Then equilibrium: Horizontal: ( R\cos\alpha = T ), Vertical: ( R\sin\alpha = W = 200 ) N.
So I = (2.5 cos50°, 5 sin50°).
Question: Trouvez les tensions ( T_1 ) et ( T_2 ) dans les câbles. Then equilibrium: Horizontal: ( R\cos\alpha = T ),
Ignore friction at the hinge.
But ( R_x = R \cos(\alpha) ), ( R_y = R \sin(\alpha) ), where ( \alpha ) = angle of ( R ) with horizontal. Ignore friction at the hinge
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°. Weight at midpoint M = (2
Now slope of AI: (\tan(\alpha) = \fracy_I - 0x_I - 0 = \frac5 \sin50°2.5 \cos50° = 2 \tan50°).