9709 P4 W12 Forces & Equilibrium — Lesson 2: Components & Resultants

💡 Lead In

Two Pulls, One Effect

Two perpendicular forces act on the same point: one due east and one due north. Their combined effect is a single force, the resultant. Set the two forces and watch the resultant grow — its size comes from Pythagoras and its direction from trigonometry.

east force $F_{1}$ = 6 N

north force $F_{2}$ = 8 N

resultant $R = \sqrt{F_{1}^{2}+F_{2}^{2}}$ = 10.0 N · direction $\theta = \tan^{-1}\!\left(\dfrac{F_{2}}{F_{1}}\right)$ = 53.1°

The resultant is the single force that has the same effect as the two original forces together.

🧠 Activating Prior Knowledge

Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.

R1. In a right-angled triangle, $\cos\theta$ equals:

$\dfrac{\text{adjacent}}{\text{hypotenuse}}$

$\dfrac{\text{opposite}}{\text{hypotenuse}}$

$\dfrac{\text{opposite}}{\text{adjacent}}$

$\dfrac{\text{hypotenuse}}{\text{adjacent}}$

Choose an option above.

$\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$.

R2. Pythagoras' theorem states that for a right-angled triangle:

$c^{2} = a^{2} + b^{2}$

$c = a + b$

$c^{2} = a^{2} - b^{2}$

$c = ab$

Choose an option above.

The square on the hypotenuse equals the sum of the squares on the other two sides.

R3. The horizontal component of a force $F$ at angle $\theta$ to the horizontal is:

$F\cos\theta$

$F\sin\theta$

$F\tan\theta$

$\dfrac{F}{\cos\theta}$

Choose an option above.

The side adjacent to the angle is horizontal, so the horizontal component is $F\cos\theta$.

📖 Key Concepts

Explore 1 — resolving a force into components.

A force $F$ at angle $\theta$ to the horizontal can be replaced by two perpendicular forces:

$F_{x} = F\cos\theta \qquad F_{y} = F\sin\theta$

Together the two components have exactly the same effect as the original force. The horizontal component uses $\cos$ because the horizontal side is adjacent to $\theta$.

Explore 2 — combining perpendicular forces into a resultant.

Two perpendicular forces form the sides of a right-angled triangle; the resultant is the hypotenuse:

$R = \sqrt{F_{1}^{2}+F_{2}^{2}} \qquad \theta = \tan^{-1}\!\left(\dfrac{F_{2}}{F_{1}}\right)$

The resultant is the single force with the same effect as the two originals combined.

Explore 3 — choosing directions and using signs.

Step 1 — fix a positive direction

Take right as positive horizontally and up as positive vertically, and record that choice.

Step 2 — resolve every force along those directions

A force pointing left or down contributes a negative component.

Step 3 — add the components

Sum the horizontal components and sum the vertical components, then combine with Pythagoras. Resolving along a slope works the same way, using directions parallel and perpendicular to the slope.

Group Match the Lines: forces, components and resultants

Pairs: tap a card on the left, then its match on the right to draw a line. When all six lines are drawn, press Check to score. Talk through each link before you commit to it.

Individual Force Vector Explorer: read the components

Set the magnitude F and the direction theta of a force. The dashed lines show the horizontal and vertical components. Sweep theta from 0 to 90 and watch how the two components trade size, then answer the challenge.

✏️ Worked Examples

Exercise 6A Q2 Worked Example

A rope pulls with a force of $50\ \text{N}$ at $40^\circ$ above the horizontal. Find the horizontal and vertical components of the pull.

Hint

Horizontal component $= 50\cos 40^\circ$; vertical component $= 50\sin 40^\circ$.

Step 1 — horizontal component

$F_{x} = 50\cos 40^\circ = 38.3\ \text{N}$.

Step 2 — vertical component

$F_{y} = 50\sin 40^\circ = 32.1\ \text{N}$.

Exercise 6A Q3 Worked Example

Forces of $9\ \text{N}$ due east and $12\ \text{N}$ due north act on a point. Find the magnitude and direction of the resultant.

Hint

Use $R = \sqrt{9^{2}+12^{2}}$ and then $\theta = \tan^{-1}(12/9)$.

Step 1 — magnitude

$R = \sqrt{9^{2}+12^{2}} = \sqrt{225} = 15\ \text{N}$.

Step 2 — direction

$\theta = \tan^{-1}\!\left(\dfrac{12}{9}\right) = 53.1^\circ$ above the east direction.

📝 Concept Quiz

Answer, then check. Options are shuffled every load. The clock starts on your first answer.

Q1. The horizontal component of a force $F$ at angle $\theta$ to the horizontal is:

$F\cos\theta$

$F\sin\theta$

$F\tan\theta$

$\dfrac{F}{\cos\theta}$

Choose an option above.

The horizontal side is adjacent to $\theta$, so the component is $F\cos\theta$.

Q2. Forces of $6\ \text{N}$ and $8\ \text{N}$ act at right angles. The magnitude of the resultant is:

$10\ \text{N}$

$14\ \text{N}$

$2\ \text{N}$

$48\ \text{N}$

Choose an option above.

$R = \sqrt{6^{2}+8^{2}} = \sqrt{100} = 10\ \text{N}$.

Q3. When resolving forces, which pair of directions is sensible?

$\text{horizontal and vertical}$

$\text{any two parallel directions}$

$\text{two directions at } 30^\circ$

$\text{along the force only}$

Choose an option above.

Components must be along two perpendicular directions; parallel directions cannot be combined by Pythagoras.

Q4. A force of $20\ \text{N}$ acts horizontally to the left. Taking right as positive, its horizontal component is:

$-20$

$20$

$0$

$10$

Choose an option above.

To the left is the negative direction, so the component is $-20\ \text{N}$.

📚 More Practice

Work each on paper, then reveal the solution to check your working.

Exercise 6A Q2 Standard

A force of $60\ \text{N}$ acts at $25^\circ$ above the horizontal. Find its horizontal and vertical components.

Reveal Solution

$F_{x} = 60\cos 25^\circ = 54.4\ \text{N}$, $F_{y} = 60\sin 25^\circ = 25.4\ \text{N}$.

Exercise 6A Q2 Standard

Forces of $5\ \text{N}$ and $12\ \text{N}$ act at right angles. Find the magnitude of the resultant.

Reveal Solution

$R = \sqrt{5^{2}+12^{2}} = \sqrt{169} = 13\ \text{N}$.

Exercise 6A Q3 Medium

A force of $100\ \text{N}$ acts at $55^\circ$ to the horizontal. Resolve it into components.

Reveal Solution

$F_{x} = 100\cos 55^\circ = 57.4\ \text{N}$, $F_{y} = 100\sin 55^\circ = 81.9\ \text{N}$.

Exercise 6A Q3 Medium

Forces of $7\ \text{N}$ due east and $24\ \text{N}$ due north act on a point. Find the magnitude and direction of the resultant.

Reveal Solution

$R = \sqrt{7^{2}+24^{2}} = 25\ \text{N}$; $\theta = \tan^{-1}(24/7) = 73.7^\circ$ above east.

Exercise 6A Q5 Hard

Three forces act on a point: $10\ \text{N}$ due east, $10\ \text{N}$ due north and $10\ \text{N}$ pointing south-west (at $225^\circ$). Find the magnitude and direction of the resultant.

Reveal Solution

Resolve: east $10 - 10\cos 45^\circ = 2.93\ \text{N}$, north $10 - 10\sin 45^\circ = 2.93\ \text{N}$. Then $R = \sqrt{2.93^{2}+2.93^{2}} = 4.14\ \text{N}$ at $45^\circ$ above east.

🎫

Exit Ticket

Check the learning objectives you feel confident about:

I can resolve a force of magnitude $F$ at angle $\theta$ into components $F\cos\theta$ and $F\sin\theta$ I can find the resultant of two perpendicular forces using Pythagoras and trigonometry I can choose sensible perpendicular directions and resolve forces along them I can use consistent sign conventions for components in each direction