Equilibre D 39un Solide Soumis A 3 Forces Exercice Corrige Pdf Exclusive Official

Magnitude of R = √(80² + 60²) = 100 N Direction: tan(θ) = R_y / R_x (in absolute) = 60/80 = 0.75 → θ ≈ 36.87° above the negative x-axis (i.e., upward left).

Equilibrium of torques: T × 2.4 = 240 → T = 100 N Using force equilibrium in x and y: Horizontal: R_x + T_x = 0 . T_x = T × (4/5) = 100 × 0.8 = 80 N (negative direction). So R_x = -80 N (to the left). Vertical: R_y + T_y – P = 0 . T_y = T × (3/5) = 60 N upward. So R_y = P – T_y = 120 – 60 = 60 N upward. Magnitude of R = √(80² + 60²) =

Introduction In the world of physics and engineering, understanding the conditions under which an object remains at rest is fundamental. One of the most classic and critical concepts in statics is "l'équilibre d'un solide soumis à trois forces" (the equilibrium of a solid subjected to three forces). Whether you are a high school student in France, a student in francophone Africa, or an engineering aspirant, mastering this principle is non-negotiable. So R_x = -80 N (to the left)