1.2 Vectors and representation of forces
After this lesson you should be able to:
- Distinguish scalars from vectors and describe a force fully by magnitude, direction, sense and point of application.
- Resolve a force into rectangular components and recombine components into magnitude and direction.
- Write a force as a magnitude times a unit vector along a given line.
Theory
A scalar has size only: mass, length, time, temperature. A vector has size and direction and adds by the parallelogram law: displacement, velocity, acceleration and force. A force is fully described by four things: its magnitude, its direction (the line along which it acts), its sense (which way along that line) and its point of application. On a drawing it is an arrow; in calculations it is a set of components.
The principle of transmissibility says that, for a rigid body, a force can slide along its own line of action without changing its external effect. A 10 kN pull on a rope tied to a block has the same effect on the block's equilibrium wherever along the rope we imagine it applied. This is only true for external effects on rigid bodies: the internal forces, and the deformation of a real member, do depend on where the force acts.
Any force in a plane can be replaced by two perpendicular components. If a force F makes an angle θ with the x-axis, its components are Fx = F cos θ and Fy = F sin θ. The reverse step recovers the magnitude from Pythagoras and the direction from the ratio of components. Signs matter: a component pointing in the negative axis direction is negative, and the angle must be placed in the correct quadrant.
Often a force acts along a member or a cable whose end coordinates are known. Then the cleanest description is F = F u, where u is the unit vector along the line, found by dividing the position-difference vector by its length. The same idea, extended to three coordinates, is the basis of 3D statics in lesson 1.10.
Derivation
- Place the tail of F at the origin with F at angle θ to the x-axis.
- Drop a perpendicular from the tip of F to the x-axis, forming a right triangle with hypotenuse F.
- By definition of cosine and sine, the adjacent side is F cos θ and the opposite side is F sin θ, so Fx = F cos θ and Fy = F sin θ.
- By Pythagoras, F² = Fx² + Fy², so F = √(Fx² + Fy²); and tan θ = Fy / Fx, with the quadrant fixed by the signs of Fx and Fy.
- For a line from A(x₁, y₁) to B(x₂, y₂): the vector AB = (x₂ − x₁, y₂ − y₁), its length d = √((x₂ − x₁)² + (y₂ − y₁)²), the unit vector u = AB / d and F = F u.
Fx = F cos θ, Fy = F sin θ (N or kN)F = √(Fx² + Fy²); θ = atan2(Fy, Fx)Unit vector along AB: u = (x₂ − x₁, y₂ − y₁) / dF = F u; components F (x₂ − x₁)/d and F (y₂ − y₁)/d
- Forces are treated as acting at points (concentrated); distributed forces are handled in lesson 1.6.
- Transmissibility applies to external effects on rigid bodies only.
- Angles are measured from the positive x-axis, anticlockwise positive, unless stated otherwise.
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; vector resolution is pure mechanics. Codes such as IS 800:2007 and IS 456:2000 assume these methods when members are analysed, but do not restate them.
Worked examples
Problem. A force of 500 N acts at 30° above the positive x-axis. Find its components. Then find the magnitude and direction of a force whose components are Fx = −3 kN and Fy = 4 kN.
- Fx = 500 cos 30° = 433.0 N; Fy = 500 sin 30° = 250.0 N.
- For the second force: F = √((−3)² + 4²) = 5 kN.
- tan⁻¹(4/3) = 53.13°; with Fx negative and Fy positive the force lies in the second quadrant, so θ = 180° − 53.13° = 126.87° from the positive x-axis.
Answer. 433.0 N and 250.0 N; 5 kN at 126.87°
Problem. In a plan view with x pointing east and y pointing north, a horizontal tie runs from a column at A(0, 0) to an anchor block at B(4 m, 3 m). The tie tension is 10 kN, acting on the column from A towards B. Write the force in components.
- Vector AB = (4 − 0, 3 − 0) = (4, 3) m.
- Length d = √(4² + 3²) = 5 m.
- Unit vector u = (4/5, 3/5) = (0.8, 0.6).
- F = 10 × (0.8, 0.6) = (8, 6) kN.
Answer. F = 8 kN along x and 6 kN along y (10 kN along AB)
Step-by-step procedure
- Sketch the force on axes and mark the angle you know.
- Decide the sign of each component from the quadrant.
- Use cos for the component adjacent to the known angle and sin for the opposite one.
- If the line of action is given by two points, form the unit vector from coordinates instead of angles.
- Check: √(Fx² + Fy²) must return the original magnitude.
Site notes
- Cable and strut forces are easiest to resolve from drawing dimensions (rise and run) rather than from measured angles, which carry rounding errors.
- Slopes are often given as ratios such as 1 vertical in 2.5 horizontal; turn them into a unit vector (1, 2.5)/√(1 + 2.5²) directly.
- When checking a temporary works sketch, verify that the arrow directions match what physically happens: cables can only pull.
- Using sin and cos the wrong way round when the angle is measured from the y-axis instead of the x-axis.
- Dropping the sign of a component in the second, third or fourth quadrant.
- Taking tan⁻¹(Fy/Fx) from a calculator without correcting the quadrant.
- Applying transmissibility to find internal forces or deflections.
- Adding magnitudes of forces that are not in the same direction.
Components can never exceed the magnitude of the force. If |Fx| or |Fy| comes out larger than F, the angle or the trigonometric function is wrong. For a 3-4-5 triangle the components are 0.6 F and 0.8 F.
Summary
- A force has magnitude, direction, sense and point of application.
- Fx = F cos θ, Fy = F sin θ; F = √(Fx² + Fy²).
- Use the unit vector along a member to write F = F u.
- Transmissibility holds for external effects on rigid bodies only.
Practice questions
Tap a question to see the answer.
1Resolve a 200 N force acting at 60° to the positive x-axis.
Fx = 200 cos 60° = 100 N; Fy = 200 sin 60° = 173.2 N.
2A force has components (6, −8) kN. Find its magnitude and direction.
F = 10 kN; it acts 53.13° below the positive x-axis (θ = 306.87°).
3A 13 kN force acts up a slope that rises 5 m in 12 m horizontal. Find its horizontal and vertical components.
Unit vector (12, 5)/13, so components are 12 kN horizontal and 5 kN vertical.
4A 1 000 N block rests on a 20° incline. Find the components of its weight parallel and perpendicular to the incline.
Parallel 1 000 sin 20° = 342.0 N; perpendicular 1 000 cos 20° = 939.7 N.
5Why can transmissibility not be used to find the stress in a bar?
Stress depends on where the force acts along the member; transmissibility only preserves external effects on a rigid body.
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.