2D Force Vectors

Resolve a force into its rectangular (x, y) components, and recombine components into a resultant.

A force is a vector: it has a magnitude and a direction, not just a size. Two forces of the same magnitude pulling in different directions do very different things to a body. Everything in statics starts from taking that seriously.

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Any 2D force can be broken into a horizontal piece and a vertical piece — its rectangular components:

F_x = F\cos\theta, \qquad F_y = F\sin\theta

Rectangular components of a force

$F$ the force's magnitude (always positive).
$\theta$ the angle from the positive x-axis, measured counterclockwise.
$F_x, F_y$ the x- and y-components — signed, can be negative.

Read F cos theta as: how much of F survives in the x-direction. At theta = 0, the whole force is along x and F_x = F. At theta = 90 degrees, none of it is, and F_x = 0. The cosine is literally measuring that overlap.

Watch the reference angle. theta here is always measured from the positive x-axis, counterclockwise. Plenty of textbook problems instead give you an angle from a different line (a cable, an incline, the y-axis) — before you plug into F cos theta / F sin theta, convert to the angle from +x. Using the wrong reference angle is the single most common error in this section, not a sign mistake.

Sign is information, not an error. A component can come out negative — that just means it points in the negative axis direction. theta = 210 degrees is in the third quadrant, so both F_x and F_y come out negative: the force points down and to the left. Don't force everything positive by hand; let cosine and sine tell you the sign.

Components recombine into Cartesian vector form using the unit vectors i and j:

\mathbf{F} = F_x\mathbf{i} + F_y\mathbf{j}

Cartesian vector notation

$\mathbf{i}, \mathbf{j}$ unit vectors along the x- and y-axis (magnitude 1).

And you can always go the other way — magnitude and direction back out of the components — with the Pythagorean theorem and inverse tangent:

F = \sqrt{F_x^2 + F_y^2}, \qquad \theta = \tan^{-1}\!\left(\frac{F_y}{F_x}\right)

Magnitude and direction from components

$\tan^{-1}$ watch the quadrant — a calculator's inverse tangent only returns -90 to 90 degrees, so check the signs of F_x and F_y to place theta in the right quadrant.

Worked example

F = 100\text{ lb},\quad \theta = 210^\circ

$F_x = F\cos\theta = 100\cos(210^\circ) = 100\left(-\tfrac{\sqrt3}{2}\right)$
Plug theta = 210 degrees directly into the component formula — no adjustment needed since it's already measured from +x.
$F_x \approx -86.60\text{ lb}$
Negative: at 210 degrees the force points into the third quadrant, so its x-component points in the -x direction.
$F_y = F\sin\theta = 100\sin(210^\circ) = 100\left(-\tfrac12\right)$
Same formula, sine instead of cosine.
$F_y = -50\text{ lb}$
Also negative, for the same reason — third quadrant, pointing down.

Answer:

F_x \approx -86.60\text{ lb}, \quad F_y = -50\text{ lb}

How to solve one of these

1.
Identify F and theta. Magnitude, and the angle measured counterclockwise from the $+x$ -axis.
2.
Convert the reference angle if needed. If the given angle is measured from something other than $+x$ (a cable, an incline, the $y$ -axis), convert it first. This is the step most errors come from.
3.
Apply the component formulas. $F_x = F\cos\theta$ , $F_y = F\sin\theta$ .
4.
Trust the sign. A negative component means that direction is negative — don't override it.
5.
Sanity-check against the quadrant. First quadrant: both positive. Second: $F_x$ negative, $F_y$ positive. Third: both negative. Fourth: $F_x$ positive, $F_y$ negative.

F_x = F\cos\theta, \quad F_y = F\sin\theta

How much of a force's magnitude lands along each axis, given its angle from the positive x-axis.

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