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What is the Resultant Force and How to Find it (with Examples)

In this article, you will learn what the resultant force (also known as net force) is, and how to find it when an object is subject to parallel forces as well as non-parallel forces with the help of examples.

What is the resultant force?

When an object is subject to several forces, the resultant force is the force that alone produces the same acceleration as all those forces.

For example, if 4 forces act on a block and cause it to accelerate 1 m/s2 south, then the resultant force is the force that, if applied alone to the block, will also make it accelerate 1 m/s2 south.

The reason why the resultant force is useful is that it allows us to think about several forces as though they were a single force. This means that to determine the effect that several forces have on an object, we only need to determine the effect that a single force has.

How to find the resultant force?

If we know the mass m of an object and the acceleration a produced by the forces that act on it, we can find the resultant force using Newton's Second Law. Indeed, according to Newton's Second Law, the force F that alone produces the acceleration a on an object of mass m is:

F = ma

This force F is our resultant force. So, we can write:

R = ma

Which indicates that the resultant force R has the same direction as a, and has magnitude equal to the product ma.

For example, if a box of 1.5 kg is subject to 5 forces which make it accelerate 2.0 m/s2 north-west, then the resultant force is directed north-west and has the magnitude equal to 1.5 kg × 2.0 m/s2 = 3.0 N.

Often, however, we know the forces that act on an object and we need to find the resultant force.

Experiments show that when an object is subject to several forces, F1, F2, ..., the resultant force R is the vector sum of those forces:

R = F1 + F2 + ...

Notice that this is not a mere sum of the magnitudes of the forces, but the sum of the forces taken as vectors, which is more involved because vectors have both a magnitude and a direction that we need to consider when doing the sum.

According to the above equation, if an object is subject to no forces, then the resultant force is zero, and if an object is subject to only one force, then the resultant force is equal to that force. These two cases are pretty simple, but what about an object subject to two or more forces? How do we perform the vector sum then?

To explain this clearly, we will now go through all the cases that can happen, from simple ones in which all the forces are parallel, to more complex ones in which the forces are not parallel, and show how to find the resultant force in each of them with the help of examples.

• Two forces acting in the same direction

Let's start with the simple case in which an object is subject to two forces that act in the same direction:

Two forces of 3N and 4N pulling to the rightN3N4

The resultant force is in the same direction as the two forces, and has the magnitude equal to the sum of the two magnitudes:

Resultant force (7N to the right)N7

• Two forces acting in opposite directions

Let's consider the case in which an object is subject to two forces that act in opposite directions.

If the two forces are equal in magnitude:

Two opposite forcesN3N3

The resultant force will be zero because two opposite forces cancel each other out.

On the other hand, if the two forces are not equal in magnitude:

Two forces: 3N to the left, 5N to the rightN3N5

The resultant force will be in the same direction as the force with the larger magnitude (the 5 N force in the example), and have the magnitude equal to the difference between the magnitudes of the two forces (in the example that would be 2 N):

Resultant force (2N rightwards)N2

• More than 2 forces parallel to one another

Let's now consider the case in which an object is subject to more than two parallel forces:

Four parallel forces: 3N and 4N to the left, 5N and 6N to the rightN3N4N5N6

To find the resultant force in this case, we first sum all the forces that go in one direction, and then all the forces that go in the other direction:

Two forces: 7N to the left and 11N to the rightN7N11

At this point, we have two forces that are in opposite directions, which is a case that we already know how to solve: the resultant force has the same direction as the force with the larger magnitude (the 11 N force), and its magnitude is equal to the difference between the two magnitudes (4 N):

Resultant Force (4N to the right)N4

• Two forces that are not parallel

In the previous cases, we have forces that are all parallel to one another. It's time to consider the case in which an object is subject to two forces that are not parallel.

For example, let's assume that we have a block subject to two forces, F1 and F2.

F1 has magnitude 50 N and is applied at a 45° angle, whereas F2 has magnitude 60 N and is applied horizontally, as shown in the free-body diagram below:

Free-body diagram of the block45°F2F1

How do we find the resultant force R in this case?

The first step is to draw coordinate axes on our free-body diagram.

Since one of the two forces is horizontal, for convenience, we choose the x-axis horizontal, and the y-axis vertical, and we place the origin at the center of our block:

The free-body diagram of the block with coordinate axes.45°F2xyF1

The next step is to determine the x and y components of all the forces that act on the block:

All the forces acting on the block have been decomposed into x and y components45°F2xyF1F1yF1x
F1x = F1 cos 45°
F2x = F2
F1y = F1 sin 45°
F2y = 0

Now comes the important part:

If we sum all the x components, we will get the x component of the resultant force:

F1x + F2x = Rx
Rx = F1x + F2x
Rx = F1 cos 45° + F2
Rx = (50 N) (cos 45°) + 60 N
Rx = 95 N

Similarly, if we sum all the y components, we will get the y component of the resultant force:

F1y + F2y = Ry
Ry = F1y + F2y
Ry = F1 sin 45° + 0
Ry = F1 sin 45°
Ry = (50 N) (sin 45°)
Ry = 35 N

At this point, we know the x and y components of R, which we can use to find the magnitude and direction of R:

Rx = 95 N
Ry = 35 N
The resultant force acting on the block, and its x and y componentsRyRxxyθR

The magnitude of R can be calculated by applying Pythagoras' Theorem:

R = Rx2 + Ry2
R = 952 + 352 N = 100 N

The angle θ that R makes with Rx can be calculated using trigonometry:

θ = tan-1 Ry
Rx
θ = tan-1 35 N = 20°
95 N

Thus, the resultant force R has magnitude 100 N and direction angle of 20°.

• More than 2 non-parallel forces

Finally, let's examine the case in which an object is subject to more than two non-parallel forces.

For example, suppose we have an object that is subject to three forces, F1, F2, and F3.

The magnitude of each force is shown below:

F1 = 10 N
F2 = 20 N
F3 = 40 N

The free-body diagram of the object looks like this:

Free-body diagram of the object60°F13FF2

We can find the resultant force R using the same process that we used in the previous case of two non-parallel forces.

First, we draw the coordinate axes on our free-body diagram:

The free-body diagram of the object with coordinate axes.F13FxyF260°

Then, we determine the x and y components of the individual forces:

All the forces acting on the object have been decomposed into x and y componentsF13FF3xF3yxyF260°
F1x = F1
F2x = 0
F3x = −F3 cos 60°
F1y = 0
F2y = F2
F3y = −F3 sin 60°

Again, the x component of the resultant force R is the sum of all x components:

Rx = F1x + F2x + F3x
Rx = F1 + 0 + (−F3 cos 60°)
Rx = F1F3 cos 60°
Rx = 10 N − (40 N) (cos 60°)
Rx = −10 N

Similarly, the y component of R is the sum of all y components:

Ry = F1y + F2y + F3y
Ry = 0 + F2 + (−F3 sin 60°)
Ry = F2F3 sin 60°
Ry = 20 N − (40 N) (sin 60°)
Ry = −15 N

Finally, let's calculate the magnitude and direction of R using its two components Rx and Ry:

Rx = −10 N
Ry = −15 N
The resultant force acting on the object, and its componentsθRRxRyxy
R = Rx2 + Ry2
R = (−10)2 + (−15)2 N = 18 N
θ = tan-1 Ry
Rx
θ = tan-1 −15 N = 56°
−10 N

To express the direction of R, we need to calculate the direction angle (i.e. the counterclockwise angle that R makes with the positive x-axis), which in our case is 180° + θ, i.e. 236°.

The process that we used in this case and in the previous one to find the resultant force when the forces are not parallel can also be used when all the forces are parallel. In fact, it can be used in any case – it's a generic process. However, in the cases of parallel forces, we recommend using the much simpler processes that we described before.

Here's a quick summary of the generic process:

  1. Draw a free-body diagram of the object
  2. Draw coordinate axes on the free-body diagram
  3. Decompose the forces acting on the object into x and y components
  4. Calculate the x and y components of the resultant force by adding the x and y components of all forces
  5. Finally, find the magnitude and direction of the resultant force by using its x and y components

A note on drawing coordinate axes on a free-body diagram: we recommend you to draw them so that one of the axes is in the same direction as the acceleration of the object. For example, if you have an object accelerating up a ramp, you should draw tilted coordinate axes with the x-axis uphill. Sometimes, however, your object may be at rest or you may not know the direction of the acceleration. In that case, place the coordinate axes so that as many forces as possible are parallel to them since this will simplify the expressions for their components.

To test your understanding, do the exercises below.

Exercises

#1

John and Rob are engaged in a tug of war. John is pulling with a force of 230 N, and Rob is pulling with a force of 215 N. Determine the magnitude and direction of the resultant force.

Solution

15 N towards John

#2

A car of 1400 kg is subject to multiple forces which produce an acceleration of 3.5 m/s2 directed north. Find the net force.

Solution

Fnet = ma

The direction of Fnet is the same as that of a (north), and the magnitude is:

Fnet = ma
Fnet = (1400 kg) (3.5 m/s2)
Fnet = 4900 N

#3

A block is pulled by two forces of 15 N and 25 N to the left, and by three forces of 10 N, 20 N, 30 N to the right. Find the magnitude and direction of the resultant force.

Solution

If you sum the forces pulling to the left, you get 40 N to the left, and if you sum the forces pulling to the right, you get 60 N to the right.

Thus, the resultant force is 20 N to the right.

#4

An apple is subject to two vertical forces: one of 40 N pulling upward, and the other of 10 N pulling downward. What is the net force acting on the apple?

Solution

30 N upward

#5

A box of 1.0 kg is in free fall (i.e. moving subject only to the force of gravity). Calculate the net force.

Solution

The net force is equal to the force of gravity because the box is subject only to that one force. Therefore, the direction is downward, and the magnitude is:

Fnet = mg
Fnet = (1.0 kg) (9.8 m/s2)
Fnet = 9.8 N

#6

A tugboat is horizontally pulled by two forces of 1450 N, each making an angle of 20° with the long axis of the tugboat, as shown in the figure (the view is from the above):

Tugboat's free-body diagram20°20°F1F2

Assuming there is no friction, what is the magnitude and direction of the resultant force acting on the tugboat?

Solution

The resultant force acting on the tugboat (directed to the right along the long axis)R
R = 2725 N

#7

A ball is subject to two forces F1 and F2. The magnitudes of the two forces are 45.0 N and 70.0 N respectively. In the figure below you can see the free-body diagram of the ball:

FBD of the ball50°F1F2

Find the magnitude and direction of the resultant force acting on the ball.

Solution

The resultant force acting on the ball, making a 40° angle with the horizontal (counterclockwise)R40°
R = 57.3 N

#8

An empty box is pulled by two men with horizontal forces, as shown below (the view is from the above):

The free-body diagram of the empty box57°26°F1F2

Assuming that F1 is 345 N and F2 is 458 N, and there is no other horizontal force acting on the box, find the magnitude and direction of the resultant force.

Solution

The resultant force acting on the empty box, making a 23° angle with the horizontal (clockwise)23°R
R = 607 N

Problems with solutions

To further test your understanding of resultant forces, see our force problems, which include problems where you need to find the resultant force acting on objects that move horizontally, move up an incline, and hang from pulleys. For each problem, we provide a step-by-step guide on how to solve it.

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