Two wrenches to single wrench; line of action

Reading of the figure: wrench 1 at O along the y axis: force 30 N along +y with a 50 N·m couple along the y axis (drawn opposite to the force, so −y). Wrench 2 at P on the z axis, 8 m from O: force 40 N along +x and a 90 N·m couple along +x.

(a) Resultant force

R = 40 i + 30 j = (40, 30, 0) N |R| = √(40² + 30²) = 50 N, at tan⁻¹(30/40) = 36.9° to the x axis, parallel to the xy plane.

Resultant moment about O

Couples: (0, −50, 0) + (90, 0, 0) = (90, −50, 0)
r × F of 40 N: (0,0,8) × (40,0,0) = (0, 320, 0)
M_O = (90, 270, 0) N·m

R·M_O = 40×90 + 30×270 = 11 700 M∥ = 11 700 / 50 = 234 N·m along R; pitch p = 11 700/2500 = 4.68 m.

(b) Line of action (axis of the wrench)

r₀ = (R × M_O)/R² R × M_O = (30×0 − 0×270, 0×90 − 40×0, 40×270 − 30×90) = (0, 0, 8100) r₀ = (0, 0, 8100/2500) = (0, 0, 3.24) m

The line of action passes through the point (0, 0, 3.24) m on the z axis (3.24 m from O) and is parallel to R = (40, 30, 0), so its equation is (x, y, z) = (0, 0, 3.24) + t (40, 30, 0).

Answer: the single equivalent wrench is a force of 50 N (40 i + 30 j) plus a couple of 234 N·m along the force; its axis passes through z = 3.24 m on the z axis, parallel to R.