Resolving forces: vectors in Year 3 without trigonometry

There is a scheduling problem that nobody explains to you, and at the classical high school, the human-sciences high school, the language high school and the art high school it repeats itself identically every year. In Year 3 physics begins: two hours a week, and already in November the book asks you to resolve a force along two perpendicular directions, that is, to use sine and cosine. Open your maths book: trigonometry sits at the end of Year 3 if all goes very well, more realistically in Year 4, and at the classical high school it arrives as late as February to April of the fourth year.

So the first serious physics test of the year asks you for a tool that your maths syllabus has not yet given you. It is not a gap of yours: it is a mismatch between two national syllabuses that neither teacher is required to resolve.

The good news is that the trigonometry needed to resolve a vector is a tiny portion of what you will study in Year 4: two definitions and a handful of numerical values, half an hour’s work. Associated angles, identities and trigonometric equations have nothing to do with Year 3 physics.

Here I set out what really comes up in the test, the minimal half hour of trigonometry, the five-step method I make everyone use, the mistakes I see repeated, and three exercises worked in full.

What really comes up in the November test

In the first term of Year 3 the physics syllabus of the classical, human-sciences, economics-and-social-sciences, language and art high schools is almost always this: quantities and measurement, SI units, scientific notation and graphs between September and October; from November to December vectors, forces, weight, Hooke’s law and friction; from January equilibrium, moments and levers; then fluids.

The November test almost always draws from four families:

1. Sum of two or more vectors that are not parallel, to be done by components.
2. Resolving a force along two axes of your choosing, typically on an inclined plane.
3. Equilibrium of a point: a body hanging from two ropes, or held still on an inclined plane.
4. Hooke’s law and friction inside one of the exercises above, usually as the last question.

The first three require sine and cosine. The fourth does not, but they use it to make the result of one of the first three come out. If the resolution holds up, the test is done.

The half hour of trigonometry you need

Take a right-angled triangle. Pick one of the two acute angles and call it $\theta$. Of the three sides, one is the hypotenuse, one is the leg adjacent to $\theta$ (the one that touches the angle) and one is the opposite leg (the one facing it).

Sine and cosine are two ratios, and that’s all:

From here, rearranging the formulas, you get the two you will really use:

That is everything. In Year 3 physics the hypotenuse is the magnitude of the vector and the two legs are its components. Resolving a force means drawing that right-angled triangle and reading it.

The values you need to know by heart are three angles — $30°$, $45°$ and $60°$ — plus the two limiting cases:

Two mental checks worth more than any formula. First: the sine and cosine of an acute angle are always between 0 and 1, so a component can never come out larger than the magnitude of the vector. If it does, you multiplied instead of dividing or pressed the wrong key. Second: as the angle grows from $0°$ to $90°$, the sine grows and the cosine falls. At $30°$ the sine is small (0.50) and the cosine large (0.87); at $60°$ it is the other way round. If you are unsure which of the two to use, imagine the angle squashing towards zero: the component that survives goes with the cosine, the one that vanishes goes with the sine.

The five-step method

This is the order I make you write on the page before you even look at the numbers. It seems pedantic, but it is exactly what separates a 5 from an 8 in a two-hour test.

1. Draw. The body as a point, all the forces as arrows starting from that point. No extra arrows: if you cannot say who exerts it, that force does not exist.
2. Choose the axes and state them. Write on the page “I take $x$ horizontal to the right, $y$ vertical upwards” — or, on the inclined plane, ”$x$ parallel to the plane pointing down the slope, $y$ perpendicular to the plane”. The choice is yours and is never wrong: what is wrong is not stating it or changing it halfway.
3. Resolve one force at a time. For each, draw the right-angled triangle with the force as the hypotenuse and the two legs along the axes. Look at where the angle is: if it is attached to the $x$ axis, then $F_x=F\cos \theta$ and $F_y=F\sin \theta$; if it is attached to the $y$ axis, the roles swap.
4. Put in the signs. Components are numbers with a sign: positive if they point in the direction of the axis, negative if they point the opposite way. This is the step where more marks are lost than any other.
5. Add by components, then recombine. The resultant has $R_x=\sum F_x$ and $R_y=\sum F_y$, and its magnitude is

which is Pythagoras, not trigonometry: you already did that in Year 2.

If the problem is about equilibrium, step 5 becomes even simpler: you impose $R_x=0$ and $R_y=0$ and solve the system. Two equations, usually two unknowns.

If you want to see how automatic it has become, the simulated physics test on vectors and forces is calibrated exactly on this block of the syllabus. It is catalogued under Year 1 because at the science high school these topics are done in Year 1, but the content is your Year 3 syllabus, identical. The correct answer is free; the step-by-step guided solution is unlocked with your first lesson.

The four mistakes that cost marks

Adding the magnitudes instead of the components. Two forces of 3 N and 4 N give 7 N only if they are parallel and in the same direction. If they are perpendicular they give 5 N, by Pythagoras. If they form any other angle, they give a value between 1 N and 7 N and it must be calculated by components. The number 7 on a November test is almost always the sign that the student drew nothing.

Swapping sine and cosine. It does not happen because you don’t know the formula, it happens because you don’t look at where the angle is. The working rule: the component near the angle goes with the cosine, the one far from it with the sine. Always check it with the angle-going-to-zero test.

Losing the signs. On the inclined plane the component of the weight parallel to the plane is negative if you chose $x$ pointing up the slope, positive if you chose it pointing down. Both choices are legitimate; writing it as positive when you declared $x$ pointing up is not.

Using the wrong angle on the inclined plane. The angle of the plane appears between the weight force and the perpendicular to the plane, not between the weight and the plane itself: hence the most widespread mistake of Year 3, swapping $P\sin \alpha$ and $P\cos \alpha$. The geometric proof is in exercise 2, and it is worth redoing once instead of learning by heart which goes where.

On this and the other recurring slips I have written a separate piece, typical physics mistakes in high school, which also covers the later years.

Exercise 1 — sum of two non-parallel forces

Two coplanar forces act on a particle: $F_1=12\text{N}$ directed horizontally to the right and $F_2=8.0\text{N}$ inclined at $60°$ above the horizontal. Find the magnitude of the resultant.

Axes: $x$ horizontal to the right, $y$ vertical upwards.

$F_1$ is already along $x$, so $F_{1x}=12\text{N}$ and $F_{1y}=0$.

$F_2$ makes $60°$ with the $x$ axis, and the angle is attached to $x$: the component along $x$ goes with the cosine.

Sum by components:

Magnitude:

Reasonableness check: the sum of the magnitudes would be 20 N, the difference 4 N. The result must lie in between, and 17.5 N does. Anyone who wrote 20 N would have added the magnitudes of two non-parallel vectors.

Exercise 2 — body at rest on an inclined plane

A block of mass $m=5.0\text{kg}$ rests on a plane inclined at $30°$ to the horizontal, and is held still. Calculate the component of the weight force parallel to the plane and the perpendicular one, with $g=9.8{\text{m/s}}^2$.

Weight: $P=mg=5.0\cdot 9.8=49\text{N}$, directed vertically downwards.

Axes: $x$ parallel to the plane, pointing down the slope; $y$ perpendicular to the plane, pointing outwards.

Now the geometry, which is the real point of the exercise. The weight is vertical; the $y$ axis is perpendicular to the inclined plane, which in turn is inclined at $\alpha =30°$ to the horizontal. Two lines perpendicular to two lines that form an angle $\alpha$ between them form the same angle $\alpha$ between themselves. So the angle between the weight and the $y$ axis is exactly $30°$: the angle of the plane is found again between the weight and the perpendicular to the plane, not between the weight and the plane.

The angle is therefore attached to the $y$ axis. So the component along $y$ goes with the cosine and the one along $x$ with the sine:

The parallel component is 25 N and is the one that tends to make the block slide; the perpendicular one is 43 N in magnitude and is negative because it points into the plane, while the $y$ axis points outwards. It is the component that the plane must balance with the normal reaction $N=43\text{N}$.

Check: $30°$ is a gentle slope, so the component pressing the block onto the plane must be much larger than the one making it slide. 43 against 25 fits. If you had got 43 N for the parallel component you would have swapped sine and cosine, and this comparison is enough to notice it, without redoing the calculations.

If this geometric step — the angle of the plane reappearing between the weight and the perpendicular — does not sink in from the drawing in the book, here is the same figure to tilt by hand: you choose the angle and the two components are rewritten before your eyes.

Last typical question: if the block stays still, what is the static friction force? It must cancel the parallel component, so $F_f=25\text{N}$ directed along the plane, up the slope. It is not calculated with $\mu N$: that formula gives the maximum static friction, not the actual one.

Exercise 3 — lamp hanging from two ropes

A lamp of mass $m=2.0\text{kg}$ is hung from the ceiling by two ropes, each making an angle of $45°$ with the vertical. Find the tension in each rope.

Weight: $P=2.0\cdot 9.8=19.6\text{N}$ downwards. By symmetry the two tensions have the same magnitude $T$.

Axes: $x$ horizontal, $y$ vertical upwards. Each tension is directed along its own rope, upwards and outwards, and makes $45°$ with the vertical, that is with the $y$ axis: the angle is attached to $y$, so the vertical component goes with the cosine.

Left rope: $T_x=-T\sin 45°$, $T_y=+T\cos 45°$. Right rope: $T_x=+T\sin 45°$, $T_y=+T\cos 45°$.

Equilibrium along $x$: the two horizontal components are opposite and cancel on their own. The equation is satisfied identically, which confirms that the symmetry was legitimate.

Equilibrium along $y$:

The result has a meaning that in the test is often worth a reasoning question: each rope carries 13.8 N, more than half the weight (9.8 N). The reason is that only part of the tension works vertically, the rest pulls sideways. And the more the ropes open towards the horizontal, the closer $\cos \theta$ gets to zero and the more the tension grows: that is why a washing line pulled almost horizontal snaps under a modest weight.

How we approach it in lessons

With students at the classical, human-sciences and art high schools I never start from “serious” trigonometry, because it is not needed and would take up three lessons. In the first hour we build the right-angled triangle and the five values in the table, and check them on a scale drawing made with a ruler: when you see that the leg measured with the ruler matches the calculated number, the formula stops being arbitrary. From there on we work only on exercises, always in the same order of the five steps, until the drawing with the axes stated becomes automatic.

The second thing we do is separate the physics from the calculation. Many Year 3 failing grades at the classical high school are not about physics: they are wrong conversions, scientific notation or powers worked out badly — first-two-years maths that, with two hours of physics a week, nobody has time to review in class. When needed we devote half a lesson to that, and the physics grade goes up without touching the physics.

If the picture is wider than just the November test, I have also written a guide to the first physics test of Year 3 and an overview of physics as a subject with what we follow for each year and track.

If instead on your track vectors are tackled with real sine and cosine, the full version, also covering equilibrium with friction and the limiting angle, is in vectors and forces in Year 3 at the classical high school.

To sum up

The trigonometry you need in November of Year 3 is two ratios and five angles. The method is always the same: draw, state the axes, resolve one force at a time looking at where the angle is, put in the signs, add by components. And two checks that save the test even when the calculations are wrong: no component can exceed the magnitude, and on the inclined plane the perpendicular component is the larger one as long as the slope is below $45°$.

If the test is in two weeks and the drawing with the axes is not yet automatic, you can catch up in a few targeted lessons: write to me and we’ll talk. The first lesson is to work out where the chain breaks — whether it is the geometry of the triangle, the signs or the first-two-years calculation — because the three problems are cured in different ways. For up-to-date rates (single lesson or 10-hour package) see the pricing page; payment is offline, by bank transfer, Satispay, PayPal or Revolut.