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Question-187789




Question Number 187789 by ajfour last updated on 21/Feb/23
Answered by mahdipoor last updated on 22/Feb/23
F_x =0⇒Tcos(θ)=(T+dT)cos(θ+dθ)  F_y =0⇒T(sinθ)+dW=(T+dT)sin(θ+dθ)  ⇒Lem I⇒   { ((0=−T.sin(θ).dθ+dT.cos(θ))),((g.dm=T.cos(θ).dθ+dT.sin(θ))) :}⇒  0=((−sin(θ))/(cos(θ)))dθ+(dT/T)⇒lnA=ln(cosθ)+ln(T)  ⇒A=Tcosθ=cte   dT=((Asin(θ).dθ)/(cos^2 (θ)))  dW=A.dθ+A((sin^2 θ)/(cos^2 θ)).dθ⇒  B+W=Aθ+(tanθ−θ)  ⇒⇒ { ((A=Tcosθ)),((B=(A−1)θ+tanθ−W)) :}     A,B=cte  for θ=α , T=T_i =(W/(2sinα))  A=(W/2)cotα , B=(A−1)α+tanα−W  −−−−−−−−−−−−  Lem I:  ⇒((d(cosθ))/dθ)=((cos(θ+dθ)−cos(θ))/dθ)=−sinθ  ⇒⇒cos(θ+dθ)=cos(θ)−sin(θ).dθ  ⇒((d(sinθ))/dθ)=((sin(θ+dθ)−sin(θ))/dθ)=cosθ  ⇒⇒sin(θ+dθ)=sin(θ)+cos(θ).dθ
$${F}_{{x}} =\mathrm{0}\Rightarrow{Tcos}\left(\theta\right)=\left({T}+{dT}\right){cos}\left(\theta+{d}\theta\right) \\ $$$${F}_{{y}} =\mathrm{0}\Rightarrow{T}\left({sin}\theta\right)+{dW}=\left({T}+{dT}\right){sin}\left(\theta+{d}\theta\right) \\ $$$$\Rightarrow{Lem}\:{I}\Rightarrow \\ $$$$\begin{cases}{\mathrm{0}=−{T}.{sin}\left(\theta\right).{d}\theta+{dT}.{cos}\left(\theta\right)}\\{{g}.{dm}={T}.{cos}\left(\theta\right).{d}\theta+{dT}.{sin}\left(\theta\right)}\end{cases}\Rightarrow \\ $$$$\mathrm{0}=\frac{−{sin}\left(\theta\right)}{{cos}\left(\theta\right)}{d}\theta+\frac{{dT}}{{T}}\Rightarrow{lnA}={ln}\left({cos}\theta\right)+{ln}\left({T}\right) \\ $$$$\Rightarrow{A}={Tcos}\theta={cte}\:\:\:{dT}=\frac{{Asin}\left(\theta\right).{d}\theta}{{cos}^{\mathrm{2}} \left(\theta\right)} \\ $$$${dW}={A}.{d}\theta+{A}\frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}.{d}\theta\Rightarrow \\ $$$${B}+{W}={A}\theta+\left({tan}\theta−\theta\right) \\ $$$$\Rightarrow\Rightarrow\begin{cases}{{A}={Tcos}\theta}\\{{B}=\left({A}−\mathrm{1}\right)\theta+{tan}\theta−{W}}\end{cases}\:\:\:\:\:{A},{B}={cte} \\ $$$${for}\:\theta=\alpha\:,\:{T}={T}_{{i}} =\frac{{W}}{\mathrm{2}{sin}\alpha} \\ $$$${A}=\frac{{W}}{\mathrm{2}}{cot}\alpha\:,\:{B}=\left({A}−\mathrm{1}\right)\alpha+{tan}\alpha−{W} \\ $$$$−−−−−−−−−−−− \\ $$$${Lem}\:\mathrm{I}: \\ $$$$\Rightarrow\frac{{d}\left({cos}\theta\right)}{{d}\theta}=\frac{{cos}\left(\theta+{d}\theta\right)−{cos}\left(\theta\right)}{{d}\theta}=−{sin}\theta \\ $$$$\Rightarrow\Rightarrow{cos}\left(\theta+{d}\theta\right)={cos}\left(\theta\right)−{sin}\left(\theta\right).{d}\theta \\ $$$$\Rightarrow\frac{{d}\left({sin}\theta\right)}{{d}\theta}=\frac{{sin}\left(\theta+{d}\theta\right)−{sin}\left(\theta\right)}{{d}\theta}={cos}\theta \\ $$$$\Rightarrow\Rightarrow{sin}\left(\theta+{d}\theta\right)={sin}\left(\theta\right)+{cos}\left(\theta\right).{d}\theta \\ $$

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