3-9. Given that the force acting on a particle has the following components: Fx = −x + y, Fy = x − y + y², F₂ = 0. Solve for the potential energy V. -
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- The magnitude of the force F = 300 N, is applied at point A and its line of action passes through B.The force F is applied at point A and its line of action passes through B. Determine the following: 1. The position vector of point B with respect to point A: rBA = ( i + j + k)m 2. The unit vector in the direction of vector F: u = ( i + j + k) Express your results using three significant figures.Find the components of the force F =(14i-5j+2k)N acting in the direction of the vector a = 4i-3k. Hence, express as vectors the parallel and perpendicular components of F relative to a
- 2-30. Three cables pull on the pipe such that they create a resultant force having a magnitude of 900 lb. If two of the cables are subjected to known forces, as shown in the figure, determine the direction @ of the third cable so that the magnitude of force F in this cable is a minimum. All forces lie in the x-y plane. What is the magnitude of F? Hint: First find the resultant of the two known forces. 600 lb 400 Ib2-33. Four concurrent forces act on the plate. Determine the magnitude of the resultant force and its direction measurelycounterclockwise from the positive x axis. Problem 2-33 30° F₁ = 38 lb F₂ = 100 lb 3 F₁ = 60 lb 5 -X 4 F₁ = 50 lbThe resultant of the force system shown is