# Net Force Calculator

Created by Davide Borchia
Last updated: Aug 30, 2022

When many forces act on the same body, the effective motion is determined by the net force: calculate it with ease with CalcTool. Our tool allows you to input up to 10 forces, and it calculates the resultant force in the blink of an eye.

Keep reading our article to learn:

• What is the net force;
• How to calculate the net force (formula with angles and formula with vector sum);
• An application of the net force equation.

## May the forces be with you: what is the net force

Imagine being in a car driving along a bend in the road when one of your friends pulls you by the arm. At that moment , three forces act on you, all of them with a specific direction and magnitude: gravity, the centripetal force, and your friend's force. From the point of view of an observer standing outside your reference frame, these forces can't be separated: only the resultant — or net — force appears.

The net force is the result of the vector sum of all the — relevant — forces acting on a body at a specific time. The result of the sum, another force, has the same effect on the motion as all the other forces together, and its action is virtually indistinguishable.

In the next section, you'll learn how to find the net force.

## How do I calculate the net force: formula with angle to calculate the resultant force

To calculate the net force, we need to perform a simple vector sum: this requires us to know:

• Magnitude; and
• Direction.

Of all forces involved in the calculation.

Take, for example a body subjected to four different forces, $\vec{F}_1$, $\vec{F}_2$, $\vec{F}_3$, and $\vec{F}_4$. As you can see, we marked the forces with the vector symbol.

We can visualize this situation in a force diagram: calculating the net force will be even more intuitive! The angles in this diagram highlight the vector nature of forces.

The resultant force is then:

$\vec{F} = \vec{F}_1+\vec{F}_2+\vec{F}_3+\vec{F}_4$

The magnitude of the calculated net force is $\left|\vec{F} \right|$. If we are only interested in the kinematics of the body, we can compute the magnitude of the acceleration of the body using Newton's second law.

For a set of $n$ forces numbered from $1$ to $n$, we can write the generalized net force equation:

$\vec{F} = \sum_{i=1}^{\infty} \vec{F}_i$

🙋 If you need to refresh vector algebra, our vector addition calculator is glad to help!

#### How to calculate the net force: formula with angles

If the vector sum is not clear enough, here you can find a step-by-step guide on how to calculate the magnitude of the net force and its direction using angles and a bit of trigonometric functions.

We will restrict the calculations to a bi-dimensional space. In this situation, a force $\vec{F}_i$ corresponds to a vector represented as such:

$\vec{F}_i = \left(F_{i,x},F_{i,y}\right)$

The force forms an angle $\theta_i$ with the reference axis (the x-axis) that allows us to write the components as:

\begin{align*} F_{i,x} &= \left\|F_i\right\|\cdot \cos{\left(\theta_i\right)}\\ F_{i,y} &= \left\|F_i\right\|\cdot \sin{\left(\theta_i\right)} \end{align*}

Apply this decomposition to any force acting on a body. In the case of three forces, we would have:

• $F_{1,x} = \left|F_1\right|\cdot \cos{\left(\theta_1\right)}$ and $F_{1,y} = \left|F_1\right|\cdot \sin{\left(\theta_1\right)}$;
• $F_{2,x} = \left|F_2\right|\cdot \cos{\left(\theta_2\right)}$ and $F_{2,y} = \left|F_2\right|\cdot \sin{\left(\theta_2\right)}$; and
• $F_{3,x} = \left|F_3\right|\cdot \cos{\left(\theta_3\right)}$ and $F_{3,y} = \left|F_3\right|\cdot \sin{\left(\theta_3\right)}$.

We can proceed now to calculate the magnitude of the net force's components on both axis:

\begin{align*} F_x &= F_{1,x}+F_{2,x}+F_{3,x}\\ F_y &= F_{1,y}+F_{2,y}+F_{3,y} \end{align*}

We apply the Pythagorean theorem to calculate the magnitude of the net force:

$\left\|\vec{F}\right\| = \sqrt{F_x^2 + F_y^2}$

We calculate the angle of the net force with the formula:

$\theta = \tan^{-1}\left(\frac{F_y}{F_x}\right)$

You can quickly expand this handy framework to an arbitrary number of forces. In the case of spaces with higher dimensionality, you will need an additional angle to find the value of the other components.

## The net force formula up in the sky: Lagrange points and halo orbits

Let's see how to find the net force in real-life situations. We will discuss a three-body problem (a pet peeve of all physicists) under the sole influence of the gravitational force, where two of the bodies are much more massive than the third.

The perfect example of this situation is the Earth orbiting the Sun and a tiny space probe somewhere around them. First, let's define the forces that come into play affecting the probe:

• The gravitational attraction of the Sun;
• The gravitational attraction of Earth; and
• The centrifugal force due to the proper motion of the probe.

All three forces act on the probe at every moment, dictating its trajectory. However, there are particular points where the resulting force is zero.

🙋 These points are unstable fixed points of the problem: if we place a probe exactly at one of them, it wouldn't move a single inch. But a minimum perturbation would throw our object on "unpredictable" (i.e., chaotic) trajectories.

We can identify five such points, called Lagrange points. Three of them lie on the line connecting the two massive bodies:

• $L_3$ lies on the other side of the more extensive body.
• $L_2$ lies behind the lesser body.
• $L_1$ lies between the bodies: there, the centripetal force sums with the attraction from the lesser body to balance the higher gravitational force from the bigger body.

There are two additional Lagrange points, $L_4$ and $L_5$. They surround the lesser body, anticipating and following it in its own orbit. There, centripetal force, attraction from the Sun, and attraction from the Earth form $60\degree$ angles between each other and are all equal in magnitude.

In $L_4$ and $L_5$ on to Earth's orbit, we can find many asteroids "attracted" by the unique properties of these points.

🔎 $L_2$ is a valuable point for astronomical satellites since it enjoys a perennial, albeit partial, Earth shadow: it comes to no surprise that the James Webb Space Telescope, which much hates sunlight, orbits $L_2$ on a graceful halo orbit.

Since the resulting force at a Lagrange point is zero, a body that lies there experiences zero net acceleration!

Davide Borchia
Force 1 (F₁)
N
Angle 1 (θ₁)
deg
Force 2 (F₂)
N
Angle 2 (θ₂)
deg You can add up to 10 different forces. Each angle is measured with respect to the positive x-axis.
Resultant force
Horizontal component (Fx)
0.97
N
Vertical component (Fy)
0.26
N
Magnitude of resultant force (F)
1
N
Direction of resultant force (θ)
15
deg
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