Answer:
Given the components of the vector F, we can calculate the angle it makes with the +x-axis:Fx = 0.140 kN
Step-by-step explanation:
Fy = -0.600 kNThe magnitude of the vector is:|F| = √(Fx^2 + Fy^2) = √(0.140^2 + (-0.600)^2) = √(0.0196 + 0.360) ≈ 0.629 kNThe angle θ is given by:tan(θ) = Fy /Fx = -0.600 / 0.140 ≈ -4.29To find the angle θ, we need to take the inverse tangent (arctangent) of the result:θ ≈ arctan(-4.29) ≈ -75.5°Since the angle is in the third quadrant, we need to add 180° to get the angle clockwise from the +x-axis:θ ≈ -75.5° + 180° ≈ 104.5°So, the vector makes an angle of approximately 104.5° with the +x-axis in the clockwise direction.