Replace two wrenches by single wrench

Reading of the figure: wrench 1 acts along the y axis through O: force 30 N along +y and a 70 N·m couple whose vector is along y (drawn pointing opposite to the force, so −y). Wrench 2 is at A on the z axis, 5 m from O: force 60 N along +x and a 40 N·m couple vector also along +x. The senses are as read from the arrows in the figure.

Step 1: resultant force

R = 30 j + 60 i = (60, 30, 0) N, |R| = √(60² + 30²) = 67.08 N

Step 2: resultant moment about O

Couples:        C1 + C2 = (0, −70, 0) + (40, 0, 0) = (40, −70, 0)
Moment of 60 N: r × F = (0,0,5) × (60,0,0) = (0, 300, 0)
M_O = (40, 230, 0) N·m

(The 30 N force passes through O, so it gives no moment.)

Step 3: components of M_O along R

R·M_O = 60×40 + 30×230 = 9300 M∥ = (R·M_O / |R|) = 9300 / 67.08 = 138.6 N·m (along R) Pitch p = M∥ / R = R·M_O / R² = 9300 / 4500 = 2.067 m

M∥ vector = (R·M_O / R²) R = 2.067 × (60, 30, 0) = (124, 62, 0) N·m

Step 4: axis of the wrench

The axis passes through the point r₀ = (R × M_O) / R². R × M_O = (30×0 − 0×230, 0×40 − 60×0, 60×230 − 30×40) = (0, 0, 12600) r₀ = (0, 0, 12600/4500) = (0, 0, 2.8) m

The axis is the line through (0, 0, 2.8) m parallel to R = (60, 30, 0). It meets the XZ plane (y = 0) at this same point, because y = 0 there.

Answer: single wrench = force R = 67.08 N (components 60 i + 30 j) with a couple of 138.6 N·m along R (pitch 2.07 m). Its axis cuts the XZ plane at x = 0, z = 2.8 m.