1.3 Resultant of coplanar concurrent forces
After this lesson you should be able to:
- Derive the parallelogram-law expressions for the resultant of two forces.
- Find the resultant of any number of coplanar concurrent forces by resolving into components.
- Interpret the resultant's magnitude and direction and check the result graphically.
Theory
Forces are concurrent when their lines of action meet at a single point, as at a pin joint of a truss, the hook of a crane or a knot in a set of ropes. Because they all pass through one point they cannot produce a moment about it, so their combined effect is a single force, the resultant, acting through that point.
For two forces, the parallelogram law gives the resultant as the diagonal of the parallelogram drawn on the two force arrows. Equivalently, placing the arrows tip to tail forms a triangle whose closing side is the resultant. For three or more forces, the tip-to-tail chain is the force polygon, and the line from the first tail to the last tip is the resultant.
Graphical methods are good for insight and quick checks, but the component method is exact and works for any number of forces. Resolve every force into x and y components, add the components algebraically, then rebuild the resultant from the two sums. This is the method used in practice and in examinations.
The resultant of two forces always lies between |P − Q| and P + Q. It equals P + Q when the forces act in the same direction, |P − Q| when they act in opposite directions, and √(P² + Q²) when they are at right angles. These limits are handy sanity checks.
Derivation
- Let P act along the x-axis and Q at angle θ to P, both from point O.
- Complete the parallelogram. Q has components Q cos θ along P and Q sin θ perpendicular to it.
- The resultant therefore has components (P + Q cos θ) along P and Q sin θ perpendicular to P.
- By Pythagoras: R² = (P + Q cos θ)² + (Q sin θ)² = P² + 2PQ cos θ + Q²(cos²θ + sin²θ) = P² + Q² + 2PQ cos θ.
- The angle α between R and P satisfies tan α = Q sin θ / (P + Q cos θ).
- For n forces: Rx = Σ Fi cos θi, Ry = Σ Fi sin θi, R = √(Rx² + Ry²), and the direction is atan2(Ry, Rx).
R = √(P² + Q² + 2PQ cos θ)tan α = Q sin θ / (P + Q cos θ), α measured from PRx = ΣFx, Ry = ΣFy, R = √(Rx² + Ry²), θR = atan2(Ry, Rx)|P − Q| ≤ R ≤ P + Q
- All forces lie in one plane and meet at one point.
- The body is treated as a particle at that point (no rotation is possible).
- Angles are measured consistently from one reference axis.
Codes and standards
Indian Standards are cited by number and clause only; no clause text, tables or figures are reproduced. Other standards are listed for comparison, not ranked.
- No design code governs this lesson. The method is used whenever member forces meet at a joint, for example in the analysis of steel trusses designed to IS 800:2007.
Worked examples
Problem. Two forces, P = 40 kN and Q = 30 kN, act at a point with an angle of 60° between them. Find the magnitude of the resultant and its angle with P.
- R² = 40² + 30² + 2 × 40 × 30 × cos 60° = 1 600 + 900 + 1 200 = 3 700.
- R = √3 700 = 60.83 kN.
- tan α = 30 sin 60° / (40 + 30 cos 60°) = 25.98 / 55 = 0.4724.
- α = 25.29°.
Answer. R = 60.83 kN at 25.29° to P
Problem. Four ropes pull on a ring used to position a precast unit: 20 kN at 0°, 30 kN at 90°, 25 kN at 150° and 15 kN at 225°, angles measured anticlockwise from the x-axis. Find the resultant pull on the ring.
- ΣFx = 20 + 30 cos 90° + 25 cos 150° + 15 cos 225° = 20 + 0 − 21.651 − 10.607 = −12.258 kN.
- ΣFy = 0 + 30 + 25 sin 150° + 15 sin 225° = 30 + 12.5 − 10.607 = 31.893 kN.
- R = √(12.258² + 31.893²) = 34.17 kN.
- tan⁻¹(31.893 / 12.258) = 68.98°; Rx is negative and Ry positive, so the resultant acts at 180° − 68.98° = 111.02° from the x-axis.
Answer. R = 34.17 kN at 111.02° from the x-axis
Step-by-step procedure
- Draw all forces from the common point with their angles from one reference axis.
- Tabulate each force with its x and y components and signs.
- Sum the columns to get Rx and Ry.
- Compute R and its direction, correcting the quadrant.
- Check against a rough force polygon sketch or the |P − Q| ≤ R ≤ P + Q limits.
Site notes
- When several slings meet at a crane hook, the resultant must be vertical and equal to the load; if your components give a horizontal residual, the load will swing until it is.
- Lifting plans compare the resultant sling force with the rated capacity of the hook and shackles, so take care with angles: sling forces rise steeply as the included angle widens.
- A table of components is easy for a checker to follow; show it in calculations.
- Using the angle between a force and the y-axis in a formula that expects the angle from the x-axis.
- Forgetting that cos θ is negative for angles between 90° and 270°.
- Using the parallelogram formula with the angle between the line of P and the reversed direction of Q.
- Reporting the direction without saying which reference it is measured from.
- Taking the arctangent without fixing the quadrant.
Two equal forces F at angle θ give R = 2F cos(θ/2). At 120° the resultant equals one force; at 60° it is 1.73 F; at 90° it is 1.41 F. Use this to check sling and rope calculations.
Summary
- Concurrent forces reduce to a single resultant through their common point.
- Two forces: R² = P² + Q² + 2PQ cos θ.
- Any number of forces: sum components, then rebuild R and its angle.
- The resultant lies between |P − Q| and P + Q.
Practice questions
Tap a question to see the answer.
1Two forces of 5 kN and 12 kN act at right angles. Find the resultant and its angle with the 5 kN force.
R = 13 kN; α = tan⁻¹(12/5) = 67.38°.
2Two equal forces F act at 120° to each other. What is the resultant?
R = 2F cos 60° = F.
3Two forces of 10 kN each act at 60°. Find the resultant.
R = 2 × 10 × cos 30° = 17.32 kN.
4Between what limits must the resultant of 8 kN and 5 kN lie?
Between 3 kN (opposite) and 13 kN (same direction).
5Three forces of 10 kN act at 0°, 120° and 240°. Find their resultant.
Rx = 10(1 − 0.5 − 0.5) = 0; Ry = 10(0 + 0.866 − 0.866) = 0; the resultant is zero.
References
Standard texts, listed in no order of merit. Our explanations, examples and diagrams are original; nothing is copied from these books.
- F. P. Beer, E. R. Johnston, D. F. Mazurek, P. J. Cornwell, B. P. Self, Vector Mechanics for Engineers: Statics and Dynamics, McGraw Hill, 12th edition, 2019.
- R. C. Hibbeler, Engineering Mechanics: Statics, Pearson, 15th edition, 2022 (published 2021).
- J. L. Meriam, L. G. Kraige, J. N. Bolton, Engineering Mechanics: Statics, Wiley, 9th edition, 2018.
- S. Timoshenko, D. H. Young, J. V. Rao, S. Pati, Engineering Mechanics (in SI units), McGraw Hill Education (India), 5th edition, 2013.