Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 45082 by ajfour last updated on 08/Oct/18

Commented by ajfour last updated on 08/Oct/18

Q.44872 (solution)

Q.44872(solution)

Answered by ajfour last updated on 09/Oct/18

Tcos θdθ+dTsin θ = λg(dx)sec θ  ..(i)  dTcos θ = Tsin θdθ     ...(ii)  ⇒ Tcos θ = T_0 cos α = (λgbsin α)cos α          = c  from (i)      d(Tsin θ)=λg(dx)sec θ  ⇒  cos θ d(((csin θ)/(cos θ)))=λgdx  ⇒ c∫_(tan α) ^(  tan θ) ((d(tan θ))/(√(1+tan^2 θ))) = λgx  let tan θ = z  ⇒ ∫_z_0  ^(  z) (dz/(√(1+z^2 ))) = ((λgx)/c)  ln (z+(√(1+z^2 )) )∣_z_0  ^z  =((λgx)/c)  ln (((z+(√(1+z^2 )))/(tan α+sec α)))= ((λgx)/c)  ⇒ z+(√(1+z^2 )) =(sec α+tan α) e^(λgx/c)    tan θ+sec θ = ke^(μx)        ....(A)   k = sec α+tan α ; μ= ((λg)/c)  Also  sec θ−tan θ = (e^(−μx) /k)    ...(B)  from eq.(A)−eq.(B):  we get    tan θ = (dy/dx) = (1/2)(ke^(𝛍x) −(e^(−𝛍x) /k))  y = ((k/(2𝛍))e^(𝛍x) +(1/(2𝛍k))e^(−𝛍x) )∣_0 ^x   y= (1/(2𝛍))[(ke^(𝛍x) +(1/(ke^(𝛍x) )))−(k+(1/k))  y=(c/(2𝛌g))[k(e^(𝛍x) −1)+(1/k)(e^(−𝛍x) −1)]       y= ((bsin αcos α)/2)[k(e^(μx) −1)+(1/k)(e^(−μx) −1)]  where k = sec 𝛂+tan 𝛂  and      𝛍= (1/(bsin 𝛂cos 𝛂))   ⇒   y =bsin 𝛂[cosh (((2x)/(bsin 2𝛂)))+sin 𝛂sinh (((2x)/(bsin 2𝛂)))−1].

Tcosθdθ+dTsinθ=λg(dx)secθ..(i)dTcosθ=Tsinθdθ...(ii)Tcosθ=T0cosα=(λgbsinα)cosα=cfrom(i)d(Tsinθ)=λg(dx)secθcosθd(csinθcosθ)=λgdxctanαtanθd(tanθ)1+tan2θ=λgxlettanθ=zz0zdz1+z2=λgxcln(z+1+z2)z0z=λgxcln(z+1+z2tanα+secα)=λgxcz+1+z2=(secα+tanα)eλgx/ctanθ+secθ=keμx....(A)k=secα+tanα;μ=λgcAlsosecθtanθ=eμxk...(B)fromeq.(A)eq.(B):wegettanθ=dydx=12(keμxeμxk)y=(k2μeμx+12μkeμx)0xy=12μ[(keμx+1keμx)(k+1k)y=c2λg[k(eμx1)+1k(eμx1)]y=bsinαcosα2[k(eμx1)+1k(eμx1)]wherek=secα+tanαandμ=1bsinαcosαy=bsinα[cosh(2xbsin2α)+sinαsinh(2xbsin2α)1].

Commented by MrW3 last updated on 09/Oct/18

Commented by MrW3 last updated on 09/Oct/18

in coordinate system uv the rope  has the equation  v=a cosh (u/a)  with a=(T_H /(λg))=b sin α cos α=(1/μ)  (dv/du)=sinh (u/a)  at (u_1 ,v_1 ):  sinh (u_1 /a)=tan α  ⇒(u_1 /a)=sinh^(−1)  (tan α)=ln (tan α+(√(1+tan^2  α)))  =ln (tan α+sec α)=ln k  v_1 =a cosh (u_1 /a)=a cosh (ln k)=a×((k+(1/k))/2)  =(1/(2μ))(k+(1/k))    u=u_1 +x  v=v_1 +y  v_1 +y=a cosh ((u_1 +x)/a)  (1/(2μ))(k+(1/k))+y=(1/μ) cosh (ln k+μx)  (1/(2μ))(k+(1/k))+y=(1/(2μ)) (e^(ln k+μx) +e^(−ln k−μx) )  (1/(2μ))(k+(1/k))+y=(1/(2μ)) (ke^(μx) +(1/k)e^(−μx) )  ⇒y=(1/(2μ)) [k(e^(μx) −1)+(1/k)(e^(−μx) −1)]

incoordinatesystemuvtheropehastheequationv=acoshuawitha=THλg=bsinαcosα=1μdvdu=sinhuaat(u1,v1):sinhu1a=tanαu1a=sinh1(tanα)=ln(tanα+1+tan2α)=ln(tanα+secα)=lnkv1=acoshu1a=acosh(lnk)=a×k+1k2=12μ(k+1k)u=u1+xv=v1+yv1+y=acoshu1+xa12μ(k+1k)+y=1μcosh(lnk+μx)12μ(k+1k)+y=12μ(elnk+μx+elnkμx)12μ(k+1k)+y=12μ(keμx+1keμx)y=12μ[k(eμx1)+1k(eμx1)]

Commented by ajfour last updated on 08/Oct/18

thanks sir, i found grave error,  and have edited quite a handful..

thankssir,ifoundgraveerror,andhaveeditedquiteahandful..

Commented by ajfour last updated on 08/Oct/18

genuine point, Sir.

genuinepoint,Sir.

Commented by MrW3 last updated on 09/Oct/18

sir you have solved the problem from  the root, it′s perfect! wonderful!

siryouhavesolvedtheproblemfromtheroot,itsperfect!wonderful!

Commented by ajfour last updated on 09/Oct/18

thank you so much Sir! shall no  more be all-afraid of hyperbolic functions.

thankyousomuchSir!shallnomorebeallafraidofhyperbolicfunctions.

Commented by MrW3 last updated on 09/Oct/18

it′s a great and ingenious trick to get  𝛉 = (𝛑/2)−2tan^(−1) (1/(ke^(μx) )) from   ((1+sin θ)/(cos θ)) = ke^(𝛍x)   I shall keep this in mind.

itsagreatandingenioustricktogetθ=π22tan11keμxfrom1+sinθcosθ=keμxIshallkeepthisinmind.

Terms of Service

Privacy Policy

Contact: info@tinkutara.com