Vectors and forces without trigonometry

End of September 2026. In Year 3 at the Italian liceo classico (classics-focused high school) physics is a subject you meet for the first time, two hours a week, and after the first weeks on quantities and measurement the programme moves in November to vectors and forces: weight, elastic force, friction. The test falls six to ten weeks from now, so between early November and early December, and in this chapter there is a timetable problem that is worth making clear straight away, because it isn’t your fault.

The classical high school paradox: physics comes before maths

To resolve a force along an inclined plane you need sine and cosine. The problem is that at the liceo classico trigonometry in maths is done at the end of Year 3, if you are lucky, or in Year 4. The physics teacher, in November, can’t wait: they give you a formula with $sin$ and $cos$ in it before you have ever seen them as a topic in their own right.

You aren’t behind. It is the programme that isn’t synchronised between the two subjects, and it happens systematically in all classical high schools, not just in your class. What you need isn’t the trigonometry chapter of the science high school (it would take weeks that you don’t have) but the bare minimum needed to use sine and cosine as a tool, without the full theory behind them. That is what you find in this post.

Scalar and vector quantities

A scalar quantity is completely defined by a number and a unit of measurement: mass, time, temperature. A vector quantity also needs a direction and a sense: velocity, acceleration, force. It is represented by an arrow (an oriented segment) whose length (to scale) is the magnitude.

The commonest mistake of those who have only just met vectors is to treat them like ordinary numbers: adding two forces of 30 N and 40 N and expecting 70 N. It works only if the two forces point in the same direction. If not, the vector sum gives a different result, often less than the sum of the magnitudes, and that is exactly the point on which the first test challenges you.

Adding vectors: tip-to-tail, parallelogram, components

There are three ways to add two vectors, and they all lead to the same result.

Tip-to-tail method. You draw the first vector, then attach the tail of the second to the tip of the first. The sum vector goes from the tail of the first to the tip of the second. It is the most intuitive method for understanding what “adding” means, but imprecise if done freehand.

Parallelogram method. You draw the two vectors from the same origin, complete the parallelogram, and the diagonal starting from the common origin is the sum. Same result as tip-to-tail, different drawing.

Components method. This is the one you will really use in a test, because it is the only one that gives an exact number without drawing to scale. Each vector is resolved along two perpendicular axes (usually $x$ and $y$ ), the $x$ components and the $y$ components are added separately, and the resultant vector is rebuilt from those two sums.

Sine and cosine: the minimum you need

In a right-angled triangle with an angle $θ$ , sine and cosine are two ratios between sides:

sin θ = \frac{opposite side}{hypotenuse}, cos θ = \frac{adjacent side}{hypotenuse}

You need nothing else for this chapter: not the addition formulas, not the unit circle, not equations. Just this ratio, and the values for three angles that come back continually in problems:

Angle	$sin θ$	$cos θ$
$30°$	$0.5$	$0.866$
$45°$	$0.707$	$0.707$
$60°$	$0.866$	$0.5$

A way not to mix them up: as the angle grows from $0°$ to $90°$ , the sine grows (from 0 to 1) and the cosine decreases (from 1 to 0). If you are in doubt which of the two to use in a problem, check that your result respects this trend and the common sense of the physical situation.

Resolving a force along two axes

A force $F$ making an angle $θ$ with the $x$ -axis has components:

F_{x} = F cos θ, F_{y} = F sin θ

The most frequent mistake isn’t getting the calculation wrong, it is swapping the two formulas: putting $sin θ$ where you need $cos θ$ . It happens especially when the angle in the problem isn’t the “standard” one relative to the horizontal axis, for example when the angle given is that of the inclined plane relative to the ground, and the force to be resolved is the weight, which is vertical. In that specific case the component of the weight parallel to the plane is $m g sin θ$ and the one perpendicular to the plane is $m g cos θ$ , the opposite of what would come instinctively if you don’t draw the triangle of forces properly. Always draw the right-angled triangle formed by the vector and its two components before writing any formula: it is the only way not to go wrong systematically.

The forces in the programme

Weight. $P = m g$ , always vertical, downwards. It is the only constant force acting everywhere on Earth (with $g \approx 9.8 m/ s^{2}$ ).

Elastic force (Hooke’s law). A spring stretched or compressed by $x$ relative to the rest position exerts a force proportional to the displacement and directed in the opposite sense:

F_{e l} = - k x

where $k$ is the spring constant, in $N/m$ . The minus sign says that the force always opposes the displacement: if you stretch the spring, the force pulls back.

Static and kinetic friction. Static friction opposes the attempt to set a stationary body in motion, and adapts: it is exactly as much as is needed to prevent motion, up to a maximum

F_{f, m a x} = μ_{s} N

where $N$ is the normal reaction of the surface (not the weight, except in the special case of a horizontal plane) and $μ_{s}$ the coefficient of static friction. If the force pushing the body exceeds this maximum, the body starts to move and kinetic friction takes over, which is exactly $F_{f} = μ_{k} N$ , not “up to”.

Tension. The force that a taut rope or string exerts on the bodies it is attached to, always directed along the string itself.

Equilibrium of a particle

A particle is in equilibrium when the vector sum of all the forces acting on it is zero:

\sum F = 0 ⟺ \sum F_{x} = 0, \sum F_{y} = 0

This is the condition you will use in almost all the exercises of the chapter: you resolve every force along two judiciously chosen axes, impose that the sum of the $x$ components is zero and the sum of the $y$ components is zero, and solve the resulting system (often of two equations).

Mistakes that cost marks

Adding the magnitudes instead of the vectors. Two forces of 30 N and 40 N never give 70 N unless they are perfectly aligned in the same sense.
Swapping sine and cosine in the resolution, especially on the inclined plane: the component of the weight along the plane is $m g sin θ$ , the perpendicular one is $m g cos θ$ . Always draw the triangle of components before writing the formula.
Using the friction coefficient with the weight instead of the normal. On an inclined plane, the normal is not equal to the weight: it is $N = m g cos θ$ , always less than the weight itself for $θ > 0$ . Friction must be calculated on that, not on $m g$ .

Three worked exercises

1. Sum of two non-aligned forces, with the components method

Two forces act on a body: $F_{1} = 50 N$ directed along the $x$ -axis, and $F_{2} = 40 N$ making an angle of $45°$ with the $x$ -axis. Find the magnitude and direction of the resultant force.

I resolve both forces:

F_{1 x} = 50 N, F_{1 y} = 0

F_{2 x} = 40 cos 45° = 40 \cdot 0.707 \approx 28.3 N, F_{2 y} = 40 sin 45° = 40 \cdot 0.707 \approx 28.3 N

I add the corresponding components:

R_{x} = 50 + 28.3 = 78.3 N, R_{y} = 0 + 28.3 = 28.3 N

The magnitude of the resultant, by Pythagoras’ theorem:

R = R_{x}^{2} + R_{y}^{2} = 78. 3^{2} + 28. 3^{2} = 6131 + 801 \approx 6932 \approx 83.3 N

And the direction, the angle that $R$ makes with the $x$ -axis:

tan α = \frac{R _{y}}{R _{x}} = \frac{28.3}{78.3} \approx 0.361 ⟹ α \approx 19.9°

The resultant force has a magnitude of about $83.3 N$ and makes an angle of about $19.9°$ with $F_{1}$ .

If you want to check for yourself whether you have understood the method, in the Year 1 physics test on vectors and forces you will find more resolution exercises among the questions: the correct answer is free, the step-by-step guided solution unlocks with the first lesson. It is catalogued under Year 1 because at the science high school this topic is done a year earlier; at the classical high school, where physics starts in Year 3, the level is the right one for you.

2. A body hanging from a spring, then a point suspended from two ropes

Part A. A mass $m = 2 kg$ hangs in equilibrium from a spring with spring constant $k = 200 N/m$ . By how much does the spring stretch?

At equilibrium, the elastic force exactly balances the weight:

k x = m g ⟹ x = \frac{m g}{k} = \frac{2 \cdot 9.8}{200} = \frac{19.6}{200} \approx 0.098 m

The spring stretches by about $9.8 cm$ .

Part B. A weight $W = m g = 19.6 N$ (the same mass as before) is suspended from a hook in the ceiling by two ropes: one makes an angle of $30°$ with the ceiling (horizontal), the other an angle of $60°$ . Find the tension in each rope.

The point where the two ropes and the weight meet is in equilibrium: the sum of the horizontal components is zero, and the sum of the vertical components balances the weight.

Along the horizontal, the horizontal components of the two tensions must be equal and opposite:

T_{1} cos 30° = T_{2} cos 60° ⟹ T_{2} = T_{1} \cdot \frac{cos 30°}{cos 60°} = T_{1} \cdot \frac{0.866}{0.5} = 1.732 T_{1}

Along the vertical, the two vertical components together support the weight:

T_{1} sin 30° + T_{2} sin 60° = W

Substituting $T_{2} = 1.732 T_{1}$ :

T_{1} \cdot 0.5 + 1.732 T_{1} \cdot 0.866 = W ⟹ 0.5 T_{1} + 1.5 T_{1} = W ⟹ 2 T_{1} = W

So $T_{1} = W /2 = 19.6/2 = 9.8 N$ , and $T_{2} = 1.732 \cdot 9.8 \approx 17.0 N$ .

Quick check: the rope more inclined to the horizontal (the one at $60°$ ) is the more “vertical” one, and indeed it carries the larger load ( $17.0 N$ against $9.8 N$ ), consistent with the fact that it supports a greater share of the weight along its own direction.

3. A block on an inclined plane: equilibrium with friction and limit angle

A block of mass $m = 5 kg$ rests on a plane inclined at $θ = 30°$ . The coefficient of static friction between block and plane is $μ_{s} = 0.7$ . Does the block stay still? And what is the maximum angle of inclination beyond which it would start to slide?

I resolve the weight $P = m g = 5 \cdot 9.8 = 49 N$ along the plane and perpendicular to the plane:

P_{∥} = m g sin θ = 49 \cdot 0.5 = 24.5 N, P_{⊥} = m g cos θ = 49 \cdot 0.866 \approx 42.4 N

Along the perpendicular to the plane the block doesn’t move, so the normal exactly balances $P_{⊥}$ : $N \approx 42.4 N$ . The maximum static friction force available is:

F_{f, m a x} = μ_{s} N = 0.7 \cdot 42.4 \approx 29.7 N

The component of the weight that would tend to make the block slide is $24.5 N$ , less than the maximum available ( $29.7 N$ ): static friction adapts and provides exactly $24.5 N$ (not its maximum), and the block stays still.

The limit angle, beyond which not even maximum friction would be enough, is found by imposing $P_{∥} = F_{f, m a x}$ , that is $m g sin θ_{l im} = μ_{s} m g cos θ_{l im}$ . The term $m g$ cancels on both sides:

sin θ_{l im} = μ_{s} cos θ_{l im} ⟹ tan θ_{l im} = μ_{s} = 0.7 ⟹ θ_{l im} \approx 35°

With $θ = 30°$ , less than $35°$ , the result is consistent with the equilibrium found above.

How we work on this chapter

The work in lessons always starts from the drawing: the triangle of components for every force, before writing a single number. It is the way never to confuse $sin θ$ with $cos θ$ , and it is what you need to arrive ready for the November test without having done a single hour of trigonometry at school. The full picture of how I work physics online, from measurements to vectors onwards, is on the physics page; the previous chapter, on quantities and significant figures, is described for another track in physics at the art high school: measurements and significant figures: same kind of starting programme, new subject, few hours a week.

If you want to train on your own before the test, the simulated Year 1 physics test on vectors and forces covers exactly these topics. It is catalogued under Year 1 because at the science high school they are tackled a year earlier, but for the Year 3 classical high school it is the right level.

If your November test is instead set without trigonometry, with just the half-hour of sine and cosine, you will find the right approach in resolving forces without trigonometry. When the equilibrium involves beams and constraints there are the constraint reactions, and when the body moves dynamics and the inclined plane.

If you want a hand before November, write to me via contacts.

Year 3 vectors and forces: no trigonometry needed