Question Number 223928 by mr W last updated on 09/Aug/25
Commented by mr W last updated on 14/Aug/25
$${a}\:{mass}\:\boldsymbol{{m}}\:{is}\:{tied}\:{on}\:{each}\:{end}\:{of}\: \\ $$$${an}\:{uniform}\:{rope}\:{with}\:{mass}\:\boldsymbol{{M}}\: \\ $$$${and}\:{length}\:\boldsymbol{{L}}.\:{in}\:{how}\:{many}\:{ways} \\ $$$${can}\:{the}\:{rope}\:{be}\:{supported}\:{on}\:{the} \\ $$$${pulleys}\:{A}\:{and}\:{B}\:{as}\:{shown}?\:{find}\:{the} \\ $$$${maximum}\:{tension}\:{in}\:{the}\:{rope}\:{in} \\ $$$${each}\:{case}. \\ $$
Answered by mr W last updated on 10/Aug/25
Commented by mr W last updated on 13/Aug/25
![CASE 1 ρ=(M/L) let μ=(m/M), ξ_1 =(k_1 /L), ξ_2 =(k_2 /L) a=(T_0 /(ρg)) k_1 +s_1 +s_2 +k_2 =L T_1 =(m+k_1 ρ)g=(μ+ξ_1 )Mg T_2 =(m+k_2 ρ)g=(μ+ξ_2 )Mg T_0 =T_1 cos θ_1 =T_2 cos θ_2 ((aMg)/L)=(μ+ξ_1 )Mg cos θ_1 =(μ+ξ_2 )Mg cos θ_2 with α=(a/L) ⇒cos θ_1 =(α/(μ+ξ_1 )) ⇒tan θ_1 =((√((μ+ξ_1 )^2 −α^2 ))/α) ⇒cos θ_2 =(α/(μ+ξ_2 )) ⇒tan θ_2 =((√((μ+ξ_2 )^2 −α^2 ))/α) a+f =a cosh (h_1 /a) a+f+k=a cosh (h_2 /a) k=a(cosh (h_2 /a)−cosh (h_1 /a)) α(cosh (h_2 /a)−cosh (h_1 /a))=(k/L) tan θ_1 =sinh (h_1 /a) ⇒cosh (h_1 /a)=(√(1+tan^2 θ_1 ))=(1/(cos θ_1 ))=((μ+ξ_1 )/α) tan θ_2 =sinh (h_2 /a) ⇒cosh (h_2 /a)=(√(1+tan^2 θ_2 ))=(1/(cos θ_2 ))=((μ+ξ_2 )/α) α(((μ+ξ_2 )/α)−((μ+ξ_1 )/α))=(k/L) ⇒ξ_2 −ξ_1 =(k/L) ...(i) s_1 =a sinh (h_1 /a)=a tan θ_1 s_2 =a sinh (h_2 /a)=a tan θ_2 s_1 +s_2 =a(tan θ_1 +tan θ_2 )=L−k_1 −k_2 α(tan θ_1 +tan θ_2 )=1−ξ_1 −ξ_2 ⇒(√((μ+ξ_1 )^2 −α^2 ))+(√((μ+ξ_2 )^2 −α^2 ))=1−ξ_1 −ξ_2 ...(ii) (h_1 /a)=cosh^(−1) ((μ+ξ_1 )/α) (h_2 /a)=cosh^(−1) ((μ+ξ_2 )/α) ((h_1 +h_2 )/a)=cosh^(−1) ((μ+ξ_1 )/α)+cosh^(−1) ((μ+ξ_2 )/α) (h/(αL))=cosh^(−1) ((μ+ξ_1 )/α)+cosh^(−1) ((μ+ξ_2 )/α) α(cosh^(−1) ((μ+ξ_1 )/α)+cosh^(−1) ((μ+ξ_2 )/α))=(h/L) ...(iii) with p=μ+ξ_1 , q=μ+ξ_2 we get q−p=(k/L) ...(I) (√(p^2 −α^2 ))+(√(q^2 −α^2 ))=1+2μ−p−q ...(II) (√(q^2 −α^2 ))=(1+2μ−p−q)−(√(p^2 −α^2 )) (1+2μ−p−q)^2 +p^2 −q^2 =2(1+2μ−p−q)(√(p^2 −α^2 )) [(1+2μ−p−q)^2 +p^2 −q^2 ]^2 =4(1+2μ−p−q)^2 (p^2 −α^2 ) α=(√(p^2 −(([(1+2μ−p−q)^2 +p^2 −q^2 ]^2 )/(4(1+2μ−p−q)^2 )))) ⇒α=(√(p^2 −(([(1+2μ−2p−(k/L))^2 −(k/L)(2p+(k/L))]^2 )/(4(1+2μ−2p−(k/L))^2 )))) we get following equation for p: α(cosh^(−1) (p/α)+cosh^(−1) ((p+(k/L))/α))=(h/L) this may have more than one root. that means there could be more than one way to support the rope on the pulleys.](https://www.tinkutara.com/question/Q223939.png)
$$\underline{\boldsymbol{{CASE}}\:\:\:\mathrm{1}} \\ $$$$\rho=\frac{{M}}{{L}} \\ $$$${let}\:\mu=\frac{{m}}{{M}},\:\xi_{\mathrm{1}} =\frac{{k}_{\mathrm{1}} }{{L}},\:\xi_{\mathrm{2}} =\frac{{k}_{\mathrm{2}} }{{L}} \\ $$$${a}=\frac{{T}_{\mathrm{0}} }{\rho{g}} \\ $$$${k}_{\mathrm{1}} +{s}_{\mathrm{1}} +{s}_{\mathrm{2}} +{k}_{\mathrm{2}} ={L} \\ $$$${T}_{\mathrm{1}} =\left({m}+{k}_{\mathrm{1}} \rho\right){g}=\left(\mu+\xi_{\mathrm{1}} \right){Mg} \\ $$$${T}_{\mathrm{2}} =\left({m}+{k}_{\mathrm{2}} \rho\right){g}=\left(\mu+\xi_{\mathrm{2}} \right){Mg} \\ $$$${T}_{\mathrm{0}} ={T}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} ={T}_{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$$\frac{{aMg}}{{L}}=\left(\mu+\xi_{\mathrm{1}} \right){Mg}\:\mathrm{cos}\:\theta_{\mathrm{1}} =\left(\mu+\xi_{\mathrm{2}} \right){Mg}\:\mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$${with}\:\alpha=\frac{{a}}{{L}} \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\alpha}{\mu+\xi_{\mathrm{1}} }\:\:\Rightarrow\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{\sqrt{\left(\mu+\xi_{\mathrm{1}} \right)^{\mathrm{2}} −\alpha^{\mathrm{2}} }}{\alpha} \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{2}} =\frac{\alpha}{\mu+\xi_{\mathrm{2}} }\:\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\sqrt{\left(\mu+\xi_{\mathrm{2}} \right)^{\mathrm{2}} −\alpha^{\mathrm{2}} }}{\alpha} \\ $$$${a}+{f}\:={a}\:\mathrm{cosh}\:\frac{{h}_{\mathrm{1}} }{{a}} \\ $$$${a}+{f}+{k}={a}\:\mathrm{cosh}\:\frac{{h}_{\mathrm{2}} }{{a}} \\ $$$${k}={a}\left(\mathrm{cosh}\:\frac{{h}_{\mathrm{2}} }{{a}}−\mathrm{cosh}\:\frac{{h}_{\mathrm{1}} }{{a}}\right) \\ $$$$\alpha\left(\mathrm{cosh}\:\frac{{h}_{\mathrm{2}} }{{a}}−\mathrm{cosh}\:\frac{{h}_{\mathrm{1}} }{{a}}\right)=\frac{{k}}{{L}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\mathrm{sinh}\:\frac{{h}_{\mathrm{1}} }{{a}}\: \\ $$$$\Rightarrow\mathrm{cosh}\:\frac{{h}_{\mathrm{1}} }{{a}}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} }=\frac{\mu+\xi_{\mathrm{1}} }{\alpha} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\mathrm{sinh}\:\frac{{h}_{\mathrm{2}} }{{a}}\: \\ $$$$\Rightarrow\mathrm{cosh}\:\frac{{h}_{\mathrm{2}} }{{a}}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{2}} }=\frac{\mu+\xi_{\mathrm{2}} }{\alpha} \\ $$$$\alpha\left(\frac{\mu+\xi_{\mathrm{2}} }{\alpha}−\frac{\mu+\xi_{\mathrm{1}} }{\alpha}\right)=\frac{{k}}{{L}} \\ $$$$\Rightarrow\xi_{\mathrm{2}} −\xi_{\mathrm{1}} =\frac{{k}}{{L}}\:\:\:…\left({i}\right) \\ $$$${s}_{\mathrm{1}} ={a}\:\mathrm{sinh}\:\frac{{h}_{\mathrm{1}} }{{a}}={a}\:\mathrm{tan}\:\theta_{\mathrm{1}} \\ $$$${s}_{\mathrm{2}} ={a}\:\mathrm{sinh}\:\frac{{h}_{\mathrm{2}} }{{a}}={a}\:\mathrm{tan}\:\theta_{\mathrm{2}} \\ $$$${s}_{\mathrm{1}} +{s}_{\mathrm{2}} ={a}\left(\mathrm{tan}\:\theta_{\mathrm{1}} +\mathrm{tan}\:\theta_{\mathrm{2}} \right)={L}−{k}_{\mathrm{1}} −{k}_{\mathrm{2}} \\ $$$$\alpha\left(\mathrm{tan}\:\theta_{\mathrm{1}} +\mathrm{tan}\:\theta_{\mathrm{2}} \right)=\mathrm{1}−\xi_{\mathrm{1}} −\xi_{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\left(\mu+\xi_{\mathrm{1}} \right)^{\mathrm{2}} −\alpha^{\mathrm{2}} }+\sqrt{\left(\mu+\xi_{\mathrm{2}} \right)^{\mathrm{2}} −\alpha^{\mathrm{2}} }=\mathrm{1}−\xi_{\mathrm{1}} −\xi_{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\frac{{h}_{\mathrm{1}} }{{a}}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{1}} }{\alpha} \\ $$$$\frac{{h}_{\mathrm{2}} }{{a}}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{2}} }{\alpha} \\ $$$$\frac{{h}_{\mathrm{1}} +{h}_{\mathrm{2}} }{{a}}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{1}} }{\alpha}+\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{2}} }{\alpha} \\ $$$$\frac{{h}}{\alpha{L}}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{1}} }{\alpha}+\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{2}} }{\alpha} \\ $$$$\alpha\left(\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{1}} }{\alpha}+\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{2}} }{\alpha}\right)=\frac{{h}}{{L}}\:\:\:…\left({iii}\right) \\ $$$${with}\:{p}=\mu+\xi_{\mathrm{1}} ,\:{q}=\mu+\xi_{\mathrm{2}} \:{we}\:{get} \\ $$$${q}−{p}=\frac{{k}}{{L}}\:\:\:…\left({I}\right) \\ $$$$\sqrt{{p}^{\mathrm{2}} −\alpha^{\mathrm{2}} }+\sqrt{{q}^{\mathrm{2}} −\alpha^{\mathrm{2}} }=\mathrm{1}+\mathrm{2}\mu−{p}−{q}\:\:…\left({II}\right) \\ $$$$ \\ $$$$\sqrt{{q}^{\mathrm{2}} −\alpha^{\mathrm{2}} }=\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)−\sqrt{{p}^{\mathrm{2}} −\alpha^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)\sqrt{{p}^{\mathrm{2}} −\alpha^{\mathrm{2}} } \\ $$$$\left[\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right]^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} \left({p}^{\mathrm{2}} −\alpha^{\mathrm{2}} \right) \\ $$$$\alpha=\sqrt{{p}^{\mathrm{2}} −\frac{\left[\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right]^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\alpha=\sqrt{{p}^{\mathrm{2}} −\frac{\left[\left(\mathrm{1}+\mathrm{2}\mu−\mathrm{2}{p}−\frac{{k}}{{L}}\right)^{\mathrm{2}} −\frac{{k}}{{L}}\left(\mathrm{2}{p}+\frac{{k}}{{L}}\right)\right]^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{2}\mu−\mathrm{2}{p}−\frac{{k}}{{L}}\right)^{\mathrm{2}} }} \\ $$$${we}\:{get}\:{following}\:{equation}\:{for}\:{p}: \\ $$$$\alpha\left(\mathrm{cosh}^{−\mathrm{1}} \:\frac{{p}}{\alpha}+\mathrm{cosh}^{−\mathrm{1}} \:\frac{{p}+\frac{{k}}{{L}}}{\alpha}\right)=\frac{{h}}{{L}} \\ $$$${this}\:{may}\:{have}\:{more}\:{than}\:{one}\:{root}. \\ $$$${that}\:{means}\:{there}\:{could}\:{be}\:{more} \\ $$$${than}\:{one}\:{way}\:{to}\:{support}\:{the}\:{rope} \\ $$$${on}\:{the}\:{pulleys}. \\ $$
Commented by mr W last updated on 10/Aug/25
$${eqn}.\:\left({i}\right)\:{means}\:{k}_{\mathrm{2}} −{k}_{\mathrm{1}} ={k},\:{i}.{e}.\:{both} \\ $$$${ends}\:{of}\:{the}\:{rope}\:{must}\:{be}\:{at}\:{the}\:{same} \\ $$$${height}. \\ $$
Commented by mr W last updated on 10/Aug/25
Commented by mr W last updated on 10/Aug/25
$${for}\:{m}/{M}=\mathrm{0}.\mathrm{2},\:{h}=\mathrm{5},\:{k}=\mathrm{2},\:{L}=\mathrm{12} \\ $$$${there}\:{are}\:{following}\:{possibilities}: \\ $$
Commented by mr W last updated on 12/Aug/25
Commented by mr W last updated on 12/Aug/25
Answered by mr W last updated on 10/Aug/25
Commented by mr W last updated on 13/Aug/25
![CASE 2 ρ=(M/L) let μ=(m/M), ξ_1 =(k_1 /L), ξ_2 =(k_2 /L) a=(T_0 /(ρg)) k_1 +s+k_2 =L T_1 =(m+k_1 ρ)g=(μ+ξ_1 )Mg T_2 =(m+k_2 ρ)g=(μ+ξ_2 )Mg T_0 =T_1 cos θ_1 =T_2 cos θ_2 ((aMg)/L)=(μ+ξ_1 )Mg cos θ_1 =(μ+ξ_2 )Mg cos θ_2 with α=(a/L) ⇒cos θ_1 =(α/(μ+ξ_1 )) ⇒tan θ_1 =((√((μ+ξ_1 )^2 −α^2 ))/α) ⇒cos θ_2 =(α/(μ+ξ_2 )) ⇒tan θ_2 =((√((μ+ξ_2 )^2 −α^2 ))/α) a+f =a cosh (h_1 /a) a+f+k=a cosh (h_2 /a) k=a(cosh (h_2 /a)−cosh (h_1 /a)) α(cosh (h_2 /a)−cosh (h_1 /a))=(k/L) tan θ_1 =sinh (h_1 /a) ⇒cosh (h_1 /a)=(√(1+tan^2 θ_1 ))=(1/(cos θ_1 ))=((μ+ξ_1 )/α) tan θ_2 =sinh (h_2 /a) ⇒cosh (h_2 /a)=(√(1+tan^2 θ_2 ))=(1/(cos θ_2 ))=((μ+ξ_2 )/α) α(((μ+ξ_2 )/α)−((μ+ξ_1 )/α))=(k/L) ⇒ξ_2 −ξ_1 =(k/L) ...(i) s_1 =a sinh (h_1 /a)=a tan θ_1 s_2 =a sinh (h_2 /a)=a tan θ_2 s=s_2 −s_1 =a(tan θ_2 −tan θ_1 )=L−k_1 −k_2 α(tan θ_2 −tan θ_1 )=1−ξ_1 −ξ_2 ⇒(√((μ+ξ_2 )^2 −α^2 ))−(√((μ+ξ_1 )^2 −α^2 ))=1−ξ_1 −ξ_2 ...(ii) (h_1 /a)=cosh^(−1) ((μ+ξ_1 )/α) (h_2 /a)=cosh^(−1) ((μ+ξ_2 )/α) ((h_2 −h_1 )/a)=cosh^(−1) ((μ+ξ_2 )/α)−cosh^(−1) ((μ+ξ_1 )/α) (h/(αL))=cosh^(−1) ((μ+ξ_2 )/α)−cosh^(−1) ((μ+ξ_1 )/α) α(cosh^(−1) ((μ+ξ_2 )/α)−cosh^(−1) ((μ+ξ_1 )/α))=(h/L) ...(iii) with p=μ+ξ_1 , q=μ+ξ_2 we get q−p=(k/L) ...(I) (√(q^2 −α^2 ))−(√(p^2 −α^2 ))=1+2μ−p−q ...(II) (√(q^2 −α^2 ))=(1+2μ−p−q)+(√(p^2 −α^2 )) q^2 −p^2 −(1+2μ−p−q)^2 =2(1+2μ−p−q)(√(p^2 −α^2 )) [q^2 −p^2 −(1+2μ−p−q)^2 ]^2 =4(1+2μ−p−q)^2 (p^2 −α^2 ) α=(√(p^2 −(([(1+2μ−p−q)^2 +p^2 −q^2 ]^2 )/(4(1+2μ−p−q)^2 )))) ⇒α=(√(p^2 −(([(1+2μ−2p−(k/L))^2 −(k/L)(2p+(k/L))]^2 )/(4(1+2μ−2p−(k/L))^2 )))) we get following equation for p: α(cosh^(−1) ((p+(k/L))/α)−cosh^(−1) (p/α))=(h/L) this may have more than one root. that means there could be more than one way to support the rope on the pulleys.](https://www.tinkutara.com/question/Q223949.png)
$$\underline{\boldsymbol{{CASE}}\:\:\:\mathrm{2}} \\ $$$$\rho=\frac{{M}}{{L}} \\ $$$${let}\:\mu=\frac{{m}}{{M}},\:\xi_{\mathrm{1}} =\frac{{k}_{\mathrm{1}} }{{L}},\:\xi_{\mathrm{2}} =\frac{{k}_{\mathrm{2}} }{{L}} \\ $$$${a}=\frac{{T}_{\mathrm{0}} }{\rho{g}} \\ $$$${k}_{\mathrm{1}} +{s}+{k}_{\mathrm{2}} ={L} \\ $$$${T}_{\mathrm{1}} =\left({m}+{k}_{\mathrm{1}} \rho\right){g}=\left(\mu+\xi_{\mathrm{1}} \right){Mg} \\ $$$${T}_{\mathrm{2}} =\left({m}+{k}_{\mathrm{2}} \rho\right){g}=\left(\mu+\xi_{\mathrm{2}} \right){Mg} \\ $$$${T}_{\mathrm{0}} ={T}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} ={T}_{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$$\frac{{aMg}}{{L}}=\left(\mu+\xi_{\mathrm{1}} \right){Mg}\:\mathrm{cos}\:\theta_{\mathrm{1}} =\left(\mu+\xi_{\mathrm{2}} \right){Mg}\:\mathrm{cos}\:\theta_{\mathrm{2}} \\ $$$${with}\:\alpha=\frac{{a}}{{L}} \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{1}} =\frac{\alpha}{\mu+\xi_{\mathrm{1}} }\:\:\Rightarrow\mathrm{tan}\:\theta_{\mathrm{1}} =\frac{\sqrt{\left(\mu+\xi_{\mathrm{1}} \right)^{\mathrm{2}} −\alpha^{\mathrm{2}} }}{\alpha} \\ $$$$\Rightarrow\mathrm{cos}\:\theta_{\mathrm{2}} =\frac{\alpha}{\mu+\xi_{\mathrm{2}} }\:\Rightarrow\mathrm{tan}\:\theta_{\mathrm{2}} =\frac{\sqrt{\left(\mu+\xi_{\mathrm{2}} \right)^{\mathrm{2}} −\alpha^{\mathrm{2}} }}{\alpha} \\ $$$${a}+{f}\:={a}\:\mathrm{cosh}\:\frac{{h}_{\mathrm{1}} }{{a}} \\ $$$${a}+{f}+{k}={a}\:\mathrm{cosh}\:\frac{{h}_{\mathrm{2}} }{{a}} \\ $$$${k}={a}\left(\mathrm{cosh}\:\frac{{h}_{\mathrm{2}} }{{a}}−\mathrm{cosh}\:\frac{{h}_{\mathrm{1}} }{{a}}\right) \\ $$$$\alpha\left(\mathrm{cosh}\:\frac{{h}_{\mathrm{2}} }{{a}}−\mathrm{cosh}\:\frac{{h}_{\mathrm{1}} }{{a}}\right)=\frac{{k}}{{L}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\mathrm{sinh}\:\frac{{h}_{\mathrm{1}} }{{a}}\: \\ $$$$\Rightarrow\mathrm{cosh}\:\frac{{h}_{\mathrm{1}} }{{a}}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} }=\frac{\mu+\xi_{\mathrm{1}} }{\alpha} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\mathrm{sinh}\:\frac{{h}_{\mathrm{2}} }{{a}}\: \\ $$$$\Rightarrow\mathrm{cosh}\:\frac{{h}_{\mathrm{2}} }{{a}}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{2}} }=\frac{\mu+\xi_{\mathrm{2}} }{\alpha} \\ $$$$\alpha\left(\frac{\mu+\xi_{\mathrm{2}} }{\alpha}−\frac{\mu+\xi_{\mathrm{1}} }{\alpha}\right)=\frac{{k}}{{L}} \\ $$$$\Rightarrow\xi_{\mathrm{2}} −\xi_{\mathrm{1}} =\frac{{k}}{{L}}\:\:\:…\left({i}\right) \\ $$$${s}_{\mathrm{1}} ={a}\:\mathrm{sinh}\:\frac{{h}_{\mathrm{1}} }{{a}}={a}\:\mathrm{tan}\:\theta_{\mathrm{1}} \\ $$$${s}_{\mathrm{2}} ={a}\:\mathrm{sinh}\:\frac{{h}_{\mathrm{2}} }{{a}}={a}\:\mathrm{tan}\:\theta_{\mathrm{2}} \\ $$$${s}={s}_{\mathrm{2}} −{s}_{\mathrm{1}} ={a}\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)={L}−{k}_{\mathrm{1}} −{k}_{\mathrm{2}} \\ $$$$\alpha\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)=\mathrm{1}−\xi_{\mathrm{1}} −\xi_{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\left(\mu+\xi_{\mathrm{2}} \right)^{\mathrm{2}} −\alpha^{\mathrm{2}} }−\sqrt{\left(\mu+\xi_{\mathrm{1}} \right)^{\mathrm{2}} −\alpha^{\mathrm{2}} }=\mathrm{1}−\xi_{\mathrm{1}} −\xi_{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\frac{{h}_{\mathrm{1}} }{{a}}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{1}} }{\alpha} \\ $$$$\frac{{h}_{\mathrm{2}} }{{a}}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{2}} }{\alpha} \\ $$$$\frac{{h}_{\mathrm{2}} −{h}_{\mathrm{1}} }{{a}}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{2}} }{\alpha}−\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{1}} }{\alpha} \\ $$$$\frac{{h}}{\alpha{L}}=\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{2}} }{\alpha}−\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{1}} }{\alpha} \\ $$$$\alpha\left(\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{2}} }{\alpha}−\mathrm{cosh}^{−\mathrm{1}} \:\frac{\mu+\xi_{\mathrm{1}} }{\alpha}\right)=\frac{{h}}{{L}}\:\:\:…\left({iii}\right) \\ $$$${with}\:{p}=\mu+\xi_{\mathrm{1}} ,\:{q}=\mu+\xi_{\mathrm{2}} \:{we}\:{get} \\ $$$${q}−{p}=\frac{{k}}{{L}}\:\:\:…\left({I}\right) \\ $$$$\sqrt{{q}^{\mathrm{2}} −\alpha^{\mathrm{2}} }−\sqrt{{p}^{\mathrm{2}} −\alpha^{\mathrm{2}} }=\mathrm{1}+\mathrm{2}\mu−{p}−{q}\:\:…\left({II}\right) \\ $$$$\sqrt{{q}^{\mathrm{2}} −\alpha^{\mathrm{2}} }=\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)+\sqrt{{p}^{\mathrm{2}} −\alpha^{\mathrm{2}} } \\ $$$${q}^{\mathrm{2}} −{p}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} =\mathrm{2}\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)\sqrt{{p}^{\mathrm{2}} −\alpha^{\mathrm{2}} } \\ $$$$\left[{q}^{\mathrm{2}} −{p}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} \right]^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} \left({p}^{\mathrm{2}} −\alpha^{\mathrm{2}} \right) \\ $$$$\alpha=\sqrt{{p}^{\mathrm{2}} −\frac{\left[\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right]^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{2}\mu−{p}−{q}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\alpha=\sqrt{{p}^{\mathrm{2}} −\frac{\left[\left(\mathrm{1}+\mathrm{2}\mu−\mathrm{2}{p}−\frac{{k}}{{L}}\right)^{\mathrm{2}} −\frac{{k}}{{L}}\left(\mathrm{2}{p}+\frac{{k}}{{L}}\right)\right]^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}+\mathrm{2}\mu−\mathrm{2}{p}−\frac{{k}}{{L}}\right)^{\mathrm{2}} }} \\ $$$${we}\:{get}\:{following}\:{equation}\:{for}\:{p}: \\ $$$$\alpha\left(\mathrm{cosh}^{−\mathrm{1}} \:\frac{{p}+\frac{{k}}{{L}}}{\alpha}−\mathrm{cosh}^{−\mathrm{1}} \:\frac{{p}}{\alpha}\right)=\frac{{h}}{{L}} \\ $$$${this}\:{may}\:{have}\:{more}\:{than}\:{one}\:{root}. \\ $$$${that}\:{means}\:{there}\:{could}\:{be}\:{more} \\ $$$${than}\:{one}\:{way}\:{to}\:{support}\:{the}\:{rope} \\ $$$${on}\:{the}\:{pulleys}. \\ $$
Commented by mr W last updated on 13/Aug/25
$${for}\:{m}/{M}=\mathrm{0}.\mathrm{5},\:{h}=\mathrm{5},\:{k}=\mathrm{2},\:{L}=\mathrm{12} \\ $$
Commented by mr W last updated on 13/Aug/25
Commented by mr W last updated on 13/Aug/25
![L−(k_1 +k_2 )=s>(√(h^2 +k^2 )) ⇒ξ_1 +ξ_2 <1−(√(((h/L))^2 +((k/L))^2 )) ξ_2 −ξ_1 =(k/L) ⇒ξ_1 <(1/2)[1−(k/L)−(√(((h/L))^2 +((k/L))^2 ))] ⇒p<(1/2)[1−(k/L)−(√(((h/L))^2 +((k/L))^2 ))]+μ s_1 =0: ⇒p_1 =α ⇒(1+2μ−2p_1 −(k/L))^2 −(k/L)(2p_1 +(k/L))=0 ⇒p_1 =(1/2)[1+2μ−(k/(2L))−(√((k/(2L))(2+4μ+(k/(2L)))))] case 1: p<p_1 case 2: p_1 <p<(1/2)[1−(k/L)−(√(((h/L))^2 +((k/L))^2 ))]+μ](https://www.tinkutara.com/question/Q223998.png)
$${L}−\left({k}_{\mathrm{1}} +{k}_{\mathrm{2}} \right)={s}>\sqrt{{h}^{\mathrm{2}} +{k}^{\mathrm{2}} } \\ $$$$\Rightarrow\xi_{\mathrm{1}} +\xi_{\mathrm{2}} <\mathrm{1}−\sqrt{\left(\frac{{h}}{{L}}\right)^{\mathrm{2}} +\left(\frac{{k}}{{L}}\right)^{\mathrm{2}} } \\ $$$$\xi_{\mathrm{2}} −\xi_{\mathrm{1}} =\frac{{k}}{{L}} \\ $$$$\Rightarrow\xi_{\mathrm{1}} <\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\frac{{k}}{{L}}−\sqrt{\left(\frac{{h}}{{L}}\right)^{\mathrm{2}} +\left(\frac{{k}}{{L}}\right)^{\mathrm{2}} }\right] \\ $$$$\Rightarrow{p}<\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\frac{{k}}{{L}}−\sqrt{\left(\frac{{h}}{{L}}\right)^{\mathrm{2}} +\left(\frac{{k}}{{L}}\right)^{\mathrm{2}} }\right]+\mu \\ $$$${s}_{\mathrm{1}} =\mathrm{0}: \\ $$$$\Rightarrow{p}_{\mathrm{1}} =\alpha \\ $$$$\Rightarrow\left(\mathrm{1}+\mathrm{2}\mu−\mathrm{2}{p}_{\mathrm{1}} −\frac{{k}}{{L}}\right)^{\mathrm{2}} −\frac{{k}}{{L}}\left(\mathrm{2}{p}_{\mathrm{1}} +\frac{{k}}{{L}}\right)=\mathrm{0} \\ $$$$\Rightarrow{p}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}+\mathrm{2}\mu−\frac{{k}}{\mathrm{2}{L}}−\sqrt{\frac{{k}}{\mathrm{2}{L}}\left(\mathrm{2}+\mathrm{4}\mu+\frac{{k}}{\mathrm{2}{L}}\right)}\right] \\ $$$${case}\:\mathrm{1}:\:{p}<{p}_{\mathrm{1}} \\ $$$${case}\:\mathrm{2}:\:{p}_{\mathrm{1}} <{p}<\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\frac{{k}}{{L}}−\sqrt{\left(\frac{{h}}{{L}}\right)^{\mathrm{2}} +\left(\frac{{k}}{{L}}\right)^{\mathrm{2}} }\right]+\mu \\ $$
Commented by mr W last updated on 15/Aug/25
Commented by mr W last updated on 14/Aug/25
$${normally}\:{we}\:{get}\:{two}\:{possibilities} \\ $$$${for}\:{case}\:\mathrm{1}\:{and}\:{only}\:{one}\:{possibility} \\ $$$${for}\:{case}\:\mathrm{2}. \\ $$
Answered by mr W last updated on 14/Aug/25
Commented by mr W last updated on 14/Aug/25
$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${T}_{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{2}} ={T}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} ={T}_{\mathrm{0}} =\rho{ga} \\ $$$${y}_{\mathrm{1}} ={a}\:\mathrm{cosh}\:\frac{{x}_{\mathrm{1}} }{{a}} \\ $$$${y}_{\mathrm{2}} ={a}\:\mathrm{cosh}\:\frac{{x}_{\mathrm{2}} }{{a}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\mathrm{sinh}\:\frac{{x}_{\mathrm{1}} }{{a}} \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\mathrm{sinh}\:\frac{{x}_{\mathrm{2}} }{{a}} \\ $$$$\mathrm{cosh}\:\frac{{x}_{\mathrm{1}} }{{a}}=\sqrt{\mathrm{1}+\mathrm{sinh}^{\mathrm{2}} \:\frac{{x}_{\mathrm{1}} }{{a}}}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} } \\ $$$$\mathrm{cosh}\:\frac{{x}_{\mathrm{2}} }{{a}}=\sqrt{\mathrm{1}+\mathrm{sinh}^{\mathrm{2}} \:\frac{{x}_{\mathrm{2}} }{{a}}}=\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{2}} } \\ $$$${y}_{\mathrm{1}} =\frac{{a}}{\mathrm{cos}\:\theta_{\mathrm{1}} }=\frac{{T}_{\mathrm{1}} }{\rho{g}} \\ $$$${y}_{\mathrm{2}} =\frac{{a}}{\mathrm{cos}\:\theta_{\mathrm{2}} }=\frac{{T}_{\mathrm{2}} }{\rho{g}} \\ $$$$\Rightarrow{T}_{\mathrm{2}} −{T}_{\mathrm{1}} =\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)\rho{g}=\rho{g}\Delta{y} \\ $$