Consider three force vectors F~ 1 with magnitude 46 N and direction 208 , F~ 2 w

Question

Consider three force vectors F~ 1 with magnitude 46 N and direction 208 , F~ 2 with magnitude 56 N and direction 263 , and F~ 3 with magnitude 88 N and direction 100 . All direction angles are measured from the positive x axis: counter-clockwise for > 0 and clockwise for < 0. What is the magnitude of the resultant vector kF~ k, where F~ = F~ 1 + F~ 2 + F~ 3 ? Answer in units of N.

Part 2: What is the direction of F~ as an angle between the limits of 180 and +180 from the positive x axis with counterclockwise as the positive angular direction? Answer in degrees.

Explanation / Answer

F1x = -46*cos(270-208) = -21.6 N

F1y = -46*sin(270-208) = -40.6 N

F2x = -56*cos(270-263) = -55.58 N

F2y = -56*sin(270-263) = -6.82 N

F3x = -88*cos(180-100) = -15.3 N

F3y = 88*sin(180-100) = 86.66 N

Fx = F1x+F2x+F3x = -21.6-55.58-15.3 = -92.5 N

Fy = -40.6-6.82+86.66 = 39.24 N

F = sqrt(Fx^2+Fy^2) = sqrt(92.5^2+39.24^2) = 100.5 N

part 2) direction is theta = tan^(-1)(39.24/92.5) = 23 deg

so finally the answer is 90+23 = 113 deg

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