Resultant of forces

Push a swing from two directions at once and it does not move along either push alone — it moves along a single combined direction that depends on both the size and the angle of the two shoves. That single equivalent push is the resultant, and finding it correctly is the very first skill needed before any equilibrium or dynamics problem can be solved. This article treats force as a vector and shows three ways to combine several concurrent forces into one resultant: the parallelogram law, the equivalent triangle law, and the more reliable method of resolving into rectangular components. It then derives the parallelogram-law formula from the cosine rule, works through the standard two-force example in full, checks a three-force equilibrium case, and closes with the exam-ready traps and formulas.

The core insight

Several concurrent forces combine like vectors, not scalars, so their resultant magnitude is $R=\sqrt{F_1^2+F_2^2+2F_1F_2\cos\theta}$ by the parallelogram law, and a system is in equilibrium exactly when this resultant is zero.

The core idea

Force has both magnitude and direction, so combining forces is vector addition, never plain arithmetic on their magnitudes.

Parallelogram law of forces. If two forces $\vec F_1$ and $\vec F_2$ act at a point and are drawn as the two adjacent sides of a parallelogram, their resultant $\vec R$ is the diagonal of that parallelogram drawn from the same point. With $\theta$ the angle between $\vec F_1$ and $\vec F_2$ , the magnitude and direction of the resultant are

R=\sqrt{F_1^2+F_2^2+2F_1F_2\cos\theta}, \qquad \tan\alpha=\frac{F_2\sin\theta}{F_1+F_2\cos\theta}

where: $R$ is the magnitude of the resultant, $F_1,F_2$ are the magnitudes of the two forces, $\theta$ is the angle between them measured at the common point, and $\alpha$ is the angle the resultant makes with $\vec F_1$ .

Triangle law of forces. Draw $\vec F_1$ , then draw $\vec F_2$ starting from the tip of $\vec F_1$ (magnitude and direction preserved). The resultant $\vec R$ is the third side of the triangle, drawn from the tail of $\vec F_1$ to the tip of $\vec F_2$ . This is just the head-to-tail method, and it gives exactly the same $R$ and $\alpha$ as the parallelogram law — the two are the same geometry viewed differently.

Resolution into rectangular components. For more than two forces, or forces at awkward angles, resolving every force along fixed $x$ - and $y$ -axes and summing component-wise is the most systematic method:

R_x=\sum F_x, \qquad R_y=\sum F_y, \qquad R=\sqrt{R_x^2+R_y^2}, \qquad \tan\theta=\frac{R_y}{R_x}

where: $R_x,R_y$ are the net components along the chosen axes, and the final $R$ is recombined from them exactly as if they were two perpendicular forces.

Figure

Two forces F1 and F2 drawn from a common point with angle theta between them completed into a parallelogram whose diagonal R is the resultant

Equilibrium of concurrent forces. A body under several concurrent forces stays at rest, or moves at constant velocity, only if their resultant is zero:

\sum \vec F=0 \quad\Longleftrightarrow\quad \sum F_x=0 \text{ and } \sum F_y=0

For exactly three concurrent, coplanar forces, this is equivalent to the geometric statement that the three vectors, taken in order, form a closed triangle — the triangle law applied in reverse as a test for zero resultant, rather than as a way to build one.

Derivation ADeriving the resultant magnitude from the cosine rule

The goal is to turn the geometric parallelogram picture of two forces $\vec F_1$ and $\vec F_2$ , inclined at angle $\theta$ , into the algebraic formula $R=\sqrt{F_1^2+F_2^2+2F_1F_2\cos\theta}$ that lets the resultant be computed from two magnitudes and one angle without drawing anything to scale.

Step 1. Place $\vec F_1$ and $\vec F_2$ tail to tail at the common point, and complete the parallelogram: draw a copy of $\vec F_2$ starting from the tip of $\vec F_1$ , so that the diagonal from the common point to the far corner is the resultant $\vec R$ . Because the sides of a parallelogram are parallel, the angle between $\vec F_1$ and the copy of $\vec F_2$ attached to its tip is $180°-\theta$ (co-interior with $\theta$ ).

Step 2. The three vectors $\vec F_1$ , the shifted copy of $\vec F_2$ , and $\vec R$ now form a triangle, with $\vec F_1$ and $\vec F_2$ as two sides and $\vec R$ as the third side, and the included angle between the sides of length $F_1$ and $F_2$ equal to $180°-\theta$ .

Step 3. Apply the cosine rule to this triangle: $R^2=F_1^2+F_2^2-2F_1F_2\cos(180°-\theta)$ . Since $\cos(180°-\theta)=-\cos\theta$ , the minus sign in front flips, giving $R^2=F_1^2+F_2^2+2F_1F_2\cos\theta$ .

Step 4. Take the positive square root, since $R$ is a magnitude: $R=\sqrt{F_1^2+F_2^2+2F_1F_2\cos\theta}$ . The direction $\alpha$ follows from resolving $\vec F_2$ along and perpendicular to $\vec F_1$ : the component of $\vec F_2$ perpendicular to $\vec F_1$ is $F_2\sin\theta$ , and the component of $\vec R$ along $\vec F_1$ is $F_1+F_2\cos\theta$ , so $\tan\alpha=\dfrac{F_2\sin\theta}{F_1+F_2\cos\theta}$ .

Worked example

Two forces of magnitude $8\ \text{N}$ and $6\ \text{N}$ act at a point, with an angle of $60°$ between them. Find the magnitude of their resultant and the angle it makes with the $8\ \text{N}$ force.

Step 1. Identify the values: $F_1=8\ \text{N}$ , $F_2=6\ \text{N}$ , $\theta=60°$ , so $\cos 60°=0.5$ and $\sin 60°\approx 0.866$ .

Step 2. Substitute into the magnitude formula: $R=\sqrt{F_1^2+F_2^2+2F_1F_2\cos\theta}=\sqrt{8^2+6^2+2(8)(6)(0.5)}=\sqrt{64+36+48}=\sqrt{148}\approx 12.17\ \text{N}$ .

Step 3. Substitute into the direction formula: $\tan\alpha=\dfrac{F_2\sin\theta}{F_1+F_2\cos\theta}=\dfrac{6(0.866)}{8+6(0.5)}=\dfrac{5.196}{11}\approx 0.4724$ , giving $\alpha\approx 25.3°$ .

The resultant has magnitude about 12.17 N, directed at approximately 25.3° from the 8 N force.

Competitive stretch

Check the same $8\ \text{N}$ , $6\ \text{N}$ pair by resolving along axes with $\vec F_1$ chosen along the $x$ -axis: $F_{1x}=8$ , $F_{1y}=0$ , and $F_{2x}=6\cos 60°=3$ , $F_{2y}=6\sin 60°\approx 5.196$ . Then $R_x=8+3=11$ , $R_y=0+5.196=5.196$ , so $R=\sqrt{11^2+5.196^2}=\sqrt{121+27}=\sqrt{148}\approx 12.17\ \text{N}$ and $\tan\alpha=5.196/11\approx0.4724$ — matching Step 2 and Step 3 exactly, since both methods describe the same physical resultant.

Common traps

Adding the two magnitudes directly, as if they were scalars — this only gives the correct resultant when the forces are exactly parallel ( $\theta=0°$ ); for any other angle the vector formula must be used.
Using the wrong angle $\theta$ in the formula — it must be the angle between the two force vectors as drawn from the common point, not the interior angle of the parallelogram or some other angle visible in a diagram.
Sign errors while resolving forces into components — a consistent choice of positive $x$ - and $y$ -directions must be maintained across every force before the components are summed.
Assuming three forces can only balance if all three have equal magnitude — equilibrium only requires that the three vectors, taken in order, close into a triangle (equivalently $\sum F_x=0$ and $\sum F_y=0$ ); equal magnitudes at $120°$ spacing is just one special case of this.

Why does the triangle law give the same resultant as the parallelogram law?

Both describe the same vector addition, only drawn differently. In the parallelogram law, $\vec F_1$ and $\vec F_2$ are drawn from a common point and $\vec R$ is the diagonal; in the triangle law, $\vec F_2$ is instead shifted to start at the tip of $\vec F_1$ , and $\vec R$ closes the triangle from the tail of $\vec F_1$ to the tip of the shifted $\vec F_2$ . Shifting a vector without changing its magnitude or direction does not change what it represents, so the triangle's third side and the parallelogram's diagonal are the exact same vector $\vec R$ .

When should I use the parallelogram formula instead of resolving into components?

The parallelogram formula is fastest for exactly two forces at a known angle, since it gives $R$ and $\alpha$ in two direct substitutions. Once there are three or more concurrent forces, or the angles are inconvenient, resolving each force into $x$ - and $y$ -components and summing is far more reliable, because it never requires tracking multiple angles between different pairs of forces at once — you just add numbers along each axis and recombine at the end.

If the resultant of two forces is zero, does that mean the forces themselves are zero?

No. A zero resultant only means the two forces are equal in magnitude and exactly opposite in direction ( $\theta=180°$ ), so they cancel — each force can still be large. This is the whole basis of equilibrium: a body can have several substantial forces acting on it and still not accelerate, provided those forces sum to zero as vectors. Checking $\sum F_x=0$ and $\sum F_y=0$ separately is the standard way to confirm this.

Resultant of forces

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads