Force to force and couple; two wrenches resultant

Reducing a force into a force and a couple

A force F acting at point A can be replaced by an equal force F acting at another point O together with a couple.

1. At O, add two equal and opposite forces F and −F, both parallel to the original force. This does not change the effect.
2. The original F at A and the −F at O form a couple of moment M_O = r × F, where r is the vector from O to A.
3. What remains is the force F at O and the couple M_O.

So F at A is equivalent to F at O plus the couple M_O = r × F (magnitude F × d, where d is the perpendicular distance from O to the line of action).

Resultant of the two wrenches

Reading of the figure:

• Wrench 1 at O along the y axis: force 40 N along +y and couple 30 N·m (vector along −y, drawn opposite to the force).
• Wrench 2 at A on the z axis, 4 m from O: force 60 N along +x and couple 40 N·m along +x.

R = (60, 40, 0) N, |R| = √(60² + 40²) = 72.11 N Direction: tan⁻¹(40/60) = 33.7° from the x axis towards y.

M_O = couples + r × F = (40, −30, 0) + (0, 0, 4) × (60, 0, 0) = (40, −30, 0) + (0, 240, 0) = (40, 210, 0) N·m

R·M_O = 60×40 + 40×210 = 10 800 Parallel part M∥ = 10 800/72.11 = 149.8 N·m, pitch p = 10 800/5200 = 2.08 m

Line of action: r₀ = (R × M_O)/R² R × M_O = (0, 0, 60×210 − 40×40) = (0, 0, 11 000) r₀ = (0, 0, 11 000/5200) = (0, 0, 2.115) m

Answer: the two wrenches reduce to one wrench: force R = 72.1 N (60 i + 40 j) with a couple of 149.8 N·m along R. The line of action of R passes through (0, 0, 2.115 m), on the z axis 2.115 m from O, and is parallel to R, lying in the horizontal plane z = 2.115 m.