Vector Forces — Resolution and Equilibrium
When chief engineer Joseph Strauss completed the Golden Gate Bridge in 1937, its two main suspension cables — each 256 mm in diameter — carried a combined resultant tension of 230 MN. Every individual hanger cable pulls at a different angle; Strauss's team resolved each force into horizontal and vertical components and applied $\sum F = 0$ to guarantee the deck remained in static equilibrium under the full 887,000-tonne load.
Practise this lesson
Printable worksheets from guided practice to exam-style questions.
Two pit crew members push a V8 Supercar from different angles to get it out of the pits. One pushes from behind at 30° left, the other at 30° right. Without doing any calculations: will the car move straight ahead, or will one side win? Where will the resultant force point?
A force of 10 N acts at 30° above the horizontal. What is the horizontal component of this force?
Know
- Resultant force as the vector sum of all forces
- 1D force addition — same direction and opposing
- $F_x = F\cos\theta$ and $F_y = F\sin\theta$
- Conditions for 2D equilibrium
Understand
- Why forces add as vectors, not scalars
- Why equilibrium requires zero net force in every direction
- Why the car pushers in the Think First produce a resultant straight ahead
Can Do
- Add 1D forces algebraically with correct sign convention
- Draw a tip-to-tail diagram for 2D forces
- Resolve any force into horizontal and vertical components
- Determine whether a 2D system is in equilibrium
Core Content
Two people tug on opposite ends of a rope. One pulls with 60 N east; the other pulls with 80 N west. The rope does not stay still — it moves west. That observable outcome tells you the forces did not simply add (160 N) but subtracted: net force = 20 N west.
Define a positive direction first. Forces in that direction are positive; forces opposing it are negative. The resultant (net force) is the algebraic sum.
where each $F$ includes its sign according to your defined positive direction.
A 1500 kg car has an engine force of 2800 N forward and friction of 450 N opposing motion. Find the net force.
Net force in 1D is the algebraic sum of all forces with their signs; always define a positive direction first, assign positive to forces in that direction and negative to opposing forces, then sum: $F_{net} = \sum F$ (N).
Pause — copy the highlighted rule into your book before moving on.
A box is pushed right with 50 N and left with 30 N. Taking right as positive, the net force is:
We just saw that 1D forces add algebraically once a sign convention is defined. That raises a question: what if forces point in different directions — how do you add them? This card answers it → resolving each force into x and y components lets you add them independently.
Any force at an angle can be split into two perpendicular components — one horizontal and one vertical — using trigonometry.
For a force $F$ acting at angle $\theta$ above the horizontal:
To add multiple forces in 2D: resolve each into components, add all x-components separately, add all y-components separately, then combine.
Two forces act on an object: $F_1 = 40\text{ N}$ at 0° (horizontal right) and $F_2 = 30\text{ N}$ at 90° (vertical upward). Find the resultant.
To resolve a force $F$ at angle $\theta$ above the horizontal: $F_x = F\cos\theta$ (horizontal component) and $F_y = F\sin\theta$ (vertical component); resultant of multiple forces: $F_R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$ at $\theta = \tan^{-1}(F_y/F_x)$.
Pause — copy the highlighted formulae into your book before moving on.
True or false: For an object to be in 2D equilibrium, the net force in the horizontal direction must be zero but the net force in the vertical direction can be non-zero.
We just saw that any force can be resolved into independent x and y components. That raises a question: how do you use those components to determine whether an object is in balance? This card answers it → 2D equilibrium requires $\sum F_x = 0$ AND $\sum F_y = 0$ simultaneously.
For equilibrium in two dimensions, the net force in every direction must be zero simultaneously.
The conditions for 2D equilibrium:
This means that if you resolve all forces into components, the sum of all x-components must equal zero AND the sum of all y-components must equal zero. Both conditions must hold.
2D equilibrium requires both $\sum F_x = 0$ AND $\sum F_y = 0$ simultaneously; each condition gives an independent equation that can be used to solve for unknown force magnitudes or directions.
Pause — copy the highlighted conditions into your book before moving on.
For 2D equilibrium, $\sum F_x = $ ______ AND $\sum F_y = $ ______.
Activities
For each force, draw a tip-to-tail vector diagram, then resolve into components:
- $F_1 = 60\text{ N}$ at 0° and $F_2 = 80\text{ N}$ at 90°. Find the resultant.
- $F = 100\text{ N}$ at 53° above horizontal. Find horizontal and vertical components.
A traffic light hangs from two cables. The left cable has tension $T_1$ at 40° to the vertical; the right cable has tension $T_2$ at 50° to the vertical. The light's weight is 200 N downward. Set up — but do not solve — the two equilibrium equations.
A 25 N force acts at 60° above the horizontal. What are the horizontal and vertical components?
Pick your answer, then rate your confidence.
UnderstandBand 3(3 marks) 1. Two forces of equal magnitude act on an object: one pointing east, one pointing north. Describe the direction and relative magnitude of the resultant compared to each individual force.
ApplyBand 4(3 marks) 2. A 150 N force acts at 37° above the horizontal. Calculate the horizontal and vertical components of this force. (Use sin 37° = 0.60, cos 37° = 0.80.)
AnalyseBand 5(4 marks) 3. A 20 kg traffic signal hangs from two cables. Cable A makes 30° with the vertical; Cable B makes 60° with the vertical. The system is in equilibrium. Set up and solve the two equilibrium equations to find tensions $T_A$ and $T_B$. (g = 9.8 m/s²)
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Short Answer — Model Answers
Q1 (3 marks): The resultant points northeast — at 45° between east and north (since the forces are equal). Its magnitude is $\sqrt{F^2+F^2} = F\sqrt{2} \approx 1.41F$, which is greater than either individual force. The resultant is always larger in magnitude than either component force when they are perpendicular.
Q2 (3 marks): $F_x = 150\cos37° = 150 \times 0.80 = 120\text{ N}$ (horizontal). $F_y = 150\sin37° = 150 \times 0.60 = 90\text{ N}$ (vertical). Check: $\sqrt{120^2+90^2} = \sqrt{22500} = 150\text{ N}$ ✓
Q3 (4 marks): Weight $= 20 \times 9.8 = 196\text{ N}$ downward. $\sum F_y = 0$: $T_A\cos30° + T_B\cos60° = 196$. $\sum F_x = 0$: $T_A\sin30° = T_B\sin60°$, so $T_A = T_B\sqrt{3}$. Substituting: $T_B\sqrt{3} \times \frac{\sqrt{3}}{2} + T_B \times \frac{1}{2} = 196$; $\frac{3T_B}{2} + \frac{T_B}{2} = 196$; $2T_B = 196$; $T_B = 98\text{ N}$; $T_A = 98\sqrt{3} \approx 170\text{ N}$.
Five timed questions on Vector Forces. Beat the boss to bank a tier.
⚔ Enter the arenaIn 1937, Joseph Strauss's team resolved every Golden Gate Bridge cable tension into components, then applied $\sum F_x = 0$ and $\sum F_y = 0$ at each anchor point — confirming the 230 MN resultant was safely directed along the cable axis. The V8 Supercar pushers in the Think First work by the same principle: symmetric 30° angles produce equal and opposite horizontal components that cancel, so only the forward components add to give a resultant straight ahead.