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Structural Analysis Fundamentals / Forces, moments and equilibrium

Forces and resolving them

Vectors, components and resultants — the arithmetic behind everything else.

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Why this matters

Resolve forces along consistent axes so that you can add their components and write the equilibrium equations.

By the end of this lesson you should be able to

  • Describe a force by its size, direction and line of action
  • Split a force into horizontal and vertical components
  • Combine several forces into a single resultant

A force is a push or a pull. To describe one fully you need three things: how big it is, which way it points, and where it acts. That last one — the line of action — matters more than people expect. The same force applied along a different line has a different turning effect, even though its size and direction are unchanged.

A force pointing at an angle is awkward to add to other forces. So we replace it with two forces along the axes that, together, do exactly the same job. That is resolving.

Resolving a force
Fx=Fcos⁡θFy=Fsin⁡θF_x = F\cos\theta \qquad F_y = F\sin\theta

What it calculates: The horizontal and vertical parts of a force F acting at an angle θ measured from the horizontal.

F
Size of the force (kN)
θ
Angle from the horizontal axis (degrees)
Fx
Horizontal component (kN)
Fy
Vertical component (kN)

This assumes

  • θ is measured from the horizontal, anticlockwise positive

In plain terms: The cosine goes with the axis the angle is measured from. If you are ever unsure, sketch the triangle — the component next to the angle uses cosine.

A single force of 25 kilonewtons drawn at 40 degrees above the horizontal, with dashed lines showing its horizontal and vertical components.xyF = 25 kN at 40°
The dashed lines are the two components. Together they have exactly the same effect as the sloping force.

Try it

Resolve two forces

Change each force and watch its horizontal and vertical parts. The body only stays still when both totals reach zero.

kN

degrees from +x

kN

degrees from +x

ForceHorizontalVertical
F₁ = 10 kN8.665.00
F₂ = 8 kN-6.934.00
Total1.739.00
NOT BALANCEDA leftover force of 9.17 kN acts at 79° — the body would accelerate that way.

Predict first

A 20 kN force acts at 60° above the horizontal. Which component is larger?

Worked example

Resolve a sloping cable force

Given

  • A cable pulls on a bracket with a force of 25 kN
  • The cable rises at 40° above the horizontal

Find

The horizontal and vertical components of the cable force.

Assumptions

  • The cable is straight and the force acts along it

    Practice

    A cable pulls with a force of 50 kN at 25° above the horizontal. What is the horizontal component?

    Practice

    Two forces act at a point: 30 kN horizontally to the right, and 40 kN vertically upwards. What is the magnitude of their resultant?

    Practice

    A force of 60 kN acts at 120° measured anticlockwise from the positive x axis. What is its horizontal component? (Watch the sign.)

    Summary

    • A force needs size, direction and line of action
    • Resolving replaces a sloping force with two perpendicular ones that do the same job
    • Fx = F cos θ and Fy = F sin θ when θ is measured from the horizontal
    • Always check by recombining the components