-- -- -- -- |.7071 0 0 -1 -.8660 0 0 0 0 | | 0 | |.7071 0 1 0 .5 0 0 0 0 | |1000| | 0 1 0 0 0 -1 0 0 0 | | 0 | | 0 0 -1 0 0 0 0 0 0 | | 0 | | 0 0 0 1 0 0 0 0 -.7071| F = | 0 | | 0 0 0 0 0 0 1 0 .7071| |-500| | 0 0 0 0 .8660 1 0 -1 0 | | 0 | | 0 0 0 0 -.5 0 -1 0 0 | | 500| | 0 0 0 0 0 0 0 1 .7071| | 0 | -- -- -- -- How this linear system of equations was derived: ------------------------------------------------ In a statically determinate trust with frictionless pin joints, the tension (Fi) in each member can be obtained from setting the sums of all forces acting horizontally or vertically at each pin equal to zero, the angle at each of the junction is indicated above): down | negative (-) left <- negative (-) is convention \-/ joint1 is fixed joint2: x sin pi/4 F1 - F4 - sin pi/3 F5 = 0 [row1] y cos pi/4 F1 + F3 + cos pi/3 F5 - 1000 = 0 [row2 after move -1000] joint3: x F2 - F6 = 0 [row3] y -F3 = 0 [row4] joint4: x F4 - cos pi/4 F9 = 0 [row5] y 500 + F7 + sin pi/4 F9 = 0 [row6 after move 500] joint5: x -F8 + F6 + sin pi/3 F5 = 0 [row7] y -500 - F7 - cos pi/3 F5 = 0 [row8 after move -500] joint6: x F8 + cos pi/4 F9 = 0 [row9] y n/a (since on rollers and no vertical motion) cos(pi/4)=sin(pi/4)=.7071 sin(pi/6)=cos(pi/3)=.5