Forces and Equilibrium

Weight is drawn first because it is never in doubt — every object has exactly one, straight down, regardless of what else is going on.

Perpendicular to the surface, not to the block or to any other force — this is the rule that changes on an inclined plane later in this section.

The angle is given, not derived — read it straight off the question, and check whether it is measured from the horizontal or the vertical before using it later.

Friction's direction is never guessed — it is read off from which way the object is being pushed or tends to slide, then drawn the other way.

The incline changes where the normal reaction and friction point, but weight is always vertical — it never tilts with the slope.

This is the detail that separates a flat-ground diagram from an inclined-plane one: 'perpendicular to the surface' now means perpendicular to a tilted line.

Friction opposes the actual motion here, since the block is already sliding — contrast this with a block merely on the point of sliding, in §06, where 'opposing the tendency to move' is the same rule applied before any motion exists.

Writing this down first avoids the single most common slip: substituting the wrong angle into the wrong ratio.

Naming the formula before substituting keeps the method visible, rather than jumping straight to a decimal.

Check the calculator is in degrees before trusting this: in radian mode cos 40 returns −0.67, and a negative horizontal component here is physically impossible, which is the warning sign to look for.

Same angle, other ratio — decide which ratio from the geometry each time rather than memorising 'horizontal is always cos', which stops being true the moment an angle is measured from the vertical.

A quick sanity check: 40° is less than 45°, so the component nearer the angle's own axis (the horizontal one) should be the larger of the two — and 7.66 > 6.43, as expected.

When forces are already at right angles, the 'resolve each one first' step is trivial — this is the special case, not the general method; the very next example, where a third force sits at 210°, needs the full version.

Pythagoras applies here because the horizontal and vertical totals are, by construction, perpendicular components of the resultant.

The resultant must be bigger than either force on its own but smaller than their sum (10 lies between 8 and 14) — a two-second check that catches a square root taken of the wrong thing.

tan⁻¹ of (opposite/adjacent) recovers the angle the resultant makes with whichever axis the horizontal total was measured along.

A direction is not an answer until it says what it is measured from — '36.9°' alone earns nothing; '36.9° above the horizontal' does.

Choosing and writing down the positive directions before any trig is what makes the minus signs later a decision rather than a guess — and a resultant question is lost far more often on signs than on trig.

A force already along an axis still gets written into the table — skipping it because 'it's obvious' is how one of the four or five contributions goes missing from the totals.

Note this force is at 90° to the 5 N one — that is why the earlier example could go straight to Pythagoras. With a third force at 210°, that shortcut is no longer available, and the general method must be used.

Work with the acute angle 30° against the nearest axis and put the sign in by hand from the picture. Typing cos 210° straight into the calculator also works, but it hides the sign, so a wrong quadrant then goes unnoticed.

Same force, second component, done as its own line — resolving one force into both components in a single step is where a stray sign or a swapped ratio usually slips through unseen.

Adding components is ordinary arithmetic precisely because they all lie along one line — this is the payoff for resolving, and the reason forces at odd angles are never added directly.

Keep the two totals in separate lines and separate directions: mixing an x-component into the y-sum is the single most costly slip in a resultant question, and it is invisible once the numbers are added.

Pythagoras is legitimate here only because the two totals are perpendicular — which they are by construction, since the axes were chosen at right angles at the start.

Sanity check: the 5 N and 3 N forces largely oppose each other horizontally, so a resultant well below 5 N is exactly what should be expected — a resultant of, say, 12 N would mean components had been added without their signs.

It is the two TOTALS that go into tan⁻¹, never an individual force's components — a classic wrong answer here comes from taking tan⁻¹ of one of the three original forces instead.

tan⁻¹ only ever returns an angle between −90° and 90°, so it cannot distinguish the first quadrant from the third, or the second from the fourth. When ΣFx is negative, add 180° to the calculator's answer; the signs of the two totals, read off as a direction, are the only reliable guide.

The 20 N force is the only one with a horizontal component too, but part (a) only needs the vertical direction, so only the vertical component is written down here.

F acts straight down, so with upward positive its only contribution is $-F$; 'no y-component' then translates directly into this equation. State the convention — the minus sign in front of F is meaningless without it.

F comes out smaller than 20 N, as it must: it only has to cancel part of the 20 N force, namely its vertical slice.

Starting part (b) by re-listing every component (rather than reusing part (a)'s intermediate numbers blindly) avoids carrying over an assumption — here F has changed, so the vertical total has too.

The vertical total is now non-zero — that is the whole difference between part (a) and part (b). In (a) it was forced to zero; here it survives as a genuine component of the resultant.

Exactly the same Pythagoras step as the invented examples — only now the two totals had to be built from resolving first.

Use the unrounded 7.32… from the calculator, not a re-typed 7.32, for every step after this — rounding at an intermediate stage is the standard way to lose the final accuracy mark.

Both totals are positive, so the resultant lies in the first quadrant and the calculator's angle needs no adjustment — but say what the angle is measured from, since 'θ = 36.2°' on its own does not name a direction.

Naming which side is which before doing anything else is the step most often skipped — and skipping it is exactly how sin and cos get swapped two lines later.

This is ordinary Pythagoras on a right-angled triangle with legs 5 and 12 — nothing to do with the force yet, just completing the geometry the ratio described.

5-12-13 is a Pythagorean triple, so a whole-number hypotenuse here is a sign the working is on track — real bank questions rarely come out this cleanly, but recognising the pattern speeds up the ones that do.

Both ratios came from the same triangle as exact fractions — nothing has been rounded, and $\theta$ itself was never found.

Same 'adjacent gets cos' rule as every other resolving example in this section — the only difference here is that cos θ is a fraction instead of a calculator decimal.

The 26 cancels neatly against the 13 because 26 = 2 × 13 — worth spotting before multiplying out, since it turns the arithmetic into 2 × 12 rather than a messier decimal multiplication.

Same force, second component, done as its own line — resolving both components in a single step is where a swapped ratio usually slips through unseen.

Check: $24^2 + 10^2 = 576 + 100 = 676 = 26^2$ — the two components rebuild the original force exactly, confirming nothing was rounded along the way.

This is the whole point of the section: α is never found as a number of degrees anywhere in this solution — every sin and cos the resolving below needs comes straight off this 8-15-17 triangle.

This is the same 'resolve every force, one at a time' routine as the three-force example earlier in this section — the only change is that exact fractions stand in for sin α and cos α instead of calculator decimals.

Every term uses only the exact fractions from the triangle above — nothing has been rounded yet, so this line is exact.

Each product simplifies to a whole number precisely because 8-15-17 is a Pythagorean triple — a decimal version of α would not have simplified this cleanly.

Keep this as its own line, entirely separate from the x-sum — mixing a horizontal term into the vertical total is the costliest slip in a resultant question, and it is invisible once the numbers are added.

Both totals are positive, which already says the resultant points up and to the right — worth noting now, before the direction is even found.

674 is not a perfect square, unlike the invented example above — real bank numbers rarely land on a clean Pythagorean triple, so don't expect one every time.

Keep the unrounded $\sqrt{674}$ on the calculator for the direction step that follows, rather than re-typing 26.0 — rounding here first is a common way to lose the final accuracy mark.

It is the two TOTALS that go into tan⁻¹, never one of the three original forces — exactly as in the invented three-force example earlier in this section.

A direction is only complete once it says what it is measured from — '15.6°' alone earns nothing; '15.6° above the positive x-axis' does.

Naming the direction on its own line, before any algebra, is what stops the two equations being written as one blur — and each resolving equation carries its own method mark.

Going force-by-force is the guard against the commonest equilibrium error: resolving in a direction and forgetting a force that does have a component in it.

Adjacent-to-the-angle takes cos — decided from the geometry, not from a memorised 'horizontal is cos', which fails the moment an angle is measured from the vertical.

Written as 12 − R cos θ = 0 it says the same thing; rearranged to R cos θ = 12 it is ready to use. Do the rearranging now rather than in the middle of the elimination later.

A second, independent direction is what turns one physical fact into two usable equations — and two unknowns, R and θ, need exactly two equations.

Same force, same angle, the other ratio — and note it is sin here purely because θ is measured from the horizontal. Measured from the vertical, these two equations would swap over.

Squaring and adding the two resolved equations is exactly Pythagoras applied to the two component equations — the same move as finding a resultant's magnitude in §02, run here to solve for an unknown instead.

R has to be at least as big as either force it is balancing, since it must cancel both at once — 15 > 12 > 9, as required. An R smaller than 12 would mean a sign had gone astray.

Dividing is cleaner than substituting R = 15 back in: it keeps θ independent of any rounding in R, so an error in one answer cannot contaminate the other.

The 12 N force is the larger, so the balancing force must lean closer to the horizontal than to the vertical — an angle under 45° is the expected shape of the answer.

Either direction could come first here, since each of the two equations contains both unknowns — but naming the direction as its own line keeps the two equations from being merged into one.

The 52 N force contributing zero horizontally is the fact to state, not skip: forgetting to check whether a force has a component in the direction being resolved is the standard way a term goes missing.

Read the angle off the figure and check what it is measured from before choosing the ratio — if the diagram had marked θ from the vertical, this component would be P sin θ instead.

This is the equilibrium condition itself: the sum of the components in the chosen direction is zero. Everything after it is algebra.

Getting the unknown alone on the left now makes the square-and-add step two lines away much tidier.

Exactly the same five moves as the horizontal equation, run in the perpendicular direction — which is why the second one can safely be written more briefly than the first.

This is the same elimination move as the invented example — square each resolved equation, add, and cos²+sin²=1 collapses the left side to P².

The left side collapses to P² only because cos²θ + sin²θ = 1 — the identity is doing real work here, so it is worth naming rather than letting θ appear to vanish by magic.

Take the positive root only: P is a magnitude, and a negative force magnitude is not an answer. (39, 52, 65 is a 3-4-5 triangle scaled by 13 — the exam writer chose these numbers deliberately.)

Vertical equation over horizontal equation gives sin/cos = tan directly. Dividing the other way round gives cot θ, and an angle of 36.9° — the complement of the right answer, and a very easy mark to lose.

The vertical force being the larger means P must lean closer to the vertical, so θ measured from the horizontal should exceed 45° — it does, which confirms the fraction was not inverted.

Reusing the same symbol for two genuinely different tensions is one of the most common ways this question type loses marks — the mark scheme requires them distinguishable from the first line.

This equation exists only because the strings are on opposite sides — that is the geometric fact doing the work. Were both strings on the same side, their horizontal pulls would add rather than cancel, and the whole solution would change.

Plus, not minus, between the two terms — the strings oppose each other horizontally but cooperate vertically. Copying the horizontal equation's minus sign into this line is the usual slip.

With two unknowns and two equations, the standard move is to isolate one unknown from the simpler equation and substitute it into the other — exactly as with any pair of simultaneous equations.