An object never feels just one force, gravity, the floor, friction, a push or a pull all act at once. A free-body diagram strips away the clutter and draws every force as an arrow on a single dot. Add those arrows up and you get the net force: the one quantity that decides whether the object stays still, speeds up, or changes direction.
A book rests on a table. Two forces act on it: gravity pulling down and the table pushing up.
Before reading: if the book is not moving, what must be true about these two forces? Write your answer and one sentence on what the net force is.
Represent the object as a dot and draw every force acting on it
A free-body diagram reduces an object to a single point and draws an arrow for each force acting on it, and only on it, so you can see the whole force picture at once.
The usual forces are: weight $W = mg$ (down), normal force $N$ (perpendicular to the surface), friction $f$ (along the surface, opposing motion), tension $T$ (along a rope), and any applied push or pull. Draw only forces that act on the object, with arrow lengths roughly to scale.
A free-body diagram represents an object as a point and draws every force acting on it as an arrow: weight $W = mg$ (down), normal $N$ (perpendicular to the surface), friction $f$ (opposing motion), tension $T$ (along ropes), and applied forces. Only forces on the object are shown.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? Which force should NOT appear on the free-body diagram of a box resting on a table?
Break angled forces into $x$ and $y$ parts before adding
We just saw how to draw all the forces. That raises a question: how do we add forces that point at awkward angles? This card answers it → resolve each into horizontal and vertical components, then add the components separately.
An angled force is split into a horizontal and a vertical part using the same trigonometry as before, so that all the $x$-parts and all the $y$-parts can each be added on their own.
For a force $F$ at angle $\theta$ to the horizontal $x$-axis:
$$F_x = F\cos\theta \qquad\qquad F_y = F\sin\theta$$
A rope pulls a sledge with $150\ \text{N}$ at $20°$ above the horizontal. Find the horizontal and vertical components of this force.
Resolve each angled force into perpendicular components $F_x = F\cos\theta$ and $F_y = F\sin\theta$ (with $\theta$ measured from the $x$-axis), so the $x$-components and $y$-components can be summed separately.
Add the highlighted formulas to your notes before the check below.
Quick check: A force of $40\ \text{N}$ acts at $30°$ above the horizontal. What is its horizontal component (to 2 s.f.)?
Pattern A, Build then combine
A $10\ \text{kg}$ box on a smooth floor is pushed with a horizontal force of $30\ \text{N}$ east while a $12\ \text{N}$ friction-like resistance acts west (take $g = 9.8$). In your book: (1) draw the free-body diagram with all four forces; (2) find the net vertical force; (3) find the net horizontal force.
Fill the gaps. For a force $F$ at angle $\theta$ to the $x$-axis, $F_x = F$ $\theta$ and $F_y = F$ $\theta$. When the net force is zero the object is in .
Dot point: Resolve vectors into perpendicular components using F_x = F cosθ, F_y = F sinθ where θ is the angle with the x-axis
Add the components, then recombine into one resultant force
We just learned to resolve forces into components. That raises a question: how do we combine all of them into the one force that matters? This card answers it → sum the components in each direction, then recombine with Pythagoras.
The net force is the single resultant of every force on the object: add all the horizontal components, add all the vertical components, then join them into one resultant.
Compute $\sum F_x$ and $\sum F_y$, then $F_{net} = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$ with direction $\theta = \tan^{-1}\!\left(\dfrac{\sum F_y}{\sum F_x}\right)$. If the net force is zero, the object is in equilibrium (next lesson); if not, it accelerates in the direction of the net force. For the stuck crate, the push is balanced by static friction so $F_{net} = 0$, until the push exceeds the maximum friction.
Net force is the vector sum of all forces: add the components, $\sum F_x$ and $\sum F_y$, then $F_{net} = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$ at $\theta = \tan^{-1}(\sum F_y / \sum F_x)$. Zero net force means no change in motion; a non-zero net force causes acceleration.
Pause, copy the highlighted method into your book before the check below.
Last check: Decide if the statement is true or false.
To find the net force you simply add together the magnitudes of all the forces acting.
Match each situation with the forces that must appear on its free-body diagram:
Dot point: Use free-body diagrams and vector analysis to determine the net force acting on an object
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A box is pulled along a smooth horizontal floor by two horizontal ropes: $40\ \text{N}$ east and $30\ \text{N}$ north. Draw a free-body diagram of the horizontal forces and determine the magnitude and direction of the net horizontal force.
1 mark: free-body diagram of the two horizontal forces · 1 mark: $F_{net} = \sqrt{40^2 + 30^2}$ · 1 mark: $F_{net} = 50\ \text{N}$ · 1 mark: direction $= \tan^{-1}(30/40) \approx 37°$ north of east
ApplyBand 5(4 marks) 2. Two forces act on a box: $40\ \text{N}$ east and $30\ \text{N}$ north. Determine the magnitude and direction of the net force, and state what a free-body diagram adds to the calculation.
1 mark: recognises the forces are perpendicular · 1 mark: magnitude $= 50\ \text{N}$ · 1 mark: direction $37^\circ$ north of east · 1 mark: the diagram isolates the object and shows every force acting ON it
Dot point: Resolve vectors into perpendicular components using F_x = F cosθ, F_y = F sinθ where θ is the angle with the x-axis
UnderstandBand 4(3 marks) 3. State the rules for drawing a correct free-body diagram, and explain why the forces the object exerts on other objects must not appear on it.
1 mark: object shown isolated, as a point or simple box · 1 mark: every force acting ON the object drawn as a labelled arrow from that point, length showing magnitude · 1 mark: reaction forces act on OTHER bodies so they cannot contribute to this object’s net force
Dot point: Use free-body diagrams and vector analysis to determine the net force acting on an object
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): The free-body diagram shows two horizontal force arrows from the box: $40\ \text{N}$ east and $30\ \text{N}$ north, which are perpendicular. The net force is their vector sum. Since they are at right angles, $F_{net} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\ \text{N}$. The direction, measured from east toward north, is $\theta = \tan^{-1}\!\left(\frac{30}{40}\right) \approx 37°$. So the net horizontal force is $50\ \text{N}$ at about $37°$ north of east, and the box will accelerate in that direction.
Q2 (4 marks): The two forces are perpendicular, so they combine by Pythagoras: $$F_{net} = \sqrt{40^2 + 30^2} = \sqrt{2500} = 50\ \text{N}$$ The direction is $$\theta = \tan^{-1}\!\left(\frac{30}{40}\right) = 37^\circ$$ north of east. A free-body diagram adds rigour rather than arithmetic: it isolates the box from its surroundings and shows every force acting ON it as an arrow from a single point, which makes it clear that no force has been forgotten and that forces the box exerts on other things are correctly excluded.
Q3 (3 marks): A free-body diagram isolates one object from everything around it, drawing it as a point or a simple box. Every force acting ON that object is then drawn as a labelled arrow starting from that point, with the arrow length representing the magnitude and the direction drawn accurately. Nothing else is included. Forces the object exerts on other objects, such as the downward push of a book on a table, are excluded because they act on a different body. The whole purpose of the diagram is to find the net force on this object, and Newton’s second law sums only the forces applied to it, so including a force that acts elsewhere would produce a net force that is simply wrong.
You can now draw a free-body diagram, resolve angled forces into components, and find the net force on an object. Next lesson uses a zero net force to state Newton's first law and the conditions for equilibrium.
↩ Back to focus-area hubAt the start a book rested on a table with gravity down and the table's normal force up. Because the book does not accelerate, these forces must be equal and opposite, so the net force is zero. A zero net force means no change in motion, exactly Newton's first law, coming up next. Revisit Card 3 if the net-force idea is still hazy.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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