Q.179. Static Equilibrium.

Question 179.

Find the resultant of the forces F1, F2, F3 and F4 in the form
R = Rx i + Ry j + Rz k.

Answer 179.

F1
= 700/√[(0.3)^2 + (-0.6)^2 + (0.2)^2] * [0.2 i - 0.6 j + 0.3 k]
= 1000 * [0.2 i - 0.6 j + 0.3 k]
= (200 i - 600 j + 300 k) N

cos^2 (θ2x)+cos^2 (θ2y)+cos^2 (θ2z) = 1
=> cos^2 (135°) +cos^2 (θ2y) +cos^2 (120°) = 1
=> 1/2 + cos^2 (θ2y) + 1/4 = 1
=> cos^2 (θ2y) = 1/4
=> cos(θ2y) = 1/2
... [because θ2y is acute from the figure]
=> θ2y = 60°

=> F2
= 1000 [cos(135°) i + cos(60°) j + cos(120°) k]
= (- 500√2 i + 500 j - 500 k) N

F3 is in YZ-plane
=> it makes 90° with x-axis and 60° with y-axis and 150° with z-axis ... [from the figure]
=> F3
= 470 [cos(90°) i + cos(60°) y + cos(150°) k]
= (235 j - 235√3 k) N

F4 = 175 i N

R = F1 + F2 + F3 + F4
= (200i - 600 j + 300 k) N + (- 500√2 i + 500 j - 500 k) N + (235 j - 235√3 k) N + (175 i) N
= (- 332 i + 135 j - 607 k) N.