you}\:{very}\:{much}. \\ $$
Answered by MrW3 last updated on 24/Oct/18
Commented by MrW3 last updated on 25/Oct/18
![T=tension in rope T_H =horizontal component of T=const M=ρ_B l p=((2lρ)/M)=((2ρ)/ρ_M ), q=((2hρ)/M)=((2hρ)/(lρ_B )), λ=tan α at point B: T_B sin α=((Mg)/2) T_H =T_B cos α=((Mg)/(2 tan α)) Eqn. of rope (catenary) in xy−system: y=a cosh (x/a) with a=(T_H /(ρg))=(M/(2ρ tan α)) (dy/dx)=tan θ=sinh (x/a) at point B: tan α=sinh (e/a) e=a sinh^(−1) (tan α) f=a cosh (e/a)=a cosh [sinh^(−1) (tan α)] at point A: f+h=a cosh ((e+l)/a) ⇒a cosh [sinh^(−1) (tan α)]+h=a cosh [(l/a)+sinh^(−1) (tan α)] ⇒cosh [sinh^(−1) (tan α)]+(h/a)=cosh [(l/a)+sinh^(−1) (tan α)] ⇒cosh [sinh^(−1) (tan α)]+(h/a)=cosh ((l/a)) cosh [sinh^(−1) (tan α)]+sinh ((l/a)) tan α ⇒[cosh ((l/a))−1] cosh [sinh^(−1) (tan α)]+sinh ((l/a)) tan α=(h/a) ⇒[cosh (((2lρ tan α)/M))−1] (√(1+tan^2 α))+sinh (((2lρ tan α)/M)) tan α=((2hρ tan α)/M) ⇒(√(1+λ^2 ))[cosh (pλ)−1]+λ[sinh (pλ) −q]=0 ...(i) ⇒λ=f(p,q) (only numerically) max. T=T_A =(T_H /(cos θ_A ))=((Mg)/(2λ))(√(1+tan^2 θ_A )) tan θ_A =sinh ((e+l)/a)=sinh [((2lρ tan α)/M)+sinh^(−1) (tan α)] =sinh (pλ+sinh^(−1) λ)=(√(1+λ^2 )) sinh (pλ)+λ cosh (pλ) tan^2 θ_A =(1+λ^2 )sinh^2 (pλ)+λ^2 cosh^2 (pλ)+2λ(√(1+λ^2 )) sinh (pλ) cosh (pλ) =((1/2)+λ^2 )cosh (2pλ)+λ(√(1+λ^2 )) sinh (2pλ)−(1/2) 1+tan^2 θ_A =(1/2)+((1/2)+λ^2 )cosh (2pλ)+λ(√(1+λ^2 )) sinh (2pλ) ⇒max. T=((Mg)/(2λ))(√((1/2)+((1/2)+λ^2 )cosh (2pλ)+λ(√(1+λ^2 )) sinh (2pλ))) ...(ii) example: h=3m, l=4m, (ρ_B /ρ)=50 ⇒λ=tan α=0.7318 from (i) ⇒max. T=1.7233×((Mg)/2) from (ii) in case that rope is massless, we have λ=tan α=(3/4)=0.75 T=(5/3)×((Mg)/2)=1.6667×((Mg)/2)](https://www.tinkutara.com/question/Q46356.png)
$${T}={tension}\:{in}\:{rope} \\ $$$${T}_{{H}} ={horizontal}\:{component}\:{of}\:{T}={const} \\ $$$${M}=\rho_{{B}} {l} \\ $$$${p}=\frac{\mathrm{2}{l}\rho}{{M}}=\frac{\mathrm{2}\rho}{\rho_{{M}} },\:{q}=\frac{\mathrm{2}{h}\rho}{{M}}=\frac{\mathrm{2}{h}\rho}{{l}\rho_{{B}} },\:\lambda=\mathrm{tan}\:\alpha \\ $$$$ \\ $$$${at}\:{point}\:{B}: \\ $$$${T}_{{B}} \:\mathrm{sin}\:\alpha=\frac{{Mg}}{\mathrm{2}} \\ $$$${T}_{{H}} ={T}_{{B}} \:\mathrm{cos}\:\alpha=\frac{{Mg}}{\mathrm{2}\:\mathrm{tan}\:\alpha} \\ $$$${Eqn}.\:{of}\:{rope}\:\left({catenary}\right)\:{in}\:{xy}−{system}: \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${with}\:{a}=\frac{{T}_{{H}} }{\rho{g}}=\frac{{M}}{\mathrm{2}\rho\:\mathrm{tan}\:\alpha} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}=\mathrm{tan}\:\theta=\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$$ \\ $$$${at}\:{point}\:{B}: \\ $$$$\mathrm{tan}\:\alpha=\mathrm{sinh}\:\frac{{e}}{{a}} \\ $$$${e}={a}\:\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right) \\ $$$${f}={a}\:\mathrm{cosh}\:\frac{{e}}{{a}}={a}\:\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right] \\ $$$$ \\ $$$${at}\:{point}\:{A}: \\ $$$${f}+{h}={a}\:\mathrm{cosh}\:\frac{{e}+{l}}{{a}} \\ $$$$\Rightarrow{a}\:\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right]+{h}={a}\:\mathrm{cosh}\:\left[\frac{{l}}{{a}}+\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right] \\ $$$$\Rightarrow\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right]+\frac{{h}}{{a}}=\mathrm{cosh}\:\left[\frac{{l}}{{a}}+\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right] \\ $$$$\Rightarrow\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right]+\frac{{h}}{{a}}=\mathrm{cosh}\:\left(\frac{{l}}{{a}}\right)\:\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right]+\mathrm{sinh}\:\left(\frac{{l}}{{a}}\right)\:\mathrm{tan}\:\alpha \\ $$$$\Rightarrow\left[\mathrm{cosh}\:\left(\frac{{l}}{{a}}\right)−\mathrm{1}\right]\:\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right]+\mathrm{sinh}\:\left(\frac{{l}}{{a}}\right)\:\mathrm{tan}\:\alpha=\frac{{h}}{{a}} \\ $$$$\Rightarrow\left[\mathrm{cosh}\:\left(\frac{\mathrm{2}{l}\rho\:\mathrm{tan}\:\alpha}{{M}}\right)−\mathrm{1}\right]\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\alpha}+\mathrm{sinh}\:\left(\frac{\mathrm{2}{l}\rho\:\mathrm{tan}\:\alpha}{{M}}\right)\:\mathrm{tan}\:\alpha=\frac{\mathrm{2}{h}\rho\:\mathrm{tan}\:\alpha}{{M}} \\ $$$$\Rightarrow\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\left[\mathrm{cosh}\:\left({p}\lambda\right)−\mathrm{1}\right]+\lambda\left[\mathrm{sinh}\:\left({p}\lambda\right)\:−{q}\right]=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\Rightarrow\lambda={f}\left({p},{q}\right)\:\left({only}\:{numerically}\right) \\ $$$$ \\ $$$${max}.\:{T}={T}_{{A}} =\frac{{T}_{{H}} }{\mathrm{cos}\:\theta_{{A}} }=\frac{{Mg}}{\mathrm{2}\lambda}\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{{A}} } \\ $$$$\mathrm{tan}\:\theta_{{A}} =\mathrm{sinh}\:\frac{{e}+{l}}{{a}}=\mathrm{sinh}\:\left[\frac{\mathrm{2}{l}\rho\:\mathrm{tan}\:\alpha}{{M}}+\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\alpha\right)\right] \\ $$$$=\mathrm{sinh}\:\left({p}\lambda+\mathrm{sinh}^{−\mathrm{1}} \:\lambda\right)=\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sinh}\:\left({p}\lambda\right)+\lambda\:\mathrm{cosh}\:\left({p}\lambda\right) \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta_{{A}} =\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)\mathrm{sinh}^{\mathrm{2}} \:\left({p}\lambda\right)+\lambda^{\mathrm{2}} \:\mathrm{cosh}^{\mathrm{2}} \:\left({p}\lambda\right)+\mathrm{2}\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sinh}\:\left({p}\lambda\right)\:\mathrm{cosh}\:\left({p}\lambda\right) \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{2}}+\lambda^{\mathrm{2}} \right)\mathrm{cosh}\:\left(\mathrm{2}{p}\lambda\right)+\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sinh}\:\left(\mathrm{2}{p}\lambda\right)−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{{A}} =\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{2}}+\lambda^{\mathrm{2}} \right)\mathrm{cosh}\:\left(\mathrm{2}{p}\lambda\right)+\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sinh}\:\left(\mathrm{2}{p}\lambda\right) \\ $$$$ \\ $$$$\Rightarrow{max}.\:{T}=\frac{{Mg}}{\mathrm{2}\lambda}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\left(\frac{\mathrm{1}}{\mathrm{2}}+\lambda^{\mathrm{2}} \right)\mathrm{cosh}\:\left(\mathrm{2}{p}\lambda\right)+\lambda\sqrt{\mathrm{1}+\lambda^{\mathrm{2}} }\:\mathrm{sinh}\:\left(\mathrm{2}{p}\lambda\right)}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$${example}:\:{h}=\mathrm{3}{m},\:{l}=\mathrm{4}{m},\:\frac{\rho_{{B}} }{\rho}=\mathrm{50} \\ $$$$\Rightarrow\lambda=\mathrm{tan}\:\alpha=\mathrm{0}.\mathrm{7318}\:{from}\:\left({i}\right) \\ $$$$\Rightarrow{max}.\:{T}=\mathrm{1}.\mathrm{7233}×\frac{{Mg}}{\mathrm{2}}\:{from}\:\left({ii}\right) \\ $$$$ \\ $$$${in}\:{case}\:{that}\:{rope}\:{is}\:{massless},\:{we}\:{have} \\ $$$$\lambda=\mathrm{tan}\:\alpha=\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0}.\mathrm{75} \\ $$$${T}=\frac{\mathrm{5}}{\mathrm{3}}×\frac{{Mg}}{\mathrm{2}}=\mathrm{1}.\mathrm{6667}×\frac{{Mg}}{\mathrm{2}} \\ $$
Commented by ajfour last updated on 25/Oct/18
$${Thank}\:{you}\:{too},\:{Sir}. \\ $$$${Very}\:{very}\:{nice}\:{question}! \\ $$