In one line
The resultant of parallel forces equals their algebraic sum, and its position is found by making its moment equal to the total moment of all the forces about a point.
Explained simply
If three boxes sit on a long plank, the plank feels one total weight, but that total must act at a certain spot so that the plank tips the same way. Finding that spot is the main job here.
Technically, for forces parallel to (say) the y-axis, \( R = \Sigma F \), and its line of action is located using Varignon's theorem: \( R \cdot \bar{x} = \Sigma (F \cdot x) \).
How it works
- Choose a sign convention: e.g. downward forces negative, upward positive; clockwise moments positive.
- Add all forces with signs: \( R = \Sigma F \). The sign tells the sense of R.
- Pick a reference point O (often the left end).
- Take moments of all forces about O: \( \Sigma M_O = \Sigma F_i x_i \).
- Distance of R from O: \( \bar{x} = \Sigma M_O / R \). Place R on the side that gives the same sense of moment.
- If \( R = 0 \) but \( \Sigma M_O \neq 0 \), the system reduces to a couple.
Key formula
\( R = \Sigma F \) (N); \( \bar{x} = \dfrac{\Sigma F_i x_i}{\Sigma F_i} \) (m), where \( x_i \) is the distance of force \( F_i \) from the reference point O.
Quick example
Downward loads of 10 kN at 0 m, 20 kN at 2 m and 30 kN at 5 m from A. \( R = 60 \) kN downward. \( \Sigma M_A = 10(0) + 20(2) + 30(5) = 190 \) kN·m. \( \bar{x} = 190/60 = 3.17 \) m from A.
Common mistakes
- Ignoring signs for unlike parallel forces, which can move R outside the span of the forces.
- Forgetting that a zero resultant with a non-zero moment means a couple, not 'no effect'.
- Measuring distances from different reference points in the same moment equation.
How MU asks it
Asked as a 5 to 6-mark numerical: 'Find the resultant of the parallel forces acting on the beam and locate it from end A'. A related form: 'replace the system by a single force and a couple at A'. Write the sign convention, \( \Sigma F \), \( \Sigma M_A \), the distance, and show R on the beam sketch.