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Question Number 51466 by mr W last updated on 27/Dec/18

Commented by mr W last updated on 27/Dec/18

A rod of length b and mass M is  hanging at both ends with two ropes  on the ceiling, see picture. One rope  is of length l_1  and mass m_1 , the other  rope is of length l_2  and mass m_2 .   Assume l_2 >l_1 .  Find the inclination angle of the rod,  when it is in equilibrium.

ArodoflengthbandmassMishangingatbothendswithtworopesontheceiling,seepicture.Oneropeisoflengthl1andmassm1,theotherropeisoflengthl2andmassm2.Assumel2>l1.Findtheinclinationangleoftherod,whenitisinequilibrium.

Commented by ajfour last updated on 28/Dec/18

let tension at lower ends be T_1 , T_2  .  _________________________       T_1 cos α = T_2 cos β       T_1 sin α+T_2 sin β = Mg       T_1 sin (α+θ)= T_2 sin (β−θ)        x_1 +x_2  = bcos θ        y_2 −y_1  = bsin θ  _________________________    Tcos φ = constant = T_1 cos α  ⇒  dT = T tan φdφ    (T +dT )sin(φ+dφ)+ρgdl = T sin φ  ⇒ T cos φdφ+dT sin φ+ρgdl= 0  ⇒ Tcos φ(1+tan^2 φ)dφ=−ρgsec φdx    −T_1 cos α∫_φ_0  ^(  α) sec φdφ = ρg∫_0 ^(  x_1 ) dx  __________________________  ⇒(T_1 cos α)ln (((sec φ_0 +tan φ_0 )/(sec α+tan α)))=ρ_1 gx_1    &  (T_2 cos β)ln (((sec ψ_0 +tan ψ_0 )/(sec β+tan β)))=ρ_2 gx_2   __________________________   similarly  Tcos φ(1+tan^2 φ)dφ=−ρgcosec φdy  ⇒(T_1 cos α)∫_φ_0  ^(  α) sec φtan φdφ=−ρg∫_0 ^(  y_(1p) ) dy  __________________________  ⇒ T_1 cos α(sec φ_0 −sec α)= ρ_1 gy_1   &   T_2 cos β(sec ψ_0 −sec β)= ρ_2 gy_2   __________________________  Finally torque eq.   (ρgdl)((dl/2))cos φ = (Tdφ)dl  ⇒ ρgdl = 2Tcos φsec^2 φdφ  __________________________  or    m_1 g = 2T_1 cos α(tan φ_0 −tan α)  &  m_2 g = 2T_2 cos β(tan ψ_0 −tan β)  __________________________  unknowns are  T_1  , T_2  , 𝛂, 𝛃, 𝛗_0  , 𝛙_0 , x_1 , x_2 , y_1 , y_2 , 𝛉 .

lettensionatlowerendsbeT1,T2._________________________T1cosα=T2cosβT1sinα+T2sinβ=MgT1sin(α+θ)=T2sin(βθ)x1+x2=bcosθy2y1=bsinθ_________________________Tcosϕ=constant=T1cosαdT=Ttanϕdϕ(T+dT)sin(ϕ+dϕ)+ρgdl=TsinϕTcosϕdϕ+dTsinϕ+ρgdl=0Tcosϕ(1+tan2ϕ)dϕ=ρgsecϕdxT1cosαϕ0αsecϕdϕ=ρg0x1dx__________________________(T1cosα)ln(secϕ0+tanϕ0secα+tanα)=ρ1gx1&(T2cosβ)ln(secψ0+tanψ0secβ+tanβ)=ρ2gx2__________________________similarlyTcosϕ(1+tan2ϕ)dϕ=ρgcosecϕdy(T1cosα)ϕ0αsecϕtanϕdϕ=ρg0y1pdy__________________________T1cosα(secϕ0secα)=ρ1gy1&T2cosβ(secψ0secβ)=ρ2gy2__________________________Finallytorqueeq.(ρgdl)(dl2)cosϕ=(Tdϕ)dlρgdl=2Tcosϕsec2ϕdϕ__________________________orm1g=2T1cosα(tanϕ0tanα)&m2g=2T2cosβ(tanψ0tanβ)__________________________unknownsareT1,T2,α,β,ϕ0,ψ0,x1,x2,y1,y2,θ.

Commented by ajfour last updated on 27/Dec/18

Commented by ajfour last updated on 28/Dec/18

mrW Sir, this is all that i could..

mrWSir,thisisallthaticould..

Commented by mr W last updated on 29/Dec/18

thank you so far sir!   i got a solution using catenary method.

thankyousofarsir!igotasolutionusingcatenarymethod.

Answered by mr W last updated on 28/Dec/18

Commented by mr W last updated on 28/Dec/18

Commented by mr W last updated on 29/Dec/18

rod AB: length b, mass M  rope 1: length l_1 , mass m_1 , ρ_1 =m_1 /l_1   rope 2: length l_2 , mass m_2 , ρ_2 =m_2 /l_2   At point A:  tension in rope 1 = T_A   axialforce in rod = N_A   T_A =((Mg cos θ)/(2 sin (α+θ)))=((Mg)/(2 (sin α+cos α tan θ)))  N_A =((Mg cos α)/(2 sin (α+θ)))=((Mg)/(2 (tan α cos θ+sin θ)))  horizontal component in rope 1 = T_(10)   T_(10) =T_A  cos α=((Mg)/(2 (tan α+tan θ)))  a_1 =(T_(10) /(ρ_1 g))=((Ml_1 )/(2(tan α+tan θ)m_1 ))  eqn. of rope 1:  y=a_1 cosh (x/a_1 )  y′=sinh (x/a_1 )  s=a_1 sinh (x/a_1 )     (length of rope upon O)  at A:  y_A =a_1 cosh (e_1 /a_1 )  y′_A =sinh (e_1 /a_1 )=tan α ⇒ (e_1 /a_1 )=sinh^(−1)  (tan α)  s_A =a_1 sinh (e_1 /a_1 )=a_1 tan α  ⇒y_A =a_1 cosh (e_1 /a_1 )=a_1 cosh [sinh^(−1)  (tan α)]  at C:  y_C =a_1 cosh ((e_1 +f_1 )/a_1 )=a_1 cosh [sinh^(−1)  (tan α)+(f_1 /a_1 )]  =y_A +h_1 =a_1 cosh [sinh^(−1)  (tan α)]+h_1   ⇒a_1 cosh [sinh^(−1)  (tan α)+(f_1 /a_1 )]−a_1 cosh [sinh^(−1)  (tan α)]=h_1   s_C =a_1 sinh ((e_1 +f_1 )/a_1 )=a_1 sinh [sinh^(−1)  (tan α)+(f_1 /a_1 )]  =s_A +l_1 =a_1 tan α+l_1   ⇒sinh [sinh^(−1)  (tan α)+(f_1 /a_1 )]=tan α+(l_1 /a_1 )  ⇒sinh^(−1)  (tan α)+(f_1 /a_1 )=sinh^(−1)  (tan α+(l_1 /a_1 ))  ⇒(f_1 /a_1 )=sinh^(−1)  (tan α+(l_1 /a_1 ))−sinh^(−1)  (tan α)  ⇒a_1 cosh [sinh^(−1)  (tan α+(l_1 /a_1 ))]−a_1 cosh [sinh^(−1)  (tan α)]=h_1   At point B:  tension in rope 2 = T_B   axialforce in rod = N_B   T_A =((Mg cos θ)/(2 sin (β−θ)))=((Mg)/(2 (sin β−cos β tan θ)))  N_B =((Mg cos β)/(2 sin (β−θ)))=((Mg)/(2 (tan β cos θ−sin θ)))  horizontal component in rope 2 = T_(20)   T_(20) =T_B  cos β=((Mg)/(2 (tan β−tan θ)))  a_2 =(T_(20) /(ρ_2 g))=((Ml_2 )/(2(tan β−tan θ)m_2 ))  eqn. of rope 2:  y=a_2 cosh (x/a_2 )  y′=sinh (x/a_2 )  s=a_2 sinh (x/a_2 )   (length of rope upon O)  at B:  y_B =a_2 cosh (e_2 /a_2 )  y′_B =sinh (e_2 /a_2 )=tan β ⇒ (e_2 /a_2 )=sinh^(−1)  (tan β)  s_B =a_2 sinh (e_2 /a_2 )=a_2 tan β  ⇒y_B =a_2 cosh (e_2 /a_2 )=a_2 cosh [sinh^(−1)  (tan β)]  at C:  y_C =a_2 cosh ((e_2 +f_2 )/a_2 )=a_2 cosh [sinh^(−1)  (tan β)+(f_2 /a_2 )]  =y_B +h_2 =a_2 cosh [sinh^(−1)  (tan β)]+h_2   ⇒a_2 cosh [sinh^(−1)  (tan β)+(f_2 /a_2 )]−a_2 cosh [sinh^(−1)  (tan β)]=h_2   s_C =a_2 sinh ((e_2 +f_2 )/a_2 )=a_2 sinh [sinh^(−1)  (tan β)+(f_2 /a_2 )]  =s_A +l_2 =a_2 tan β+l_2   ....... to be continued ......

rodAB:lengthb,massMrope1:lengthl1,massm1,ρ1=m1/l1rope2:lengthl2,massm2,ρ2=m2/l2AtpointA:tensioninrope1=TAaxialforceinrod=NATA=Mgcosθ2sin(α+θ)=Mg2(sinα+cosαtanθ)NA=Mgcosα2sin(α+θ)=Mg2(tanαcosθ+sinθ)horizontalcomponentinrope1=T10T10=TAcosα=Mg2(tanα+tanθ)a1=T10ρ1g=Ml12(tanα+tanθ)m1eqn.ofrope1:y=a1coshxa1y=sinhxa1s=a1sinhxa1(lengthofropeuponO)atA:yA=a1coshe1a1yA=sinhe1a1=tanαe1a1=sinh1(tanα)sA=a1sinhe1a1=a1tanαyA=a1coshe1a1=a1cosh[sinh1(tanα)]atC:yC=a1coshe1+f1a1=a1cosh[sinh1(tanα)+f1a1]=yA+h1=a1cosh[sinh1(tanα)]+h1a1cosh[sinh1(tanα)+f1a1]a1cosh[sinh1(tanα)]=h1sC=a1sinhe1+f1a1=a1sinh[sinh1(tanα)+f1a1]=sA+l1=a1tanα+l1sinh[sinh1(tanα)+f1a1]=tanα+l1a1sinh1(tanα)+f1a1=sinh1(tanα+l1a1)f1a1=sinh1(tanα+l1a1)sinh1(tanα)a1cosh[sinh1(tanα+l1a1)]a1cosh[sinh1(tanα)]=h1AtpointB:tensioninrope2=TBaxialforceinrod=NBTA=Mgcosθ2sin(βθ)=Mg2(sinβcosβtanθ)NB=Mgcosβ2sin(βθ)=Mg2(tanβcosθsinθ)horizontalcomponentinrope2=T20T20=TBcosβ=Mg2(tanβtanθ)a2=T20ρ2g=Ml22(tanβtanθ)m2eqn.ofrope2:y=a2coshxa2y=sinhxa2s=a2sinhxa2(lengthofropeuponO)atB:yB=a2coshe2a2yB=sinhe2a2=tanβe2a2=sinh1(tanβ)sB=a2sinhe2a2=a2tanβyB=a2coshe2a2=a2cosh[sinh1(tanβ)]atC:yC=a2coshe2+f2a2=a2cosh[sinh1(tanβ)+f2a2]=yB+h2=a2cosh[sinh1(tanβ)]+h2a2cosh[sinh1(tanβ)+f2a2]a2cosh[sinh1(tanβ)]=h2sC=a2sinhe2+f2a2=a2sinh[sinh1(tanβ)+f2a2]=sA+l2=a2tanβ+l2.......tobecontinued......

Commented by mr W last updated on 29/Dec/18

⇒sinh [sinh^(−1)  (tan β)+(f_2 /a_2 )]=tan β+(l_2 /a_2 )  ⇒sinh^(−1)  (tan β)+(f_2 /a_2 )=sinh^(−1)  (tan β+(l_2 /a_2 ))  ⇒(f_2 /a_2 )=sinh^(−1)  (tan β+(l_2 /a_2 ))−sinh^(−1)  (tan β)  ⇒a_2 cosh [sinh^(−1)  (tan β+(l_2 /a_2 ))]−a_2 cosh [sinh^(−1)  (tan β)]=h_2     N_A =N_B :  ⇒((Mg)/(2 (tan α cos θ+sin θ)))=((Mg)/(2 (tan β cos θ−sin θ)))  ⇒tan α cos θ+sin θ=tan β cos θ−sin θ  ⇒tan θ=((tan β−tan α)/2)    ...(iii)  ⇒a_1 =((Ml_1 )/((tan α+tan β)m_1 ))⇒(l_1 /a_1 )=(((tan α+tan β)m_1 )/M)  ⇒a_2 =((Ml_2 )/((tan α+tan β)m_2 ))⇒(l_2 /a_2 )=(((tan α+tan β)m_2 )/M)  f_1 +f_2 =b cos 𝛉:  ⇒a_1 {sinh^(−1)   (tan α+(l_1 /a_1 ))−sinh^(−1)  (tan α)}+a_2 {sinh^(−1)  (tan β+(l_2 /a_2 ))−sinh^(−1)  (tan β)}=b cos θ  ⇒((Ml_1 )/((tan α+tan β)m_1 b)){sinh^(−1)  [(1+(m_1 /M))tan α+(m_1 /M) tan β]−sinh^(−1)  (tan α)}+((Ml_2 )/((tan α+tan β)m_2 b)){sinh^(−1)  ((m_2 /M) tan α+(1+(m_2 /M))tan β)−sinh^(−1)  (tan β)}=(2/(√(4+(tan β−tan α)^2 )))  ⇒((Ml_1 )/(m_1 b)){sinh^(−1)  [(1+(m_1 /M))tan α+(m_1 /M) tan β]−sinh^(−1)  (tan α)}+((Ml_2 )/(m_2 b)){sinh^(−1)  ((m_2 /M) tan α+(1+(m_2 /M))tan β)−sinh^(−1)  (tan β)}=((2 (tan β+tan α))/(√(4+(tan β−tan α)^2 )))   h_2 −h_1 =b sin 𝛉:  ⇒a_2 {cosh [sinh^(−1)  (tan β+(l_2 /a_2 ))]−cosh [sinh^(−1)  (tan β)]}−a_1 {cosh [sinh^(−1)  (tan α+(l_1 /a_1 ))]−cosh [sinh^(−1)  (tan α)]}=b sin θ  ⇒((Ml_2 )/((tan α+tan β)m_2 b)){cosh [sinh^(−1)  ((m_2 /M) tan α+(1+(m_2 /M))tan β)]−cosh [sinh^(−1)  (tan β)]}−((Ml_1 )/((tan α+tan β)m_1 b)){cosh [sinh^(−1)  ((1+(m_1 /M))tan α+(m_1 /M) tan β)]−cosh [sinh^(−1)  (tan α)]}=((tan β−tan α)/(√(4+(tan β−tan α)^2 )))  ⇒((Ml_2 )/(m_2 b)){cosh [sinh^(−1)  ((m_2 /M) tan α+(1+(m_2 /M))tan β)]−cosh [sinh^(−1)  (tan β)]}−((Ml_1 )/(m_1 b)){cosh [sinh^(−1)  ((1+(m_1 /M))tan α+(m_1 /M) tan β)]−cosh [sinh^(−1)  (tan α)]}=((tan^2  β−tan^2  α)/(√(4+(tan β−tan α)^2 )))  with λ_1 =(l_1 /b), λ_2 =(l_2 /b), μ_1 =(m_1 /M), μ_2 =(m_2 /M)  (𝛌_2 /𝛍_2 ){sinh^(−1) (𝛍_2 tan 𝛂+(1+𝛍_2 )tan 𝛃)−sinh^(−1) (tan 𝛃)}+(𝛌_1 /𝛍_1 ){sinh^(−1) [(1+𝛍_1 )tan 𝛂+𝛍_1 tan 𝛃]−sinh^(−1) (tan 𝛂)}=((2 (tan 𝛃+tan 𝛂))/(√(4+(tan 𝛃−tan 𝛂)^2 )))     ...(i)  (𝛌_2 /𝛍_2 ){cosh [sinh^(−1) (𝛍_2 tan 𝛂+(1+𝛍_2 )tan 𝛃)]−cosh [sinh^(−1) (tan 𝛃)]}−(𝛌_1 /𝛍_1 ){cosh [sinh^(−1) ((1+𝛍_1 )tan 𝛂+𝛍_1 tan 𝛃)]−cosh [sinh^(−1) (tan 𝛂)]}=((tan^2  𝛃−tan^2  𝛂)/(√(4+(tan 𝛃−tan 𝛂)^2 )))     ...(ii)  we obtain tan α and tan β from (i) and (ii),  and then θ from (iii).

sinh[sinh1(tanβ)+f2a2]=tanβ+l2a2sinh1(tanβ)+f2a2=sinh1(tanβ+l2a2)f2a2=sinh1(tanβ+l2a2)sinh1(tanβ)a2cosh[sinh1(tanβ+l2a2)]a2cosh[sinh1(tanβ)]=h2NA=NB:Mg2(tanαcosθ+sinθ)=Mg2(tanβcosθsinθ)tanαcosθ+sinθ=tanβcosθsinθtanθ=tanβtanα2...(iii)a1=Ml1(tanα+tanβ)m1l1a1=(tanα+tanβ)m1Ma2=Ml2(tanα+tanβ)m2l2a2=(tanα+tanβ)m2Mf1+f2=bcosθ:a1{sinh1(tanα+l1a1)sinh1(tanα)}+a2{sinh1(tanβ+l2a2)sinh1(tanβ)}=bcosθMl1(tanα+tanβ)m1b{sinh1[(1+m1M)tanα+m1Mtanβ]sinh1(tanα)}+Ml2(tanα+tanβ)m2b{sinh1(m2Mtanα+(1+m2M)tanβ)sinh1(tanβ)}=24+(tanβtanα)2Ml1m1b{sinh1[(1+m1M)tanα+m1Mtanβ]sinh1(tanα)}+Ml2m2b{sinh1(m2Mtanα+(1+m2M)tanβ)sinh1(tanβ)}=2(tanβ+tanα)4+(tanβtanα)2h2h1=bsinθ:a2{cosh[sinh1(tanβ+l2a2)]cosh[sinh1(tanβ)]}a1{cosh[sinh1(tanα+l1a1)]cosh[sinh1(tanα)]}=bsinθMl2(tanα+tanβ)m2b{cosh[sinh1(m2Mtanα+(1+m2M)tanβ)]cosh[sinh1(tanβ)]}Ml1(tanα+tanβ)m1b{cosh[sinh1((1+m1M)tanα+m1Mtanβ)]cosh[sinh1(tanα)]}=tanβtanα4+(tanβtanα)2Ml2m2b{cosh[sinh1(m2Mtanα+(1+m2M)tanβ)]cosh[sinh1(tanβ)]}Ml1m1b{cosh[sinh1((1+m1M)tanα+m1Mtanβ)]cosh[sinh1(tanα)]}=tan2βtan2α4+(tanβtanα)2withλ1=l1b,λ2=l2b,μ1=m1M,μ2=m2Mλ2μ2{sinh1(μ2tanα+(1+μ2)tanβ)sinh1(tanβ)}+λ1μ1{sinh1[(1+μ1)tanα+μ1tanβ]sinh1(tanα)}=2(tanβ+tanα)4+(tanβtanα)2...(i)λ2μ2{cosh[sinh1(μ2tanα+(1+μ2)tanβ)]cosh[sinh1(tanβ)]}λ1μ1{cosh[sinh1((1+μ1)tanα+μ1tanβ)]cosh[sinh1(tanα)]}=tan2βtan2α4+(tanβtanα)2...(ii)weobtaintanαandtanβfrom(i)and(ii),andthenθfrom(iii).

Commented by ajfour last updated on 29/Dec/18

Long story Sir, i have saved; shall  take print out later.

LongstorySir,ihavesaved;shalltakeprintoutlater.

Commented by mr W last updated on 29/Dec/18

thanks for viewing sir!  i′m not able to make the equations  more simple.

thanksforviewingsir!imnotabletomaketheequationsmoresimple.

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