How this instrument works
When two forces act on the same point but point in different directions, they combine into a single equivalent force called the resultant. Draw the two force vectors tail to tip, or by the parallelogram rule, and the resultant is the vector that closes the triangle — the same geometric construction used to add any two vectors, not just forces. Force behaves this way because it is a vector quantity: two 30-newton pulls on a ring do not add up to 60 newtons unless they happen to line up exactly.
The cosine term in the formula is what the geometry demands. In a triangle formed by F1, F2, and the resultant, the angle entered sits between the two forces, and the law of cosines gives the third side as F_net = √(F1² + F2² + 2·F1·F2·cosθ). At θ = 0°, cosθ = 1 and the forces add directly to F1 + F2, since they point the same way. At θ = 180°, cosθ = −1 and the formula subtracts, returning the difference between the magnitudes, since one force cancels part of the other. At exactly 90°, cosθ = 0 and the whole cross term vanishes, leaving the familiar Pythagorean sum — the right-triangle shortcut is a special case of this formula, not a separate rule.
The formula only holds for two forces meeting at a single point in one plane. Add a third force, or let the two forces act on different points of a rigid body, and this shortcut stops being enough — resolving each vector into x and y components and summing those becomes necessary, or a full free-body diagram is needed. A common error is adding the two magnitudes outright, as if force behaved like mass on a scale; that answer is only correct when the angle between the forces happens to be zero.
- Enter the magnitude of the first pull or push, in newtons, in the Force 1 field.
- Enter the second force's magnitude, in newtons, in the Force 2 field — swapping which force goes where does not change the result.
- Set the Angle between the two forces field to the angle measured between their directions, in degrees.
- Read the net force from the Resultant (net) force field — the single equivalent force the pair produces, in newtons.
Worked example — two forces at 90°, 30 N and 40 N
Picture a mooring ring pulled by two lines at a right angle: one line pulling north at 30 N in the Force 1 field, the other pulling east at 40 N in Force 2, with the Angle between the two forces field set to 90° — exactly π/2 radians, 1.5707963267949, once converted for the cosine. Because θ is 90°, cosθ is 0 and the cross term 2·F1·F2·cosθ drops out entirely, leaving F_net = √(F1² + F2²).
Substituting gives F_net = √(30² + 40²) = √(900 + 1600) = √2500 = 50.0 N — the familiar 3-4-5 right triangle scaled up by ten. The resultant does not point north or east; it points along the diagonal between the two lines, at the angle a rigger would read straight off a protractor laid over them. Move the angle away from 90° and the shortcut disappears — the full law-of-cosines term is what keeps the answer correct at 45° or 120° as well.
Questions
Why is 90° a special case of this formula?
At 90° the cosine term equals zero, so the general law-of-cosines formula F_net = √(F1²+F2²+2F1F2cosθ) collapses to the plain Pythagorean sum √(F1²+F2²). It is not a separate rule, just the same formula with cosθ = 0. Any other angle needs the full cross term, since cosθ is then nonzero and the resultant is no longer simply the hypotenuse of a right triangle.
Can I just add the two forces together?
Only if they point in exactly the same direction, where θ = 0° and cosθ = 1, making F_net = F1 + F2. At any other angle, adding the raw magnitudes overstates the result, because part of each force is spent working against or across the other rather than along a single line. The formula's cross term corrects for exactly this overlap.
What happens when the two forces point in opposite directions?
At θ = 180°, cosθ = −1 and the cross term subtracts the full product 2F1F2, so the formula reduces to F_net = |F1 − F2| — the smaller force is simply overwhelmed by the difference. Two forces of 30 N and 40 N pulling directly against each other leave a net force of exactly 10 N, still directed toward the stronger pull.
Does the order of Force 1 and Force 2 matter?
No. The formula is symmetric in F1 and F2 — swapping the two values produces an identical net force, since squaring and the cross term treat both forces the same way. What matters is the angle between them, measured consistently as the angle you would see drawing both force vectors from the same point.
What if I have three or more forces instead of two?
This formula only combines two concurrent forces at a time. For three or more, resolve each force into perpendicular x and y components, sum the components separately, then find the magnitude of that combined vector with the Pythagorean theorem. You can also chain this calculator: combine two forces into a resultant, then combine that resultant with the third force.
Why does a right angle give such a clean 50 N answer?
Because 30, 40, and 50 form a Pythagorean triple, the same 3-4-5 pattern scaled by ten. At 90°, cosθ is zero, so F_net = √(F1²+F2²) = √(900+1600) = √2500 = 50 N exactly, with no rounding. Change either force or the angle even slightly and the resultant stops landing on a whole number, though the formula itself stays exact.