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