As shown in the figure A 0 0 6 B 2 3 0 C 3 2 0 represent thr
Solution
Answer -
Since the coordinates of Points A, B and C are (0,0,6), (2,-3,0) and (3,2,0) in the three dimensions on a flag pole and the values of Force along B and C are;
FB = 560 N, FC = 700 N.
The values of FB along three points can be given as;
Along A Point = 560*0 x +560*0 y +560*6 k = 3360 k
Along B Point = 560*2 x - 560*3 y +560*0 k = 1120 x - 1680 y
Along C Point = 560*3 x +560*2 y +560*0 k = 1680 x +1120 y
Thus, the value of FB in cartesian vector form;
FB = 3360 k +1120 x -1680 y +1680 x +1120 y = 2800x - 560y + 3360k
Now, The values of FC along three points can be given as;
Along A Point = 700*0 x +700*0 y +700*6 k = 4200 k
Along B Point = 700*2 x - 700*3 y +700*0 k = 1400 x - 2100 y
Along C Point = 700*3 x +700*2 y +700*0 k = 2100 x +1400 y
Thus, the value of FC in cartesian vector form;
FB = 4200 k +1400 x -2100 y +2100 x +1400 y = 3500x - 700y + 4200k
