Meriam, Kraige & Bolton — Statics of Particles: Problem 2_155

⚡ Mecademy AIENG정역학 · ch2  Problem Statement 2/155 During a drilling operation, the small robotic device is subjected to an 800-N force at point C as shown. Replace this force by an equivalent force–couple system at point O. Problem 2/155 ● 1. Analysis of Force Vector The force has a magnitude of and acts at point . Based on the diagram, the force is directed from the workpiece towards the robotic arm along the axis of the drill bit (reaction force). The geometry of the drill bit is defined by two angles: A tilt of from the vertical (-axis). A rotation of in the horizontal plane from the -axis towards the -axis. Since the force acts from the ground towards the tool head (point ), it has a positive - component and negative and components (assuming the drill points "forward" in the direction). Let's define the unit vector along the force's line of action: Substituting the given values: The resultant force vector at is: ● 2. Calculation of Moment F F800 NC 30 ∘ z 20 ∘ xy FBz xy +x,+yu F F = x −Fsin30cos20 ∘∘ F = y −Fsin30sin20 ∘∘ F = z Fcos30 ∘ F = x −800⋅sin(30)⋅ ∘ cos(20)= ∘ −800⋅0.5⋅0.9397=−375.88 N F = y −800⋅sin(30)⋅ ∘ sin(20)= ∘ −800⋅0.5⋅0.3420=−136.81 N F = z 800⋅cos(30)= ∘ 800⋅0.8660=692.82 N O R=(−375.9i−136.8j+692.8k) N M O To find the equivalent couple at point , we calculate the moment of force about . By the principle of transmissibility, since the force acts along the axis of the drill bit which passes through joint , we can use the position vector of point to cal