How is the resultant of a general coplanar (non-concurrent) force system found and located?

In one line

Find the resultant's magnitude and direction from \( \Sigma F_x \) and \( \Sigma F_y \), then locate its line of action using \( \Sigma M \) about a point.

Explained simply

When forces act at different points on a body in different directions, they both push it and turn it. The resultant must copy both effects: the same net push and the same turning.

Technically, a general coplanar system reduces to a single resultant R (unless it is a pure couple). R is given by its components, and its perpendicular distance d from a point O is \( d = |\Sigma M_O| / R \).

How it works

1. Resolve every force: compute \( \Sigma F_x \) and \( \Sigma F_y \).
2. \( R = \sqrt{(\Sigma F_x)^2 + (\Sigma F_y)^2} \), angle \( \theta = \tan^{-1}|\Sigma F_y/\Sigma F_x| \) with quadrant.
3. Take moments of all forces about a convenient point O (corner of plate or end of beam), using Varignon's theorem for inclined forces.
4. Perpendicular distance \( d = |\Sigma M_O|/R \).
5. Or find where R cuts the x-axis: \( x = |\Sigma M_O / \Sigma F_y| \) (or the y-axis: \( y = |\Sigma M_O/\Sigma F_x| \)).
6. Place R so its moment about O has the same sense (clockwise or anticlockwise) as \( \Sigma M_O \).

Key formula

\( d = \dfrac{|\Sigma M_O|}{R} \), \( x\text{-intercept} = \dfrac{|\Sigma M_O|}{|\Sigma F_y|} \). \( \Sigma M_O \) in N·m, R in N, d in m.

Quick example

On a plate: \( \Sigma F_x = 30 \) N, \( \Sigma F_y = -40 \) N, \( \Sigma M_O = -120 \) N·m (clockwise). \( R = 50 \) N at 53.13° below +x. \( d = 120/50 = 2.4 \) m; x-intercept \( = 120/40 = 3 \) m. A downward-right force cutting the +x axis 3 m from O gives a clockwise moment about O, so R cuts the x-axis at x = +3 m.

Common mistakes

• Getting the right magnitude but placing R on the wrong side of O; always check the sense of the moment.
• Forgetting the moment of a given couple when taking \( \Sigma M_O \) (couples add directly to the moment sum).
• Confusing perpendicular distance d with the axis intercept.

How MU asks it

One of the most frequent 6 to 8-mark numericals: 'Find the resultant of the force system acting on the plate/lamina and locate where it cuts AB (or the x-axis)'. Also asked as 'reduce to an equivalent force-couple system at A'. Full marks need: sign convention, component table, R with angle, \( \Sigma M \) with sense, the location, and a final diagram.