Resultant Force Vector Calculator

Calculate the resultant force using the parallelogram law:
\[ R = \sqrt{F_1^2 + F_2^2 + 2\,F_1\,F_2\cos\theta} \] and the angle between \( F_1 \) and the resultant: \[ \alpha = \arctan\!\left(\frac{F_2\sin\theta}{F_1+F_2\cos\theta}\right) \]

* Enter forces in newtons (N) and angle in degrees.

Step 1: Enter Parameters

Example: 10 N

Example: 15 N

Example: 60°

Equations used:
\( R = \sqrt{F_1^2 + F_2^2 + 2\,F_1\,F_2\cos\theta} \)
\( \alpha = \arctan\!\left(\frac{F_2\sin\theta}{F_1+F_2\cos\theta}\right) \)


Practical Example:
For \( F_1 = 10 \) N, \( F_2 = 15 \) N, and \( \theta = 60° \):
\[ R \approx \sqrt{10^2 + 15^2 + 2\times10\times15\times\cos60°} \approx \sqrt{100+225+300\times0.5} \approx \sqrt{100+225+150} \approx \sqrt{475} \approx 21.79\, \text{N} \] and \[ \alpha \approx \arctan\!\left(\frac{15\times\sin60°}{10+15\times\cos60°}\right) \approx \arctan\!\left(\frac{15\times0.866}{10+15\times0.5}\right) \approx \arctan\!\left(\frac{12.99}{17.5}\right) \approx 36.87°. \]

Resultant Force Vector Calculator (In-Depth Explanation)

Resultant Force Vector Calculator (In-Depth Explanation)

Whenever multiple forces act on an object, the overall or “net” force is determined by adding all the force vectors together. This final or combined force is called the resultant force. Because force is a vector, we have to add them using vector rules, not just ordinary arithmetic. In simpler terms, each force can push or pull in different directions, and we must account for the direction as well as the magnitude (size) of each force.

The process to find the resultant force may feel a bit abstract, so let’s break it down carefully. We generally use a coordinate system—often an $x$-axis (horizontal) and a $y$-axis (vertical)—to simplify the math. Each force can be broken down into components along these axes. Then we sum up all the horizontal components to get the net $x$-component, and we sum up all the vertical components to get the net $y$-component.

1. Vector Summation: The Core Concept

Suppose we have $n$ different forces acting on an object: $$\vec{F}_1, \quad \vec{F}_2, \quad \dots, \quad \vec{F}_n.$$ Each of these forces has both a magnitude (how strong the force is) and a direction (which way it pushes).

To add them properly, we can rewrite each force in terms of its $x$ (horizontal) and $y$ (vertical) components. Symbolically, for force $i$:

$$\vec{F}_i = \Bigl(F_{i,x},\; F_{i,y}\Bigr).$$

This simply means that $\vec{F}_i$ is described by two numbers: how much force acts in the $x$-direction ($F_{i,x}$) and how much force acts in the $y$-direction ($F_{i,y}$).

If you already know the force’s magnitude $|\vec{F}_i|$ (for example, 10 newtons) and its angle $\theta_i$ (the direction measured relative to the positive $x$-axis), you can convert from magnitude-angle form to component form using the standard trigonometry:

$$F_{i,x} = |\vec{F}_i|\cos(\theta_i), \quad F_{i,y} = |\vec{F}_i|\sin(\theta_i).$$

This relies on the idea that the adjacent side of a right triangle (the $x$-component) is found using cosine, and the opposite side (the $y$-component) is found using sine, given the hypotenuse is the force vector’s magnitude $|\vec{F}_i|$.

Once we have all forces expressed as $(F_{i,x}, F_{i,y})$, we can sum them up one by one:

$$\vec{F}_\mathrm{net} = \sum_{i=1}^n \vec{F}_i \;=\; \Bigl(\sum_{i=1}^n F_{i,x},\; \sum_{i=1}^n F_{i,y}\Bigr).$$

In other words, we add all the $x$-components together to get one final horizontal component, and we add all the $y$-components together to get one final vertical component:

$$F_{\mathrm{net},x} = \sum_{i=1}^n F_{i,x}, \quad F_{\mathrm{net},y} = \sum_{i=1}^n F_{i,y}.$$

Conceptually, this is just the same as having $n$ arrows on a plane, each pointing in different directions, and shifting them “tip to tail” so we can see where the final arrow ends up. The final arrow from the start to the end is the resultant.

2. Finding Magnitude and Direction of the Resultant

After summing up components, we typically want to know how strong (in newtons, for instance) the net force is, and in which direction (the angle with respect to the $+x$-axis).

To find the magnitude of the resultant force vector, we use the Pythagorean theorem:

$$|\vec{F}_\mathrm{net}| = \sqrt{\bigl(F_{\mathrm{net},x}\bigr)^2 + \bigl(F_{\mathrm{net},y}\bigr)^2}.$$

This formula is essentially describing the length of the hypotenuse of a right triangle whose sides are $F_{\mathrm{net},x}$ and $F_{\mathrm{net},y}$.

Next, we find the angle $\theta_\mathrm{net}$ by using the inverse tangent ($\tan^{-1}$) to figure out the ratio of vertical to horizontal force:

$$\theta_\mathrm{net} = \tan^{-1}\!\Bigl(\frac{F_{\mathrm{net},y}}{F_{\mathrm{net},x}}\Bigr).$$

However, we must be careful with the signs of $F_{\mathrm{net},x}$ and $F_{\mathrm{net},y}$. If $F_{\mathrm{net},x}$ is negative (meaning the net force is pointing to the left), or $F_{\mathrm{net},y}$ is negative (the net force is pointing downward), we may need to adjust the angle. In practical terms, many scientific calculators or software functions like atan2(y, x) handle this quadrant check automatically.

Example: Summing Two Forces Step by Step

Imagine you have two forces acting on a point mass on a frictionless surface:

  • Force 1: Magnitude $10\,\mathrm{N}$ at angle $30^\circ$ from the $+x$-axis
  • Force 2: Magnitude $15\,\mathrm{N}$ at angle $120^\circ$ from the $+x$-axis

Step 1) Convert each force to components.

For Force 1:

$$F_{1,x} = 10\cos(30^\circ) \;\approx\; 10\times0.866 \;=\; 8.66\,\mathrm{N},$$ $$F_{1,y} = 10\sin(30^\circ) \;\approx\; 10\times0.500 \;=\; 5.00\,\mathrm{N}.$$

This means Force 1 effectively pushes $8.66\,\mathrm{N}$ in the $x$-direction (to the right) and $5.00\,\mathrm{N}$ in the $y$-direction (upwards).

For Force 2:

$$F_{2,x} = 15\cos(120^\circ) \;\approx\; 15\times(-0.500)\;=\;-7.50\,\mathrm{N},$$ $$F_{2,y} = 15\sin(120^\circ) \;\approx\; 15\times0.866 \;=\; 12.99\,\mathrm{N}.$$

Here, we see $F_{2,x}$ is negative, meaning it points left ($-7.50\,\mathrm{N}$ along the $x$-axis), and $12.99\,\mathrm{N}$ upward along the $y$-axis.

Step 2) Sum the components to get the net $x$ and $y$.

$$F_{\mathrm{net},x} = 8.66 + (-7.50) = 1.16\,\mathrm{N},$$ $$F_{\mathrm{net},y} = 5.00 + 12.99 = 17.99\,\mathrm{N}.$$

So, looking at these net values:

  • The net horizontal force is $+1.16\,\mathrm{N}$ (a small push to the right).
  • The net vertical force is $+17.99\,\mathrm{N}$ (a larger push upward).

Step 3) Calculate the resultant force’s magnitude and direction.

$$|\vec{F}_\mathrm{net}| = \sqrt{(1.16)^2 + (17.99)^2} \;\approx\; \sqrt{1.35 + 323.64} \;\approx\; \sqrt{324.99} \;\approx\; 18.03\,\mathrm{N}.$$

The resultant force is about $18.03\,\mathrm{N}$ in total. Next, let’s find its angle:

$$\theta_\mathrm{net} = \tan^{-1}\Bigl(\frac{17.99}{1.16}\Bigr) \;\approx\; \tan^{-1}(15.51) \;\approx\; 86.3^\circ.$$

This angle means the resultant vector is pointing almost straight up (since $90^\circ$ would be exactly vertical).

So if we drew both forces on a diagram and then combined them, the end result is a single force of around $18\,\mathrm{N}$, nearly vertical, pushing the object up and slightly to the right.

Conceptual Recap & Tips:
  • Components Are Key: By splitting each force into $x$- and $y$-parts, you simplify the addition into straightforward sums.
  • Magnitude from Pythagorean Theorem: Once you have the net $x$ and $y$, you treat it like a right triangle to find the resultant’s length.
  • Angle from Arctangent: The ratio of $F_{\mathrm{net},y}$ over $F_{\mathrm{net},x}$ tells you how steep the resulting vector is, but watch your quadrant.
  • Any Number of Forces: This process repeats for $n$ forces—just more summations in $x$ and $y$. The final approach remains the same.

This method of vector addition is foundational in physics and engineering: from analyzing structures under multiple loads, to computing net thrust in rocket motors, to evaluating the net aerodynamic force on an aircraft. Mastering these steps gives you a powerful toolkit for many real-world applications.