Coordinate Systems & Components

A coordinate system is laid over a situation that does not care about it. A block slides, a ball flies and a rope pulls exactly as before whether zero sits on the floor or on the table, so the physics must live in what no drawing of axes can reach: the distance between two points, the length of a vector, the angle between two directions. Components are the price of turning that geometry into arithmetic, and the price is dependence. Move the origin and every position reads differently; point the axes another way and the same arrow gets a new pair of numbers, each sign reporting only which way along one particular axis the arrow leans. A change of description is therefore never free-form. It follows rules strict enough that lengths, distances and angles survive every legitimate change, which is why Newton's laws are written for vectors rather than for components in one favoured set of axes. The freedom ends where the axes stop being a drawing and become a moving object. Axes carried by a train measure different velocities, and axes that accelerate or keep turning change which forces seem to act: here we change fixed coordinates in the same frame; moving or continuously rotating frames require additional care.

Before a single component exists you have chosen four things: where the origin sits, which directions the axes point, which way along each counts as positive, and what unit one step represents. In three dimensions the axes are also made right-handed by convention, so with x to the right and y up the page, z points out of the page towards you. None of these choices is part of the physics. A lamp hanging 2.40 m above the floor is at y = 2.40 m with the origin on the floor and y up, at y = 1.65 m with the origin on a 0.75 m tabletop, and at y = −1.65 m if you keep that origin but call down positive. Three numbers, one lamp. Choose to make later work short: put the origin where something starts, stops or balances, and lay an axis along the direction things actually move.

A 10.0 m displacement aimed 30.0° above +x has components (8.66 m, 5.00 m). Turn the axes 30.0° counterclockwise and the same displacement reads (10.0 m, 0.00 m) — still 10.0 m long.

Drop perpendiculars from the tail and the tip of a vector onto the x-axis. The stretch of axis between the two feet, counted positive if it runs towards +x and negative if it runs the other way, is the component Aₓ. Trigonometry supplies its size: with θ measured counterclockwise from +x, Aₓ = A cos θ and A_y = A sin θ, and the signs take care of themselves in every quadrant. A 250 N pull at θ = 135° has Fₓ = 250 cos 135° = −177 N and F_y = 250 sin 135° = +177 N, so it leans left and up. A component is not a magnitude. It carries the vector's unit, it can be zero, and a minus sign means direction, not smallness: Fₓ = −177 N pulls exactly as hard along the x-axis as +177 N would. The pair of signed components fixes the vector completely; the magnitude alone never does.

Real problems rarely hand you θ from +x. They give a bearing such as 35° east of south, a rope angle measured from the vertical, or a slope angle measured from the ground. Two routes both work. Convert the angle to θ from +x and let cosine and sine carry the signs, or use the angle as given, give the cosine to the axis the angle is measured from — the adjacent side — and read each sign off the sketch. A rope under 180 N of tension, leaning 40.0° to the left of vertical, pulls up with 180 cos 40.0° = 138 N and sideways with 180 sin 40.0° = 116 N; the sketch says the sideways part points left, so Fₓ = −116 N and F_y = +138 N. The first route agrees: θ = 90.0° + 40.0° = 130.0°, and 180 cos 130.0° = −116 N. Cosine belongs to the adjacent axis, never automatically to x.

Given Aₓ and A_y, the length comes from Pythagoras, A = √(Aₓ² + A_y²), and it is the same whatever the signs. The direction is where a calculator misleads. arctan(A_y/Aₓ) only returns angles between −90° and +90°, so it cannot tell (+6.0, −2.5) from (−6.0, +2.5): both ratios are −0.417. For a displacement of (−6.0 m, +2.5 m) the length is √(36.0 + 6.25) = 6.5 m, and the calculator offers −22.6°, an arrow pointing right and down. The signs say left and up, the second quadrant, so the direction is −22.6° + 180° = 157.4° from +x. The two-argument function atan2(A_y, Aₓ) does that bookkeeping for you; by hand, sketch the signs before you trust the angle. Notice that the pair (A, θ) is itself a description of the same arrow — polar rather than Cartesian.

Move the origin to the point (x₀, y₀) of the old system and every position changes by the same constant: x′ = x − x₀ and y′ = y − y₀. A cart that rolls from x = 2.0 m to x = 5.5 m in 1.4 s, described from a new origin at x₀ = 7.0 m, starts at x′ = −5.0 m and ends at x′ = −1.5 m. Both positions turned negative, yet the cart did not reverse: Δx′ = −1.5 − (−5.0) = +3.5 m, exactly Δx, and its average velocity is still 2.5 m s⁻¹. Calculus says why in one line. x₀ is a constant, so dx′/dt = dx/dt − 0: velocities agree, and differentiating again, so do accelerations. Positions depend on where you put zero; displacements, velocities and accelerations do not. The new origin must be a fixed point, though: an origin moving at velocity u gives dx′/dt = dx/dt − u, subtracting its own velocity from every velocity you record.

Rotate the axes counterclockwise through φ. A vector at θ from the old +x axis sits at θ − φ from the new one, so Aₓ′ = A cos(θ − φ) and A_y′ = A sin(θ − φ); the difference formulas turn these into Aₓ′ = Aₓ cos φ + A_y sin φ and A_y′ = −Aₓ sin φ + A_y cos φ. A velocity of 12.0 m s⁻¹ at 50.0° has components (7.71, 9.19) m s⁻¹. Turn the axes through 20.0° and it becomes (7.71 cos 20.0° + 9.19 sin 20.0°, −7.71 sin 20.0° + 9.19 cos 20.0°) = (10.39, 6.00) m s⁻¹. Different numbers, same arrow: 10.39² + 6.00² ≈ 144 = 12.0². That is what makes tilted axes worth the trouble. On a slope, aim x′ along the surface and a sliding block has a single velocity component instead of two, so one equation carries the motion.

The minus sign is not part of g. Many students carry g = −9.81 m s⁻² into every problem as a fact about gravity. Then they choose up as positive, write a_y = −g, substitute their negative g and get a_y = +9.81 m s⁻²: a dropped ball that speeds up towards the ceiling. Others choose down as positive, write a_y = g, and keep the minus anyway. The error is filing a sign under a magnitude. g is the size of the free-fall acceleration, 9.81 m s⁻², and a size is never negative. The sign belongs to a component, and a component's sign is set by the axis you drew: a_y = −g with y up, a_y = +g with y down, aₓ′ = −g sin α for a block on a frictionless slope with x′ pointing up it. Draw the axis, read each sign from the drawing, and substitute every magnitude as a positive number.