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Question Number 187789 by ajfour last updated on 21/Feb/23

	Answered by mahdipoor last updated on 22/Feb/23
	$F_x =0⇒Tcos(θ)=(T+dT)cos(θ+dθ) F_y =0⇒T(sinθ)+dW=(T+dT)sin(θ+dθ) ⇒Lem I⇒ { ((0=−T.sin(θ).dθ+dT.cos(θ))),((g.dm=T.cos(θ).dθ+dT.sin(θ))) :}⇒ 0=((−sin(θ))/(cos(θ)))dθ+(dT/T)⇒lnA=ln(cosθ)+ln(T) ⇒A=Tcosθ=cte dT=((Asin(θ).dθ)/(cos^2 (θ))) dW=A.dθ+A((sin^2 θ)/(cos^2 θ)).dθ⇒ B+W=Aθ+(tanθ−θ) ⇒⇒ { ((A=Tcosθ)),((B=(A−1)θ+tanθ−W)) :} A,B=cte for θ=α , T=T_i =(W/(2sinα)) A=(W/2)cotα , B=(A−1)α+tanα−W −−−−−−−−−−−− Lem I: ⇒((d(cosθ))/dθ)=((cos(θ+dθ)−cos(θ))/dθ)=−sinθ ⇒⇒cos(θ+dθ)=cos(θ)−sin(θ).dθ ⇒((d(sinθ))/dθ)=((sin(θ+dθ)−sin(θ))/dθ)=cosθ ⇒⇒sin(θ+dθ)=sin(θ)+cos(θ).dθ$
	$F_{x} = 0 \Rightarrow T c o s (θ) = (T + d T) c o s (θ + d θ)$ $F_{y} = 0 \Rightarrow T (s i n θ) + d W = (T + d T) s i n (θ + d θ)$ $\Rightarrow L e m I \Rightarrow$ ${\begin{cases} 0 = - T . s i n (θ) . d θ + d T . c o s (θ) \\ g . d m = T . c o s (θ) . d θ + d T . s i n (θ) \end{cases} \Rightarrow$ $0 = \frac{- s i n (θ)}{c o s (θ)} d θ + \frac{d T}{T} \Rightarrow l n A = l n (c o s θ) + l n (T)$ $\Rightarrow A = T c o s θ = c t e d T = \frac{A s i n (θ) . d θ}{{c o s}^{2} (θ)}$ $d W = A . d θ + A \frac{{s i n}^{2} θ}{{c o s}^{2} θ} . d θ \Rightarrow$ $B + W = A θ + (t a n θ - θ)$ $\Rightarrow\Rightarrow {\begin{cases} A = T c o s θ \\ B = (A - 1) θ + t a n θ - W \end{cases} A, B = c t e$ $f o r θ = α, T = T_{i} = \frac{W}{2 s i n α}$ $A = \frac{W}{2} c o t α, B = (A - 1) α + t a n α - W$ $- - - - - - - - - - - -$ $L e m I :$ $\Rightarrow \frac{d (c o s θ)}{d θ} = \frac{c o s (θ + d θ) - c o s (θ)}{d θ} = - s i n θ$ $\Rightarrow\Rightarrow c o s (θ + d θ) = c o s (θ) - s i n (θ) . d θ$ $\Rightarrow \frac{d (s i n θ)}{d θ} = \frac{s i n (θ + d θ) - s i n (θ)}{d θ} = c o s θ$ $\Rightarrow\Rightarrow s i n (θ + d θ) = s i n (θ) + c o s (θ) . d θ$
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