Question-187675

Question Number 187675 by mr W last updated on 20/Feb/23

Commented by mr W last updated on 20/Feb/23

$${an}\:{uniform}\:{rope}\:{of}\:{length}\:{L}\:{and}\: \\ $$$${mass}\:{M}\:{is}\:{fixed}\:{on}\:{both}\:{ends}\:{at}\:{a} \\ $$$${distance}\:{d}\:{as}\:{shown}.\:{a}\:{small}\:{object} \\ $$$${of}\:{mass}\:{m}\:{moves}\:{very}\:{slowly}\:{along} \\ $$$${the}\:{rope}\:{from}\:{one}\:{end}\:{to}\:{the}\:{other}. \\ $$$${find}\:{the}\:{locus}\:{of}\:{the}\:{small}\:{object} \\ $$$${and}\:{the}\:{total}\:{distance}\:{it}\:{moved}. \\ $$

Answered by mr W last updated on 22/Feb/23

Commented by mr W last updated on 22/Feb/23

rope without small object y=a cosh (x/a) at x=(d/2): (L/2)=a sinh (d/(2a)) ⇒((sinh (d/(2a)))/(d/(2a)))=(L/d) we can solve for (d/(2a)) and get a.

$$\underline{\boldsymbol{{rope}}\:\boldsymbol{{without}}\:\boldsymbol{{small}}\:\boldsymbol{{object}}} \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${at}\:{x}=\frac{{d}}{\mathrm{2}}: \\ $$$$\frac{{L}}{\mathrm{2}}={a}\:\mathrm{sinh}\:\frac{{d}}{\mathrm{2}{a}} \\ $$$$\Rightarrow\frac{\mathrm{sinh}\:\frac{{d}}{\mathrm{2}{a}}}{\frac{{d}}{\mathrm{2}{a}}}=\frac{{L}}{{d}} \\ $$$${we}\:{can}\:{solve}\:{for}\:\frac{{d}}{\mathrm{2}{a}}\:{and}\:{get}\:{a}. \\ $$

Commented by mr W last updated on 24/Feb/23

small object at distance d_1 to left end and deflection f T_1 cos θ_1 =T_2 cos θ_2 =T_0 T_2 sin θ_2 −T_1 sin θ_1 =mg T_0 (tan θ_2 −tan θ_1 )=mg T_0 =((mg)/(tan θ_2 −tan θ_1 )) a=(T_0 /(ρg))=((mgL)/(Mg(tan θ_2 −tan θ_1 )))=((μL)/(tan θ_2 −tan θ_1 )) with μ=(m/M) for rope L_1 : y=a cosh (x/a) at point C: tan θ_1 =sinh (x_C /a) ⇒(x_C /a)=sinh^(−1) (tan θ_1 ) y_C =a cosh (x_C /a)=a (√(1+tan^2 θ_1 ))=(a/(cos θ_1 )) at point A: y_C +f=a cosh ((x_C −d_1 )/a) ⇒(f/a)=cosh (sinh^(−1) (tan θ_1 )−(d_1 /a))−(1/(cos θ_1 )) L_1 =a(sinh (x_C /a)−sinh ((x_C −d_1 )/a)) ⇒(L_1 /a)=tan θ_1 −sinh (sinh^(−1) (tan θ_1 )−(d_1 /a)) for rope L_2 : y=a cosh (x/a) at point C: tan θ_2 =sinh (x_C /a) ⇒(x_C /a)=sinh^(−1) (tan θ_2 ) y_C =a cosh (x_C /a)=a (√(1+tan^2 θ_2 ))=(a/(cos θ_2 )) at point B: y_C +f=a cosh ((x_C +d_2 )/a) ⇒(f/a)=cosh (sinh^(−1) (tan θ_2 )+(d_2 /a))−(1/(cos θ_2 )) L_2 =a(sinh ((x_C +d_2 )/a)−sinh (x_C /a)) ⇒(L_2 /a)=sinh (sinh^(−1) (tan θ_2 )+(d_2 /a))−tan θ_2 d_2 =d−d_1 L_1 +L_2 =L cosh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )d_1 )/(μL))]−(1/(cos θ_1 ))=cosh [sinh^(−1) (tan θ_2 )+(((tan θ_2 −tan θ_1 )(d−d_1 ))/(μL))]−(1/(cos θ_2 )) sinh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )d_1 )/(μL))]−sinh [sinh^(−1) (tan θ_2 )+(((tan θ_2 −tan θ_1 )(d−d_1 ))/(μL))]+(1+(1/μ))(tan θ_2 −tan θ_1 )=0 (f/a)=cosh (sinh^(−1) (tan θ_1 )−(d_1 /a))−(1/(cos θ_1 )) with ξ=(d_1 /d), η=(f/d), λ=((μL)/d), μ=(m/M) cosh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )ξ)/λ)]−(1/(cos θ_1 ))=cosh [sinh^(−1) (tan θ_2 )+(((tan θ_2 −tan θ_1 )(1−ξ))/λ)]−(1/(cos θ_2 )) ...(i) sinh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )ξ)/λ)]−sinh [sinh^(−1) (tan θ_2 )+(((tan θ_2 −tan θ_1 )(1−ξ))/λ)]+(1+(1/μ))(tan θ_2 −tan θ_1 )=0 ...(ii) η=(λ/(tan θ_2 −tan θ_1 )){cosh [sinh^(−1) (tan θ_1 )−(((tan θ_2 −tan θ_1 )ξ)/λ)]−(1/(cos θ_1 ))} ...(iii) for given ξ∈[0,1] we can solve (i) and (ii) for θ_1 and θ_2 and then get η from (iii). η=η(ξ) is the locus of the small object.

$$\underline{\boldsymbol{{small}}\:\boldsymbol{{object}}\:\boldsymbol{{at}}\:\boldsymbol{{distance}}\:\boldsymbol{{d}}_{\mathrm{1}} \:\boldsymbol{{to}}\:\boldsymbol{{left}}\:} \\ $$$$\underline{\boldsymbol{{end}}\:\boldsymbol{{and}}\:\boldsymbol{{deflection}}\:\boldsymbol{{f}}} \\ $$$$ \\ $$$${T}_{\mathrm{1}} \:\mathrm{cos}\:\theta_{\mathrm{1}} ={T}_{\mathrm{2}} \:\mathrm{cos}\:\theta_{\mathrm{2}} ={T}_{\mathrm{0}} \\ $$$${T}_{\mathrm{2}} \:\mathrm{sin}\:\theta_{\mathrm{2}} −{T}_{\mathrm{1}} \:\mathrm{sin}\:\theta_{\mathrm{1}} ={mg} \\ $$$${T}_{\mathrm{0}} \left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)={mg} \\ $$$${T}_{\mathrm{0}} =\frac{{mg}}{\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} } \\ $$$${a}=\frac{{T}_{\mathrm{0}} }{\rho{g}}=\frac{{mgL}}{{Mg}\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)}=\frac{\mu{L}}{\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} } \\ $$$${with}\:\mu=\frac{{m}}{{M}} \\ $$$$ \\ $$$$\underline{\boldsymbol{{for}}\:\boldsymbol{{rope}}\:\boldsymbol{{L}}_{\mathrm{1}} :} \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${at}\:{point}\:{C}: \\ $$$$\mathrm{tan}\:\theta_{\mathrm{1}} =\mathrm{sinh}\:\frac{{x}_{{C}} }{{a}} \\ $$$$\Rightarrow\frac{{x}_{{C}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right) \\ $$$${y}_{{C}} ={a}\:\mathrm{cosh}\:\frac{{x}_{{C}} }{{a}}={a}\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{\mathrm{1}} }=\frac{{a}}{\mathrm{cos}\:\theta_{\mathrm{1}} } \\ $$$${at}\:{point}\:{A}: \\ $$$${y}_{{C}} +{f}={a}\:\mathrm{cosh}\:\frac{{x}_{{C}} −{d}_{\mathrm{1}} }{{a}} \\ $$$$\Rightarrow\frac{{f}}{{a}}=\mathrm{cosh}\:\left(\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right)−\frac{{d}_{\mathrm{1}} }{{a}}\right)−\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} } \\ $$$${L}_{\mathrm{1}} ={a}\left(\mathrm{sinh}\:\frac{{x}_{{C}} }{{a}}−\mathrm{sinh}\:\frac{{x}_{{C}} −{d}_{\mathrm{1}} }{{a}}\right) \\ $$$$\Rightarrow\frac{{L}_{\mathrm{1}} }{{a}}=\mathrm{tan}\:\theta_{\mathrm{1}} −\mathrm{sinh}\:\left(\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right)−\frac{{d}_{\mathrm{1}} }{{a}}\right) \\ $$$$ \\ $$$$\underline{\boldsymbol{{for}}\:\boldsymbol{{rope}}\:\boldsymbol{{L}}_{\mathrm{2}} :} \\ $$$${y}={a}\:\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$${at}\:{point}\:{C}: \\ $$$$\mathrm{tan}\:\theta_{\mathrm{2}} =\mathrm{sinh}\:\frac{{x}_{{C}} }{{a}} \\ $$$$\Rightarrow\frac{{x}_{{C}} }{{a}}=\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{2}} \right) \\ $$$${y}_{{C}} ={a}\:\mathrm{cosh}\:\frac{{x}_{{C}} }{{a}}={a}\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta_{\mathrm{2}} }=\frac{{a}}{\mathrm{cos}\:\theta_{\mathrm{2}} } \\ $$$${at}\:{point}\:{B}: \\ $$$${y}_{{C}} +{f}={a}\:\mathrm{cosh}\:\frac{{x}_{{C}} +{d}_{\mathrm{2}} }{{a}} \\ $$$$\Rightarrow\frac{{f}}{{a}}=\mathrm{cosh}\:\left(\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{2}} \right)+\frac{{d}_{\mathrm{2}} }{{a}}\right)−\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{2}} } \\ $$$${L}_{\mathrm{2}} ={a}\left(\mathrm{sinh}\:\frac{{x}_{{C}} +{d}_{\mathrm{2}} }{{a}}−\mathrm{sinh}\:\frac{{x}_{{C}} }{{a}}\right) \\ $$$$\Rightarrow\frac{{L}_{\mathrm{2}} }{{a}}=\mathrm{sinh}\:\left(\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{2}} \right)+\frac{{d}_{\mathrm{2}} }{{a}}\right)−\mathrm{tan}\:\theta_{\mathrm{2}} \\ $$$${d}_{\mathrm{2}} ={d}−{d}_{\mathrm{1}} \\ $$$${L}_{\mathrm{1}} +{L}_{\mathrm{2}} ={L} \\ $$$$\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right)−\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right){d}_{\mathrm{1}} }{\mu{L}}\right]−\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} }=\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{2}} \right)+\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)\left({d}−{d}_{\mathrm{1}} \right)}{\mu{L}}\right]−\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{2}} } \\ $$$$\mathrm{sinh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right)−\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right){d}_{\mathrm{1}} }{\mu{L}}\right]−\mathrm{sinh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{2}} \right)+\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)\left({d}−{d}_{\mathrm{1}} \right)}{\mu{L}}\right]+\left(\mathrm{1}+\frac{\mathrm{1}}{\mu}\right)\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)=\mathrm{0} \\ $$$$\frac{{f}}{{a}}=\mathrm{cosh}\:\left(\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right)−\frac{{d}_{\mathrm{1}} }{{a}}\right)−\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} } \\ $$$${with}\:\xi=\frac{{d}_{\mathrm{1}} }{{d}},\:\eta=\frac{{f}}{{d}},\:\lambda=\frac{\mu{L}}{{d}},\:\mu=\frac{{m}}{{M}} \\ $$$$\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right)−\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)\xi}{\lambda}\right]−\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} }=\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{2}} \right)+\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)\left(\mathrm{1}−\xi\right)}{\lambda}\right]−\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{2}} }\:\:\:…\left({i}\right) \\ $$$$\mathrm{sinh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right)−\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)\xi}{\lambda}\right]−\mathrm{sinh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{2}} \right)+\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)\left(\mathrm{1}−\xi\right)}{\lambda}\right]+\left(\mathrm{1}+\frac{\mathrm{1}}{\mu}\right)\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\eta=\frac{\lambda}{\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} }\left\{\mathrm{cosh}\:\left[\mathrm{sinh}^{−\mathrm{1}} \:\left(\mathrm{tan}\:\theta_{\mathrm{1}} \right)−\frac{\left(\mathrm{tan}\:\theta_{\mathrm{2}} −\mathrm{tan}\:\theta_{\mathrm{1}} \right)\xi}{\lambda}\right]−\frac{\mathrm{1}}{\mathrm{cos}\:\theta_{\mathrm{1}} }\right\}\:\:\:…\left({iii}\right) \\ $$$${for}\:{given}\:\xi\in\left[\mathrm{0},\mathrm{1}\right]\:{we}\:{can}\:{solve}\:\left({i}\right)\:{and} \\ $$$$\left({ii}\right)\:{for}\:\theta_{\mathrm{1}} \:{and}\:\theta_{\mathrm{2}} \:{and}\:{then}\:{get}\:\eta\: \\ $$$${from}\:\left({iii}\right).\:\eta=\eta\left(\xi\right)\:{is}\:{the}\:{locus}\:{of}\:{the} \\ $$$${small}\:{object}. \\ $$

Commented by mr W last updated on 24/Feb/23