Consider three force vectors F~ 1 with magnitude 46 N and direction 150 ◦?
Consider three force vectors F~ 1 with magnitude 46 N and direction 150 ◦ , F~ 2 with magnitude 32 N and direction −160 ◦ , and F~ 3 with magnitude 10 N and direction 160 ◦ . All direction angles θ are measured from the positive x axis: counter-clockwise for θ > 0 and clockwise for θ < 0. What is the magnitude of the resultant vector F~ = F~ 1 + F~ 2 + F~ 3 ? Answer in units of N
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Take the components of each force - sin and cos - so you get x and y vectors for each force. Add the x's together and the y's together. Square the results and add them. Take the root of that number and it is your answer.
i'm no longer one hundred% particular if I understand your question thoroughly, yet from what i'm assuming is you're attempting to discover the perspective of FR? if so then your FRx = -37.766N, FRy = 89.56N, so which you're perspective of FR could be sixty 8.23 clockwise from detrimental x axis or 111.seventy six counter clockwise from useful x axis.