Equilibre D 39-un Solide Soumis A 3 Forces Exercice Corrige Pdf 100%

Now slope of AI: (\tan(\alpha) = \fracy_I - 0x_I - 0 = \frac5 \sin50°2.5 \cos50° = 2 \tan50°).

Then equilibrium: Horizontal: ( R\cos\alpha = T ), Vertical: ( R\sin\alpha = W = 200 ) N. Now slope of AI: (\tan(\alpha) = \fracy_I -

So ( R = \frac200\sin\alpha = \frac200\sin 67.2° \approx \frac2000.922 \approx 216.9 , N). Intersection I: Horizontal line through B: y_B = 5 sin50°

Given the intersection I, distances: Let’s put coordinates: A = (0,0), B = (5 cos50°, 5 sin50°). Weight at midpoint M = (2.5 cos50°, 2.5 sin50°). Rope at B, horizontal left. Intersection I: Horizontal line through B: y_B = 5 sin50°. Vertical through M: x_M = 2.5 cos50°. Given the intersection I

Forces in y-direction: [ R_y = W = 200 , N ]

Question: Trouvez les tensions ( T_1 ) et ( T_2 ) dans les câbles.