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Question Number 66527 by mr W last updated on 16/Aug/19

Commented by mr W last updated on 16/Aug/19

the perimeter of a rope loop is L. now it is hanged on two pins A and B. the distance between the pins is b. all contact is frictionless. (L>2b) if the lowest point of the loop is h below the pins. find h in terms of b and L.

$${the}\:{perimeter}\:{of}\:{a}\:{rope}\:{loop}\:{is}\:{L}. \\ $$$${now}\:{it}\:{is}\:{hanged}\:{on}\:{two}\:{pins}\:{A}\:{and}\:{B}. \\ $$$${the}\:{distance}\:{between}\:{the}\:{pins}\:{is}\:{b}.\: \\ $$$${all}\:{contact}\:{is}\:{frictionless}. \\ $$$$\left({L}>\mathrm{2}{b}\right) \\ $$$${if}\:{the}\:{lowest}\:{point}\:{of}\:{the}\:{loop}\:{is}\:{h} \\ $$$${below}\:{the}\:{pins}. \\ $$$${find}\:{h}\:{in}\:{terms}\:{of}\:{b}\:{and}\:{L}. \\ $$

Answered by mr W last updated on 17/Aug/19

Commented by mr W last updated on 17/Aug/19

Commented by mr W last updated on 17/Aug/19

when the rope loop is hanging on the pins, it consists of 2 parts S_1 and S_2 . S_1 +S_2 =L generally for a rope between A and B: mass of rope of unit length= ρ tension in rope at point A (B)= T length of rope from A to B =S T_0 =T cos θ a=(T_0 /(ρg))=((T cos θ)/(ρg)) y=a cosh (x/a) y′=sinh (x/a) at point B: y′=sinh (b/(2a))=tan θ=t ⇒(b/(2a))=sinh^(−1) t=ln (t+(√(1+t^2 ))) (S/2)=a sinh (b/(2a))=a tan θ=at ⇒S=2at=((2a)/b)×bt=((bt)/(sinh^(−1) t))=((bt)/(ln (t+(√(1+t^2 ))))) ⇒(T/(ρg))=(a/(cos θ))=((2a)/b)×((b(√(1+tan^2 θ)))/2)=((b(√(1+t^2 )))/(2 ln (t+(√(1+t^2 ))))) h=a(1+cosh (b/(2a)))=(b/2)×((2a)/b)(1+cosh (b/(2a))) ⇒((2h)/b)=((2a)/b)(1+cosh (b/(2a)))=((1+(√(1+t^2 )))/(ln (t+(√(1+t^2 ))))) for rope part 1: t_1 =tan θ_1 S_1 =((bt_1 )/(ln (t_1 +(√(1+t_1 ^2 ))))) (T_1 /(ρg))=((b(√(1+t_1 ^2 )))/(2 ln (t_1 +(√(1+t_1 ^2 ))))) for rope part 2: t_2 =tan θ_2 S_2 =((bt_2 )/(ln (t_2 +(√(1+t_2 ^2 ))))) (T_2 /(ρg))=((b(√(1+t_2 ^2 )))/(2 ln (t_2 +(√(1+t_2 ^2 ))))) T_1 =T_2 ((b(√(1+t_1 ^2 )))/(2 ln (t_1 +(√(1+t_1 ^2 )))))=((b(√(1+t_2 ^2 )))/(2 ln (t_2 +(√(1+t_2 ^2 ))))) ⇒((√(1+t_1 ^2 ))/(ln (t_1 +(√(1+t_1 ^2 )))))=((√(1+t_2 ^2 ))/(ln (t_2 +(√(1+t_2 ^2 ))))) ...(i) S_1 +S_2 =L ((bt_1 )/(ln (t_1 +(√(1+t_1 ^2 )))))+((bt_2 )/(ln (t_2 +(√(1+t_2 ^2 )))))=L let λ=(L/b) ⇒ (t_1 /(ln (t_1 +(√(1+t_1 ^2 )))))+(t_2 /(ln (t_2 +(√(1+t_2 ^2 )))))=λ ...(ii) t_1 =t_2 =t is always a solution of (i), ⇒ (t/(ln (t+(√(1+t^2 )))))=(λ/2) in this case both rope parts are equal: S_1 =S_2 =(L/2) in the diagram the green curve shows the equation (i). the red curve shows the equation (ii) for different values of λ. we can see that there is only one solution for the shape of the rope if λ≤2.5155, with t_1 =t_2 , i.e. S_1 =S_2 . but for λ>2.5155 there are two possible shapes, one with t_1 =t_2 , i.e. S_1 =S_2 and the other one with t_1 ≠t_2 , i.e. S_1 ≠S_2 . for a given value of λ=(L/b), numerical solutions of eqn. (i) and (ii) can be obtained.

$${when}\:{the}\:{rope}\:{loop}\:{is}\:{hanging}\:{on}\:{the} \\ $$$${pins},\:{it}\:{consists}\:{of}\:\mathrm{2}\:{parts}\:{S}_{\mathrm{1}} \:{and}\:{S}_{\mathrm{2}} . \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} ={L} \\ $$$$ \\ $$$${generally}\:{for}\:{a}\:{rope}\:{between}\:{A}\:{and}\:{B}: \\ $$$${mass}\:{of}\:{rope}\:{of}\:{unit}\:{length}=\:\rho \\ $$$${tension}\:{in}\:{rope}\:{at}\:{point}\:{A}\:\left({B}\right)=\:{T} \\ $$$${length}\:{of}\:{rope}\:{from}\:{A}\:{to}\:{B}\:={S} \\ $$$${T}_{\mathrm{0}} ={T}\:\mathrm{cos}\:\theta \\ $$$${a}=\frac{{T}_{\mathrm{0}} }{\rho{g}}=\frac{{T}\:\mathrm{cos}\:\theta}{\rho{g}} \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${y}'=\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$${at}\:{point}\:{B}: \\ $$$${y}'=\mathrm{sinh}\:\frac{{b}}{\mathrm{2}{a}}=\mathrm{tan}\:\theta={t} \\ $$$$\Rightarrow\frac{{b}}{\mathrm{2}{a}}=\mathrm{sinh}^{−\mathrm{1}} \:{t}=\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$$$\frac{{S}}{\mathrm{2}}={a}\:\mathrm{sinh}\:\frac{{b}}{\mathrm{2}{a}}={a}\:\mathrm{tan}\:\theta={at} \\ $$$$\Rightarrow{S}=\mathrm{2}{at}=\frac{\mathrm{2}{a}}{{b}}×{bt}=\frac{{bt}}{\mathrm{sinh}^{−\mathrm{1}} \:{t}}=\frac{{bt}}{\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\frac{{T}}{\rho{g}}=\frac{{a}}{\mathrm{cos}\:\theta}=\frac{\mathrm{2}{a}}{{b}}×\frac{{b}\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta}}{\mathrm{2}}=\frac{{b}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$${h}={a}\left(\mathrm{1}+\mathrm{cosh}\:\frac{{b}}{\mathrm{2}{a}}\right)=\frac{{b}}{\mathrm{2}}×\frac{\mathrm{2}{a}}{{b}}\left(\mathrm{1}+\mathrm{cosh}\:\frac{{b}}{\mathrm{2}{a}}\right) \\ $$$$\Rightarrow\frac{\mathrm{2}{h}}{{b}}=\frac{\mathrm{2}{a}}{{b}}\left(\mathrm{1}+\mathrm{cosh}\:\frac{{b}}{\mathrm{2}{a}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$ \\ $$$${for}\:{rope}\:{part}\:\mathrm{1}: \\ $$$${t}_{\mathrm{1}} =\mathrm{tan}\:\theta_{\mathrm{1}} \\ $$$${S}_{\mathrm{1}} =\frac{{bt}_{\mathrm{1}} }{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)} \\ $$$$\frac{{T}_{\mathrm{1}} }{\rho{g}}=\frac{{b}\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)} \\ $$$${for}\:{rope}\:{part}\:\mathrm{2}: \\ $$$${t}_{\mathrm{2}} =\mathrm{tan}\:\theta_{\mathrm{2}} \\ $$$${S}_{\mathrm{2}} =\frac{{bt}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)} \\ $$$$\frac{{T}_{\mathrm{2}} }{\rho{g}}=\frac{{b}\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)} \\ $$$$ \\ $$$${T}_{\mathrm{1}} ={T}_{\mathrm{2}} \\ $$$$\frac{{b}\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}=\frac{{b}\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{2}\:\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }}{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}=\frac{\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }}{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}\:\:\:\:...\left({i}\right) \\ $$$${S}_{\mathrm{1}} +{S}_{\mathrm{2}} ={L} \\ $$$$\frac{{bt}_{\mathrm{1}} }{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}+\frac{{bt}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}={L} \\ $$$${let}\:\lambda=\frac{{L}}{{b}} \\ $$$$\Rightarrow\:\frac{{t}_{\mathrm{1}} }{\mathrm{ln}\:\left({t}_{\mathrm{1}} +\sqrt{\mathrm{1}+{t}_{\mathrm{1}} ^{\mathrm{2}} }\right)}+\frac{{t}_{\mathrm{2}} }{\mathrm{ln}\:\left({t}_{\mathrm{2}} +\sqrt{\mathrm{1}+{t}_{\mathrm{2}} ^{\mathrm{2}} }\right)}=\lambda\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${t}_{\mathrm{1}} ={t}_{\mathrm{2}} ={t}\:{is}\:{always}\:{a}\:{solution}\:{of}\:\left({i}\right), \\ $$$$\Rightarrow\:\frac{{t}}{\mathrm{ln}\:\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)}=\frac{\lambda}{\mathrm{2}} \\ $$$${in}\:{this}\:{case}\:{both}\:{rope}\:{parts}\:{are}\:{equal}:\: \\ $$$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} =\frac{{L}}{\mathrm{2}} \\ $$$$ \\ $$$${in}\:{the}\:{diagram}\:{the}\:{green}\:{curve}\:{shows} \\ $$$${the}\:{equation}\:\left({i}\right).\:{the}\:{red}\:{curve}\:{shows} \\ $$$${the}\:{equation}\:\left({ii}\right)\:{for}\:{different}\:{values} \\ $$$${of}\:\lambda.\:{we}\:{can}\:{see}\:{that}\:{there}\:{is}\:{only}\:{one} \\ $$$${solution}\:{for}\:{the}\:{shape}\:{of}\:{the}\:{rope}\:{if} \\ $$$$\lambda\leqslant\mathrm{2}.\mathrm{5155},\:{with}\:{t}_{\mathrm{1}} ={t}_{\mathrm{2}} ,\:{i}.{e}.\:{S}_{\mathrm{1}} ={S}_{\mathrm{2}} . \\ $$$${but}\:{for}\:\lambda>\mathrm{2}.\mathrm{5155}\:{there}\:{are}\:{two} \\ $$$${possible}\:{shapes},\:{one}\:{with}\:{t}_{\mathrm{1}} ={t}_{\mathrm{2}} ,\:{i}.{e}. \\ $$$${S}_{\mathrm{1}} ={S}_{\mathrm{2}} \:{and}\:{the}\:{other}\:{one}\:{with}\:{t}_{\mathrm{1}} \neq{t}_{\mathrm{2}} , \\ $$$${i}.{e}.\:{S}_{\mathrm{1}} \neq{S}_{\mathrm{2}} . \\ $$$${for}\:{a}\:{given}\:{value}\:{of}\:\lambda=\frac{{L}}{{b}},\:{numerical} \\ $$$${solutions}\:{of}\:{eqn}.\:\left({i}\right)\:{and}\:\left({ii}\right)\:{can}\:{be} \\ $$$${obtained}. \\ $$

Commented by mr W last updated on 17/Aug/19

Commented by mr W last updated on 17/Aug/19

Commented by Cmr 237 last updated on 17/Aug/19

please i want the name of that application

$${please}\:{i}\:{want}\:{the}\:{name}\: \\ $$$$\:{of}\:{that}\:{application} \\ $$

Commented by mr W last updated on 17/Aug/19

for the graph i used an app called Grapher.

$${for}\:{the}\:{graph}\:{i}\:{used}\:{an}\:{app}\:{called} \\ $$$${Grapher}. \\ $$

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