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 \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