So ( R = \frac200\sin\alpha = \frac200\sin 67.2° \approx \frac2000.922 \approx 216.9 , N).
But ( R_x = R \cos(\alpha) ), ( R_y = R \sin(\alpha) ), where ( \alpha ) = angle of ( R ) with horizontal.
Forces in x-direction: [ R_x = T \quad (\textsince R \text has a horizontal component toward the right) ] So ( R = \frac200\sin\alpha = \frac200\sin 67
Question: Trouvez les tensions ( T_1 ) et ( T_2 ) dans les câbles.
Forces in y-direction: [ R_y = W = 200 , N ] Forces in y-direction: [ R_y = W =
Numerically: (\tan50° \approx 1.1918) → ( \tan\alpha \approx 2.3836) → ( \alpha \approx 67.2°) above horizontal? That seems too steep. Let's check: I is above and left of A? No, A is at origin, I has x positive (2.5cos50°=1.607), y positive (5sin50°=3.83). So R points up-right? But rope pulls left, so hinge must pull right-up to balance. Yes, so R angle ≈ 67° from horizontal upward right.
Ignore friction at the hinge.
Then equilibrium: Horizontal: ( R\cos\alpha = T ), Vertical: ( R\sin\alpha = W = 200 ) N.