Year 11 Physics Module 2 ⏱ ~35 min 5 MC · 3 Short Answer Lesson 2 of 15

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.

Today's hook: In 1937, Strauss's Golden Gate Bridge cables carried 230 MN of resultant tension, yet each individual cable pulls at a different angle. How did engineers combine hundreds of angled forces into one safe resultant?

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Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Printable worksheets from guided practice to exam-style questions.

BuildFoundations & guided practice ApplyApplication practice MasterMastery challenge ExamExam-style questions

Before you read, predict

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?

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Learning Intentions

goals

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

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Key Terms

vocab

Resultant forcethe single vector that has the same effect as all the individual forces combined; found by vector addition

Componentthe projection of a vector onto an axis; a force at angle θ has $F_x = F\cos\theta$ and $F_y = F\sin\theta$

Equilibrantthe force equal in magnitude but opposite in direction to the resultant; it produces equilibrium

Tip-to-tail diagrama method of adding vectors by placing the tail of each vector at the tip of the previous one; the resultant joins start to finish

2D equilibriumcondition where $\sum F_x = 0$ AND $\sum F_y = 0$ simultaneously

Misconceptions to fix

When two forces act at angles, you just add their magnitudes to find the resultant.Forces are vectors, direction matters. You must use vector addition (tip-to-tail or component method). Scalar addition only works for forces in the same direction.

Two forces can balance each other even if they act in the same direction.For equilibrium, the vector sum must equal zero. Two forces can only cancel if they are equal in magnitude AND exactly opposite in direction.

Cross-lesson links: L01 introduced contact vs field forces and Newton's Laws, the same forces you now add as vectors. The equilibrium condition ($\sum F = 0$) applied here feeds directly into L03 (inclined planes), L04 (friction), and the energy work-integral in L06. Mastering vector addition here prevents sign errors in every subsequent dynamics calculation.

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D Force Addition

Core Content

1D Force Addition

+5 XP

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.

$$F_{net} = F_1 + F_2 + F_3 + \ldots$$

where each $F$ includes its sign according to your defined positive direction.

Worked Example 11D Addition

A 1500 kg car has an engine force of 2800 N forward and friction of 450 N opposing motion. Find the net force.

Positive direction = forward

Establishes sign convention before substituting any values.

$F_{engine} = +2800\text{ N}$; $F_{friction} = -450\text{ N}$

Friction opposes forward motion, so it is negative.

$F_{net} = +2800 + (-450) = +2350\text{ N}$ (forward)

Positive result confirms net force is in the defined positive (forward) direction.

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:

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Vector Resolution, Force Components

+5 XP

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:

$$F_x = F\cos\theta \qquad F_y = F\sin\theta$$

To add multiple forces in 2D: resolve each into components, add all x-components separately, add all y-components separately, then combine.

$$F_{resultant} = \sqrt{F_x^2 + F_y^2} \qquad \theta = \tan^{-1}\!\left(\frac{F_y}{F_x}\right)$$

Worked Example 2Component Addition

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.

$F_{1x} = 40\cos0° = 40\text{ N}$; $F_{1y} = 40\sin0° = 0\text{ N}$

A horizontal force has no vertical component.

$F_{2x} = 30\cos90° = 0\text{ N}$; $F_{2y} = 30\sin90° = 30\text{ N}$

A vertical force has no horizontal component.

$\sum F_x = 40\text{ N}$; $\sum F_y = 30\text{ N}$

$F_R = \sqrt{40^2+30^2} = \sqrt{2500} = 50\text{ N}$; $\theta = \tan^{-1}(30/40) = 36.9°$ above horizontal

This is a 3-4-5 triangle scaled by 10. The resultant is 50 N at about 37° above horizontal.

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.

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D Equilibrium

2D Equilibrium

+5 XP

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:

$$\sum F_x = 0 \quad \text{AND} \quad \sum F_y = 0$$

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.

Bridge Application: A suspension bridge cable has tension pulling diagonally up-and-inward, weight pulling down, and the road deck force. Engineers ensure $\sum F_x = 0$ (no horizontal drift) and $\sum F_y = 0$ (no vertical acceleration) simultaneously.

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 = $ ______.

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Vector Diagrams and Component Resolution

Activities

Vector Diagrams and Component Resolution

ApplyBand 3

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.

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D Equilibrium Analysis

2D Equilibrium Analysis

AnalyseBand 4

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?

✓

Show what you have learned

Quick recall, Vector Forces

+5 XP

Pick your answer, then rate your confidence.

Short Answer, 10 marks

+5 XP

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)

Show all answers

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: $T_B\sqrt{3} \times \frac{\sqrt{3}}{2} + T_B \times \frac{1}{2} = 196$; $\frac{3T_B}{2} + \frac{T_B}{2} = 196$; $2T_B = 196$; $T_B = 98\text{ N}$; $T_A = 98\sqrt{3} \approx 170\text{ N}$.

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Retrieve, reflect and finish

Check what actually stuck

Take the full module quiz

quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

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How did your thinking change?

In 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.

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