The Weight, Springs & Friction topic met forces one at a time and drew them on free-body diagrams. Most bodies feel several forces at once, and what decides how a body moves is their combined effect, the resultant force. When the forces lay along one line, as with the two pulls on the Friction page, combining them was a matter of adding and subtracting. This page shows how to combine forces that point in any directions, with the parallelogram rule and the triangle rule, and a shortcut for forces at right angles. Then it runs the idea backwards and splits one force into two perpendicular parts, which is the first step in every slope question.
Only one question in the four papers so far has been purely about adding forces: April 2026 Q6, two opposite forces along a line, worked through below. The ideas on this page carry much more than that. The range of a resultant, which the next page builds from them, has been asked in every paper. And splitting a force into parts is the first step in the slope questions that three of the four papers have asked, mostly on 30° slopes. So this page is short on direct exam questions and long on tools you will use all through mechanics.
Forces add as vectors
Tie two strings to a small ring and pull on them with two spring balances, one reading 3 N and the other 4 N. Pull both the same way and the ring feels 7 N. Pull in opposite directions and it feels 1 N, towards the 4 N pull. Pull at right angles and you can replace both balances with a single one that does exactly the same job, holding the ring in the same place; it reads 5 N. The same two forces gave three different results, so their sizes alone do not settle the combined effect. Their directions matter too.
A quantity that has a size and a direction, and combines with others in a way that takes the directions into account, is a vector. Force is one, like the displacement and velocity of the Describing Motion topic. The single force that has the same effect as several forces acting together is their resultant, and finding it is called adding, or composing, the forces.
Resultant force. The resultant of several forces acting on a body is the single force that would have the same effect on the body as all of them together. It replaces them; it is not one more force acting on the body.
Forces acting at one point. On this page every force acts at one point, as on the free-body diagrams of the last topic, where the body was treated as a particle at its centre. Forces like this are called concurrent forces (共点力): their lines of action meet at one point. The papers have asked only about concurrent forces. Turning effects, which need forces acting at different points of a body, have not come up.
Try it! · Check-in question
Two forces act at the same point on a crate. Which statement describes their resultant?
The resultant is defined by its effect: one force that could replace the two and change the crate's motion in exactly the same way. So the answer is B. A is true only when the two forces point the same way; at an angle the resultant is smaller than the sum, and in opposite directions it is the difference. C ignores the smaller force, which always counts unless it is zero. D treats the resultant as an extra force. It replaces the two forces, so adding it on top of them would count the same pushes twice.
Forces along one line
The simplest case is the one you have already met. When two forces act along the same line, draw their arrows end to end along that line, each to scale. If they point the same way, the second arrow carries on from the first, and the resultant is the sum of their sizes, in their common direction. If they point opposite ways, the second arrow runs back over the first, and the resultant is the difference, in the direction of the larger force.
Forces along one line, drawn end to end to scale. Pointing the same way, 15 N and 5 N give 20 N. Pointing opposite ways, the 5 N arrow runs back over the 15 N one, and the resultant is 10 N, along the larger force. These are the numbers of April 2026 Q6.
With more than two forces, or to keep track of the direction, choose one direction along the line as positive. Give each force a + or − sign, add the signed values, and read the direction from the sign of the answer, just as you did with displacements and velocities in the Describing Motion topic.
Forces along one line. Same direction: F = F₁ + F₂, in their common direction. Opposite directions: F = |F₁ − F₂|, in the direction of the larger force. For any number of forces along one line, choose a positive direction and add the forces with their signs; the sign of the result gives the direction.
Worked example · two opposite forces along one line (April 2026, Q6)
Two forces of magnitudes 15 N and 5 N act in opposite directions. The magnitude of their resultant force is ( )
A. 15 N B. 10 N C. 5 N D. 20 N
Step 1: "in opposite directions" means the two forces lie along one line and point opposite ways. Take the direction of the 15 N force as positive, so the forces are +15 N and −5 N.
Step 2: F = 15 + (−5) = 10 N. The answer is B. The sign is positive, so the resultant points the same way as the 15 N force, although this question asks only for the magnitude.
Each wrong option is a habit. D, 20 N, adds the sizes, which is right only when the forces point the same way. A, 15 N, answers with the larger force, as if the smaller one simply lost. C, 5 N, is the smaller force. Notice too that 10 N is the smallest resultant that 15 N and 5 N can ever give, at any angle; the next page builds a rule on that.
Exam wording. The Chinese text of April 2026 Q6 reads 大小为 15 N 和 5 N 的两个力方向相反，它们的合力大小为. Here 方向相反 means "in opposite directions" and 合力 is the resultant force, which some English papers translate as "combined force". Each of the forces being combined is a 分力, a component force.
Try it! · Check-in question
In a tug of war, the team on the left pulls the rope with a total force of 1200 N and the team on the right pulls with 1350 N. What is the resultant of the two pulls on the rope?
The two pulls lie along the rope and point opposite ways. Take right as positive: +1350 N and −1200 N, so F = 1350 − 1200 = +150 N. The plus sign means to the right, the direction of the larger pull, so the answer is D. A adds the sizes, as if both teams pulled the same way. B has the right size but the wrong direction: the resultant points along the larger force. C answers with the larger pull alone, as if the left team's pull did nothing.
Forces at an angle: the parallelogram rule
When two forces act at an angle, adding their sizes no longer works, so we need a rule that uses the directions. It comes from experiment. Fix one end of a rubber band to a board, and pull the other end, a small ring, with two spring balances at an angle until the ring sits on a marked point O. Record the two readings and the two directions. Then pull the ring to the same point O with a single balance. That one force has the same effect as the two, so it is their resultant. Now draw all three forces to scale from O. A pattern appears: the two forces form two sides of a parallelogram, and the single force lies along the diagonal from O, with the diagonal's length.
The pattern holds for any two forces at any angle, and it is the rule for adding any vectors, displacements and velocities included.
Parallelogram rule. To add two forces F₁ and F₂ acting at a point O, draw them to scale from O and complete the parallelogram that has these two arrows as sides. The diagonal from O is the resultant F: its length gives the size and its direction gives the direction.
Look again at the parallelogram. Its far side, from the tip of F₁ to the far corner, is parallel to F₂ and just as long: it is F₂ moved along. So you can leave out half of the drawing. Draw F₁, then draw F₂ starting at the tip of F₁, "tip to tail". The resultant runs from the tail of F₁ to the tip of F₂ and closes the triangle. This is the triangle rule, and it gives exactly the same resultant as the parallelogram.
Left: the resultant F is the diagonal from O of the parallelogram built on F₁ and F₂. Right: the far side of that parallelogram is F₂ moved along, so drawing F₂ from the tip of F₁ gives the same resultant as the third side of a triangle.
The triangle rule works for any number of forces. Draw them one after another, each starting where the last one ended, and the resultant runs from the start of the first arrow to the end of the last. Forces along one line are the flat case of this: the "triangle" lies along the line, which is exactly what the figure in the last section showed.
Triangle rule. Draw the forces tip to tail, in any order. The resultant goes from the tail of the first arrow to the tip of the last. Mind its direction: the same line drawn the other way round is the force that would balance the others, not their resultant.
Try it! · Check-in question
Two forces F₁ and F₂ act at a point O. To add them with the triangle rule, you draw F₁ starting at O and then draw F₂ starting at the tip of F₁. Which arrow is the resultant?
The resultant closes the triangle, running from the tail of the first arrow, O, to the tip of the last, so the answer is A. It is the same arrow as the diagonal from O in the parallelogram. B lies along the right line and has the right length, but it points the wrong way: that arrow is the force that would balance F₁ and F₂, not their resultant. C is the other diagonal of the parallelogram, which measures the difference between the two forces, not their sum. D adds the sizes along one line, which is right only if F₂ points the same way as F₁.
Two forces at right angles
When the two forces are perpendicular, the parallelogram is a rectangle, and the triangle from the triangle rule has a right angle between F₁ and F₂. The resultant is its hypotenuse, so Pythagoras gives its size. The angle α that the resultant makes with F₁ comes from the tangent: in the triangle, the side opposite α is F₂ and the side next to it is F₁, so tan α = F₂/F₁.
Two perpendicular forces. The resultant of two forces at right angles is F = √(F₁² + F₂²), at an angle α to F₁ where tan α = F₂/F₁. It is longer than either force, as the diagonal of a rectangle is longer than its sides.
Worked example · two forces at right angles
Two forces act at a point: 6 N to the east and 8 N to the north. Find the size and direction of their resultant.
The forces are perpendicular, so F = √(6² + 8²) = √(36 + 64) = √100 = 10 N.
Direction: tan α = 8/6 = 4/3, where α is measured from east towards north. That is about 53°, so the resultant points a little nearer north than east, towards the larger force.
Check: 10 N is more than either force, and less than their sum, 14 N, which they would give only if they pointed the same way.
For perpendicular forces the parallelogram is a rectangle, so the resultant is the hypotenuse of a right-angled triangle with sides 6 N and 8 N: F = 10 N, at an angle α to F₁ with tan α = 4/3, about 53°.
Pythagoras only for a right angle. √(F₁² + F₂²) is the resultant only when the forces are perpendicular. The exam's range questions give two or three forces with no angle at all, and the setters like numbers such as 3, 4 and 5 N or 6, 8 and 10 N, so that a student who assumes a right angle finds a matching wrong option. The next page deals with those questions. Until then, use Pythagoras only when the question says "perpendicular" or "at right angles".
Try it! · Check-in question
Two forces of 5 N and 12 N act at a point, at right angles to each other. What is the size of their resultant?
At right angles, F = √(5² + 12²) = √(25 + 144) = √169 = 13 N, so the answer is C. A adds the sizes, which is right only for forces in the same direction; 17 N is the largest resultant these two forces could ever give. B subtracts them, the smallest they could give, which needs opposite directions. D subtracts the squares, √(144 − 25). For perpendicular forces the squares add, and the resultant comes out longer than either force, as the diagonal of a rectangle must.
As the angle opens
For any angle θ between the two forces, use the triangle from the triangle rule. Its sides are F₁, F₂ and F. Drawn from the same point, the forces open by θ; drawn tip to tail, they meet at the next angle of the parallelogram, 180° − θ. So the angle inside the triangle between F₁ and F₂ is 180° − θ, and the cosine rule from maths gives
Two forces at any angle. Two forces F₁ and F₂ with an angle θ between them have a resultant F = √(F₁² + F₂² + 2F₁F₂ cos θ).
Check the formula against the cases you know. At θ = 0, cos θ = 1 and F² = (F₁ + F₂)², so F = F₁ + F₂. At θ = 90°, cos θ = 0 and the formula becomes Pythagoras. At θ = 180°, cos θ = −1 and F² = (F₁ − F₂)², so F = |F₁ − F₂|. In between, cos θ falls steadily as θ grows, so the resultant shrinks steadily from the sum down to the difference. The table shows this for two pairs of forces.
Angle θ
3 N and 4 N
10 N and 10 N
0°
7 N
20 N
60°
√37 ≈ 6.1 N
10√3 ≈ 17.3 N
90°
5 N
10√2 ≈ 14.1 N
120°
√13 ≈ 3.6 N
10 N
180°
1 N
0
Two entries deserve a second look. 3 N and 4 N give 5 N only at exactly 90°; at every other angle they give something else between 1 N and 7 N. And two equal forces at 120° give a resultant exactly as large as each of them. The figure shows why: their parallelogram is a rhombus, and the diagonal cuts it into two equilateral triangles, so the diagonal is as long as the sides.
Two 10 N forces 120° apart. The diagonal makes 60° with each side, so it splits the rhombus into two equilateral triangles, and the resultant is 10 N, as large as each force, pointing midway between them.
Two equal forces F. At 60° apart their resultant is √3F, at 90° it is √2F, and at 120° it is F, always pointing midway between the two forces. Wider than 120°, the resultant is smaller than each force; at 180° it is 0.
Worked example · two forces at 60°
Two forces of 3 N and 5 N act at a point with an angle of 60° between them. Find the size of their resultant.
Check: at 90° the two forces would give √34 ≈ 5.8 N. An angle of 60° is narrower than 90°, so the resultant should be larger than that, and 7 N is. It also lies between 5 − 3 = 2 N and 5 + 3 = 8 N.
As θ runs from 0° to 180°, the resultant takes every value from F₁ + F₂ down to |F₁ − F₂|, and no other. That is the range rule of the next page. In a question that asks which values the resultant can or cannot take, you need neither the angle nor this formula, only the two ends.
Try it! · Check-in question
Two forces, each of 6 N, act at a point. Their resultant is also 6 N. What is the angle between the two forces?
Put F₁ = F₂ = F = 6 N into F² = F₁² + F₂² + 2F₁F₂ cos θ: 36 = 36 + 36 + 72 cos θ, so 72 cos θ = −36, cos θ = −½ and θ = 120°. The answer is A: the rhombus splits into two equilateral triangles. B is the angle inside those triangles, between each force and the resultant, not the angle between the two forces; at 60° apart the resultant would be 6√3 ≈ 10.4 N. C gives 6√2 ≈ 8.5 N by Pythagoras. D puts the forces in opposite directions, where they cancel and the resultant is 0.
Resolving a force into two parts
Adding turns several forces into one. Resolving runs the other way: it replaces one force by two forces, its components, that together have the same effect. By the parallelogram rule, any two forces whose parallelogram has F as its diagonal would do, so a force can be split in endless ways. The useful choice is almost always two perpendicular directions, picked to suit the question: horizontal and vertical for a body on a level floor, along and across the surface for a body on a slope.
Take a pull F at an angle θ above the horizontal. Its horizontal part and its vertical part are the two shorter sides of a right-angled triangle with F as the hypotenuse. The horizontal side is next to θ, so it is F cos θ. The vertical side is opposite θ, so it is F sin θ. As a check, (F cos θ)² + (F sin θ)² = F²(cos²θ + sin²θ) = F², so by Pythagoras the two parts rebuild F.
An angled pull F split into two parts. The horizontal part is the side next to θ, F cos θ; the vertical part is the side opposite θ, F sin θ. Put end to end, the two parts rebuild F.
Components of a force. A force F at an angle θ to a chosen direction has a component F cos θ along that direction and F sin θ perpendicular to it. For a pull at θ above the horizontal, Fx = F cos θ horizontally and Fy = F sin θ vertically. The components replace F; they are not extra forces.
Which part gets the cosine? The cosine goes with the side next to the angle you are given, and the sine with the side opposite it. So if a question measures θ from the vertical instead, the vertical part becomes F cos θ. When in doubt, test an extreme: at θ = 0 the pull lies along the horizontal, so the horizontal part must be the whole of F. Indeed F cos 0 = F, and the vertical part is F sin 0 = 0.
Why split a force at all? Because forces along one line are easy to combine. Once the angled pull is split, the horizontal and the vertical directions can be handled separately. On a level floor a body neither rises nor sinks, so the upward forces balance the downward ones. Gravity, mg, pulls down. The floor's normal force N and the vertical part of the pull, F sin θ, act upwards. So N + F sin θ = mg, and N = mg − F sin θ. This makes exact a note from the Drawing the Forces on a Body page: the upward part of an angled pull holds up some of the weight, so the floor pushes less than mg.
Worked example · the parts of an angled pull
A child pulls a 10 kg sledge across level snow with a rope at 37° above the horizontal. The tension in the rope is 25 N. Find the horizontal and vertical parts of the pull, and the normal force from the snow. (sin 37° = 0.6, cos 37° = 0.8, g = 10 m/s²)
Horizontal part: F cos 37° = 25 × 0.8 = 20 N, forwards.
Vertical part: F sin 37° = 25 × 0.6 = 15 N, upwards.
Weight: mg = 10 × 10 = 100 N. The sledge neither rises nor sinks, so N = mg − F sin 37° = 100 − 15 = 85 N.
Check: 20² + 15² = 400 + 225 = 625 = 25², so the two parts rebuild the 25 N pull.
Angles the papers use. Chinese papers often use 37° and give sin 37° = 0.6 and cos 37° = 0.8, the ratios of a 3-4-5 triangle. January 2026 Q21 pulled a box along level ground with a 50 N force at 37° to the horizontal and gave cos 37° ≈ 0.8. Only the horizontal part, 50 × 0.8 = 40 N, acts along the ground, and the Work & Energy topic uses it to find the work done. Know the 30° values by heart as well, since 30° is the papers' favourite slope: sin 30° = ½ and cos 30° = √3/2 ≈ 0.87.
Try it! · Check-in question
A rope pulls a 5 kg box along a level floor with a force of 20 N at 30° above the horizontal. The box stays on the floor. What is the normal force from the floor on the box? (g = 10 m/s²)
The vertical part of the pull is 20 sin 30° = 20 × ½ = 10 N, upwards. The box neither rises nor sinks, so N + 10 = mg = 50 N, and N = 40 N. The answer is D. A is N = mg out of habit, which ignores the upward part of the pull. B adds the 10 N instead of subtracting it; the rope pulls upwards, so it takes some of the weight off the floor. C subtracts 20 cos 30° = 10√3 N, the horizontal part, as if it pointed up. The vertical side of the triangle is opposite the 30° angle, so it takes the sine.
Resolving the weight on a slope
On a slope the natural directions are along the surface and perpendicular to it, because, as the Drawing the Forces on a Body page showed, the normal force and friction already lie along them. Only gravity does not: it stays vertical. So split mg into two parts, one along the slope and one into it.
First find the angle. The slope's surface is tilted by θ from the horizontal. Turn the horizontal and the surface each through 90°, and you get the vertical and the line perpendicular to the surface, still θ apart. So the angle between mg, which is vertical, and the perpendicular to the slope is θ, the angle of the slope itself. In the figure it is marked at the centre of the block, between mg and the arrow into the slope.
The part into the slope is next to θ, so it is mg cos θ. The part along the slope is opposite θ, so it is mg sin θ, pointing down the slope. Check with the extremes. On a level floor, θ = 0, the part along the surface is mg sin 0 = 0, and the whole weight presses into the floor, as it should. On a very steep slope, with θ close to 90°, sin θ is close to 1, and nearly all of the weight pulls the block along the surface, like a block dropping beside a wall.
The weight resolved along and across a slope. The angle between mg and the perpendicular to the slope equals the slope's angle θ, so the part into the slope, next to θ, is mg cos θ, and the part down the slope, opposite θ, is mg sin θ.
Weight on a slope. On a slope at θ to the horizontal, the weight mg has a component mg sin θ down the slope and a component mg cos θ perpendicular to the slope, pressing into it. When no other force acts across the slope, the normal force balances the second part: N = mg cos θ.
Worked example · the weight on a 30° slope
A 4 kg block sits on a slope at 30° to the horizontal. Find the components of its weight along the slope and perpendicular to it, and the normal force on the block. (g = 10 m/s²)
Weight: mg = 4 × 10 = 40 N, vertically downward.
Along the slope: mg sin 30° = 40 × ½ = 20 N, down the slope.
Perpendicular to the slope: mg cos 30° = 40 × √3/2 = 20√3 N ≈ 34.6 N, into the surface.
Normal force: nothing else acts across the slope, so N = mg cos 30° = 20√3 N ≈ 34.6 N, less than the 40 N the block would feel on a level floor.
The 20 N down the slope is what friction has to hold if the block stays at rest. The last page of this topic, Equilibrium on an Incline, takes that up.
Parts, not extra forces. The components replace mg; they are not added to it. On a free-body diagram, draw mg once, or draw its two parts in its place, but never both as full forces, or the weight is counted twice. And when a question asks about gravity itself, use the whole of mg: the impulse of gravity on a block sliding down a slope for a time t is mgt, not mg sin θ × t. December 2025 Q16 and April 2026 Q18 tested exactly that.
Try it! · Check-in question
A 6 kg block rests on a slope at 30° to the horizontal. What are the components of its weight along the slope and perpendicular to it? (g = 10 m/s²)
The weight is mg = 6 × 10 = 60 N. Along the slope: mg sin 30° = 60 × ½ = 30 N. Perpendicular to it: mg cos 30° = 60 × √3/2 = 30√3 N ≈ 52 N. The answer is B. A swaps sine and cosine; test it with a level floor, θ = 0, where the part along the surface must vanish: sin 0 = 0, so the sine belongs along the slope. C keeps the whole weight pressing into the slope, as on a level floor, but part of the weight now pulls along the slope, so less is left to press into it. D uses the mass, 6, in place of the weight, mg = 60 N.
You will meet this again. The 30° slope is the papers' favourite setting, and each slope question starts by resolving mg. December 2025 Q9 and April 2026 Q10 let a block slide down a smooth 30° slope, where mg sin 30° sets the acceleration (the Newton's Laws topic). March 2026 Q9 needs the friction on a block resting on a slope, which equals mg sin θ, and March 2026 Q46 pulls a block up a rough slope with a rope; both are on the last page of this topic. Adding forces at an angle comes back with electric forces: January 2026 Q29 combines the forces of two charges on a third with the parallelogram rule.
Your turn! · Practice questions
Practice · Question 1
Three horizontal forces act on a cart along one straight line: 7 N to the right, 4 N to the left and 9 N to the left. What is their resultant?
Take right as positive: +7 − 4 − 9 = −6 N. The minus sign means to the left, so the answer is C. Grouping gives the same result: 13 N to the left against 7 N to the right leaves 6 N to the left. A has the right size but drops the sign, and the sign is what gives the direction. B adds all three sizes and ignores their directions. D gives the 4 N force the wrong sign, 7 + 4 − 9 = 2, as if it pulled to the right.
Practice · Question 2
Two perpendicular forces act at a point, and their resultant is 15 N. One of the two forces is 9 N. What is the other?
For perpendicular forces F² = F₁² + F₂², so F₂² = 15² − 9² = 225 − 81 = 144, and F₂ = 12 N. The answer is D. A subtracts the sizes, 15 − 9, as if the forces lay along one line. B adds them, 15 + 9. C adds the squares, √(225 + 81) ≈ 17.5 N, which would make one force longer than the resultant; but the resultant is the hypotenuse, the longest side of the right-angled triangle.
Practice · Question 3
Two forces of 8 N and 5 N act at a point with an angle of 120° between them. What is the size of their resultant?
F² = 8² + 5² + 2 × 8 × 5 × cos 120° = 64 + 25 + 80 × (−½) = 89 − 40 = 49, so F = 7 N, and the answer is A. It lies between 8 − 5 = 3 N and 8 + 5 = 13 N, as it must. B takes cos 120° as +½, which gives the resultant for an angle of 60°. C is Pythagoras, √(64 + 25), right only at 90°. D adds the sizes, right only at 0°. Since 120° is wider than 90°, the answer must be less than √89 ≈ 9.4 N, which rules out B and D straight away.
Practice · Question 4
A gardener pushes a 20 kg lawnmower along level ground. The push acts along the handle: 100 N at 37° below the horizontal. What is the normal force from the ground on the mower? (sin 37° = 0.6, cos 37° = 0.8, g = 10 m/s²)
The push points down into the ground, so its vertical part, 100 sin 37° = 60 N, acts downwards, on top of the weight, mg = 200 N. The mower neither sinks nor rises, so N = 200 + 60 = 260 N, and the answer is C. A subtracts the 60 N, as for a rope pulling upwards; a push down the handle presses the mower into the ground instead. B is N = mg, which leaves out the push altogether. D adds the horizontal part, 100 cos 37° = 80 N, instead of the vertical one; the horizontal part pushes the mower along the ground, not into it.
Practice · Question 5
A block is placed on a slope at 30° to the horizontal. The component of its weight along the slope is 15 N. What is the mass of the block? (g = 10 m/s²)
The component along the slope is mg sin 30° = ½mg, so ½mg = 15 N, mg = 30 N and m = 30 ÷ 10 = 3 kg. The answer is B. A treats the 15 N as the whole weight: 15 ÷ 10 = 1.5 kg. C uses the cosine, mg cos 30° = 15 N, which gives mg = 10√3 N and m = √3 kg; but the cosine gives the part perpendicular to the slope. D multiplies by sin 30° instead of dividing by it: 15 × ½ ÷ 10 = 0.75 kg, a block whose along-slope part would be only 3.75 N.
Things to remember
Concept
What to know
Resultant force
The single force with the same effect as all the forces together (合力). It replaces them; it is not one more force on the body.
Forces along one line
Same direction: add. Opposite directions: subtract, and the resultant points along the larger force. With several forces, choose a positive direction and add with signs. April 2026 Q6: 15 − 5 = 10 N.
Parallelogram rule
Draw both forces from one point and complete the parallelogram; the diagonal from that point is the resultant, in size and direction.
Triangle rule
Draw the forces tip to tail; the resultant runs from the first tail to the last tip. The same line pointing the other way is the force that would balance them.
At right angles
F = √(F₁² + F₂²), with tan α = F₂/F₁. Use it only when the question says the forces are perpendicular: 3 N and 4 N give 5 N at 90° and at no other angle.
At any angle
F = √(F₁² + F₂² + 2F₁F₂ cos θ). As θ opens from 0° to 180°, the resultant shrinks from F₁ + F₂ to |F₁ − F₂|.
Two equal forces F
√3F at 60°, √2F at 90°, F at 120°, 0 at 180°; the resultant points midway between them.
Components
A force F at θ to a direction has F cos θ along it and F sin θ perpendicular to it. The cosine goes with the side next to θ. The components replace F.
Angled force on a floor
A pull at θ above the horizontal: N = mg − F sin θ. A push at θ below the horizontal: N = mg + F sin θ.
Weight on a slope
mg sin θ down the slope, mg cos θ into it, and N = mg cos θ when nothing else acts across the slope. Test with θ = 0: the along-slope part must vanish.
Values to know
sin 30° = ½, cos 30° = √3/2; sin 37° = 0.6, cos 37° = 0.8 (the papers give these).
