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.