(Note: For the purpose of this article, we illustrate the content. In a real scenario, a download link or email capture would appear here.) The equilibrium of a solid under three forces is a cornerstone of mechanics. Understanding it requires both conceptual clarity and rigorous practice. With our exclusive PDF of corrected exercises , you move from passive reading to active mastery. Each exercise is a building block toward solving real-world static problems with confidence.
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).
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| Feature | Benefit | |---------|---------| | Detailed diagrams | Visualize forces and concurrency point | | Step-wise reasoning | Learn the methodology, not just the answer | | Both graphical & analytical solutions | Understand the link between geometry and algebra | | Real exam-style problems | Prepare for BAC or entrance exams | | Exclusive content | No generic problems – carefully curated |
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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 torque from T = T × 2.4 m (counterclockwise, balancing P). With our exclusive PDF of corrected exercises ,
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.