Lesson Explainer: Resultant of Two Forces Mathematics

In this explainer, we will learn how to find the resultant of two forces acting on one point and how to find the direction of the resultant.

A force acting on a body is represented by vector โƒ‘๐น. When two forces act on a body, we call their resultant the force that describes their combined effect.

Definition: Resultant Force

When two forces, โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, act on a body at the same point, the combined effect of these two forces is the same as the effect of a single force, called the resultant force.

The resultant force, โƒ‘๐‘…, is given by โƒ‘๐‘…=โƒ‘๐น+โƒ‘๐น.๏Šง๏Šจ

The vector equality โƒ‘๐‘…=โƒ‘๐น+โƒ‘๐น๏Šง๏Šจ can be represented in two ways, as illustrated in the following diagram.

As โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and โƒ‘๐‘… are three sides of a triangle, we can use either the law of sines or the law of cosines in the triangle to find the resultant of the two forces, the angles between the resultant and the forces, or any other unknown.

Let ๐›ผ be the angle between forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, ๐œƒ๏Šง the angle between โƒ‘๐‘… and โƒ‘๐น๏Šง, and ๐œƒ๏Šจ the angle between โƒ‘๐‘… and โƒ‘๐น๏Šจ, as shown in the diagram below.

The law of sines in this triangle gives us ๐น๐œƒ=๐น๐œƒ=๐‘…(180โˆ’๐›ผ),๏Šง๏Šจ๏Šจ๏Šงโˆ˜sinsinsin where ๐น๏Šง, ๐น๏Šจ, and ๐‘… are the magnitudes of โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and โƒ‘๐‘… respectively.

As sinsin(180โˆ’๐‘ฅ)=๐‘ฅโˆ˜ for all ๐‘ฅ, we find the relationship given in the following box.

Property: Law of Sines in a Triangle Formed by Two Forces and Their Resultant

We have ๐น๐œƒ=๐น๐œƒ=๐‘…๐›ผ,๏Šง๏Šจ๏Šจ๏Šงsinsinsin where ๐น๏Šง, ๐น๏Šจ, and ๐‘… are the magnitudes of โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and โƒ‘๐‘…, respectively, ๐›ผ is the angle between forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, ๐œƒ๏Šง is the angle between โƒ‘๐‘… and โƒ‘๐น๏Šง, and ๐œƒ๏Šจ is the angle between โƒ‘๐‘… and โƒ‘๐น๏Šจ.

Applying the law of cosines in our triangle now, we find that ๐‘…=๐น+๐นโˆ’2๐น๐น(180โˆ’๐›ผ).๏Šจ๏Šง๏Šจ๏Šจ๏Šจ๏Šง๏Šจโˆ˜cos

As coscos(180โˆ’๐‘ฅ)=โˆ’๐‘ฅโˆ˜ for all ๐‘ฅ, we find the relationship given in the following box.

Property: Law of Cosines in a Triangle Formed by Two Forces and Their Resultant

We have ๐‘…=๐น+๐น+2๐น๐น๐›ผ,๏Šจ๏Šง๏Šจ๏Šจ๏Šจ๏Šง๏Šจcos where ๐น๏Šง, ๐น๏Šจ, and ๐‘… are the magnitudes of โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and โƒ‘๐‘…, respectively, ๐›ผ is the angle between forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, ๐œƒ๏Šง is the angle between โƒ‘๐‘… and โƒ‘๐น๏Šง, and ๐œƒ๏Šจ is the angle between โƒ‘๐‘… and โƒ‘๐น๏Šจ

Let us start with an example in which the magnitude of the resultant of two forces acting at a point is determined.

Example 1: Finding the Magnitude of the Resultant of Two Forces

Two forces of magnitudes 35 N and 91 N are acting at a particle. Given that the resultant is perpendicular to the first force, find the magnitude of the resultant.

Answer

It will be convenient to assume that the first force acts horizontally. Let us call this force โƒ‘๐น๏Šง and the other force โƒ‘๐น๏Šจ. The resultant of these forces, โƒ‘๐น+โƒ‘๐น๏Šง๏Šจ, acts vertically as it is perpendicular to โƒ‘๐น๏Šง, as shown in the following figure.

The force โƒ‘๐น๏Šจ can be represented by an arrow with its tail at the head of โƒ‘๐น๏Šง and its head at the head of โƒ‘๐น+โƒ‘๐น๏Šง๏Šจ, as shown in the following figure.

The resultant force โƒ‘๐‘… is given by โƒ‘๐‘…=โƒ‘๐น+โƒ‘๐น.๏Šง๏Šจ

As โƒ‘๐น๏Šง and โƒ‘๐‘… are perpendicular, we see that the two forces and their resultant form a right triangle. Therefore, applying the Pythagorean theorem gives โ€–โ€–โƒ‘๐นโ€–โ€–+โ€–โ€–โƒ‘๐‘…โ€–โ€–=โ€–โ€–โƒ‘๐นโ€–โ€–.๏Šง๏Šจ๏Šจ๏Šจ๏Šจ

It is worth noting that the Pythagorean theorem is just a special case of the law of cosines.

Substituting in the values of โ€–โ€–โƒ‘๐นโ€–โ€–๏Šง and โ€–โ€–โƒ‘๐นโ€–โ€–๏Šจ, we find that 35+โ€–โ€–โƒ‘๐‘…โ€–โ€–=91โ€–โ€–โƒ‘๐‘…โ€–โ€–=91โˆ’35=7056โ€–โ€–โƒ‘๐‘…โ€–โ€–=โˆš7056=84.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏ŠจN

Note that as the magnitude of a vector is always positive, โˆ’84 N is not a valid solution.

The magnitude of the resultant of the forces is 84 N.

Let us now look at an example in which the direction of the line of action of the resultant of two forces acting at a point is determined.

Example 2: Finding the Direction of the Resultant of Two Forces Acting at the Same Point

Two perpendicular forces of magnitudes 88 N and 44 N act at a point. Their resultant makes an angle ๐œƒ with the 88 N force. Find the value of sin๐œƒ.

Answer

It will be convenient to assume that one of the forces acts horizontally. Let us call this force โƒ‘๐น๏Šง and the other force โƒ‘๐น๏Šจ, as shown in the following figure.

By choosing to make โƒ‘๐น๏Šง correspond to the line adjacent to ๐œƒ, we have chosen this force to be the 88-newton force. The magnitude of โƒ‘๐น๏Šจ is 44 newtons; therefore, the magnitude of โƒ‘๐น๏Šจ is half that of โƒ‘๐น๏Šง. The magnitude of the resultant of the forces, ๐‘…, can be expressed as ๐‘…=๏„๐น+๐น=๏„Ÿ๐น+๏€ฝ๐น2๏‰.๏Šง๏Šจ๏Šจ๏Šจ๏Šง๏Šจ๏Šง๏Šจ

We can see from this that ๐น=๐น+๏€ฝ๐น2๏‰=๐น+๐น4=๏€ผ54๏ˆ๐น.๏Šจ๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ๏Šง๏Šจ

Taking square roots, we have that ๐น=โˆš5๐น2.๏Šง

Applying the law of sines in the triangle gives sinsin๐œƒ๐น=90๐น.๏Šจโˆ˜

As sin90=1โˆ˜, we have sinsin๐œƒ=๐น๐น๐œƒ==1โˆš5=โˆš55.๏Šจ๏Œฅ๏Šจโˆš๏Šซ๏Œฅ๏Šจ๏Ž ๏Ž 

We have, therefore, that sin๐œƒ=โˆš55.

Let us now look at an example in which the magnitude and direction of the line of action of the resultant of two perpendicular forces are known and the magnitudes of the forces must be determined.

Example 3: Finding Two Forces given the Magnitude and Direction of Their Resultant

Two perpendicular forces, โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, act at a point. Their resultant, โƒ‘๐‘…, has magnitude 188 N and makes an angle of 60โˆ˜ with โƒ‘๐น๏Šง. Find the magnitudes of โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ.

Answer

The perpendicular forces, โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, and their resultant are shown in the following figure.

We see that โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ are perpendicular and the resultant โƒ‘๐น+โƒ‘๐น๏Šง๏Šจ makes an angle of 60โˆ˜ with โƒ‘๐น๏Šง. As we have a right triangle, we have coscosN60=โ€–โ€–โƒ‘๐นโ€–โ€–188โ€–โ€–โƒ‘๐นโ€–โ€–=18860=12ร—188=94โˆ˜๏Šง๏Šงโˆ˜ and sinsinN60=โ€–โ€–โƒ‘๐นโ€–โ€–188โ€–โ€–โƒ‘๐นโ€–โ€–=18860=โˆš32ร—188=94โˆš3.โˆ˜๏Šจ๏Šจโˆ˜

โƒ‘๐น๏Šง has a magnitude of 94 N, and โƒ‘๐น๏Šจ has a magnitude of 94โˆš3 N.

Let us now look at an example involving two nonperpendicular forces.

Example 4: Finding a Missing Force given Information About the Resultant Force

The angle between forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ is 112โˆ˜, and the measure of the angle between their resultant and โƒ‘๐น๏Šจ is 56โˆ˜. If the magnitude of โƒ‘๐น๏Šง is 28 N, what is the magnitude of โƒ‘๐น๏Šจ?

Answer

The following figure shows the forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ and their resultant โƒ‘๐น+โƒ‘๐น๏Šง๏Šจ. The forces act at a point ๐‘ƒ.

The resultant forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ form a parallelogram whose diagonal through ๐‘ƒ is the resultant.

The angle, ๐œƒ, between โƒ‘๐น๏Šง and the resultant of โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ is given by ๐œƒ=112โˆ’56=56.โˆ˜

We can now add this angle and its alternate interior angle in our diagram as shown.

Applying the law of sines in the triangle formed by โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and โƒ‘๐น+โƒ‘๐น๏Šง๏Šจ, we find that โ€–โ€–โƒ‘๐นโ€–โ€–56=โ€–โ€–โƒ‘๐นโ€–โ€–56,๏Šงโˆ˜๏Šจโˆ˜sinsin that is, โ€–โ€–โƒ‘๐นโ€–โ€–=โ€–โ€–โƒ‘๐นโ€–โ€–.๏Šง๏Šจ

The magnitude of โƒ‘๐น๏Šง is given as 28 N, so the magnitude of โƒ‘๐น๏Šจ is also 28 N.

Let us look at our last example where the direction of one of the forces is reversed.

Example 5: Finding the Magnitude of Two Identical Forces given Their Resultant at Two Cases

Two forces, both of magnitude ๐น N, act at the same point. The magnitude of their resultant is 90 N. When the direction of one of the forces is reversed, the magnitude of their resultant is 90 N. Determine the value of ๐น.

Answer

Let us represent the first situation.

When we add two forces, โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, the resultant is the diagonal of the parallelogram formed by โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, with its tail being the point of application of โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ. If the two forces have the same magnitude, then the parallelogram is a rhombus, and the two forces and their resultant form an isosceles triangle, as shown in the following diagram.

Applying the law of cosines, we find that ๐‘…=๐น+๐น+2๐น๐น๐›ผ,๏Šจ๏Šง๏Šจ๏Šจ๏Šจ๏Šง๏Šจcos with ๐น=โ€–โ€–โƒ‘๐นโ€–โ€–๏Šง๏Šง, ๐น=โ€–โ€–โƒ‘๐นโ€–โ€–๏Šจ๏Šจ, and ๐‘…=โ€–โ€–โƒ‘๐‘…โ€–โ€–.

Since ๐น=๐น=๐น๏Šง๏Šจ, we have ๐‘…=2๐น+2๐น๐›ผ.๏Šจ๏Šจ๏Šจcos

If we now reverse the direction of one of the forces (for symmetry reasons, it does not matter which force has its direction reversed; we will get the same result), the resultant will still be the diagonal of a rhombus congruent to the previous one, but it will be the other diagonal, and the angle between forces โˆ’โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ will be 180โˆ’๐›ผโˆ˜.

The magnitude of โˆ’โƒ‘๐น๏Šง is the same as the magnitude of โƒ‘๐น๏Šง, ๐น.

Applying the law of cosines in the triangle formed by โˆ’โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and their resultant gives us ๐‘…โ€ฒ=2๐น+2๐น(180โˆ’๐›ผ),๏Šจ๏Šจ๏Šจโˆ˜cos that is, ๐‘…โ€ฒ=2๐นโˆ’2๐น๐›ผ.๏Šจ๏Šจ๏Šจcos

We are told that the magnitude of the resultant is the same in both cases, 90 N. Hence, we have ๐‘…=๐‘…โ€ฒ=90,N which means that 2๐น+2๐น๐›ผ=2๐นโˆ’2๐น๐›ผ=90.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจcoscos

This is true only if cos๐›ผ=0, that is, if ๐›ผ=90โˆ˜. Forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ are, thus, perpendicular.

Hence, we have 2๐น=90๐น=๏„ž902๐น=90๏„ž12๐น=45โˆš2.๏Šจ๏Šจ๏ŠจN

It is worth noting that, in the previous example, we could have concluded that the two forces are perpendicular with simple geometric considerations: the diagonals in a rhombus have the same length only if the rhombus is a square.

Let us now summarize what has been learned in these examples.

Key Points

  • The resultant, โƒ‘๐‘…, of two forces, โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, acting on a body at the same point is a single force thatis given by โƒ‘๐‘…=โƒ‘๐น+โƒ‘๐น.๏Šง๏Šจ
  • The combined effect of โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ is the same as the effect of only โƒ‘๐‘….
  • โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and โƒ‘๐‘… are three sides of a triangle or two adjacent sides and a diagonal of a parallelogram.
  • Applying the law of sines in the triangle formed by two forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ and their resultant, โƒ‘๐‘…, gives ๐น๐œƒ=๐น๐œƒ=๐‘…๐›ผ,๏Šง๏Šจ๏Šจ๏Šงsinsinsin where ๐น๏Šง, ๐น๏Šจ, and ๐‘… are the magnitudes of โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and โƒ‘๐‘…, respectively, ๐›ผ is the angle between forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, ๐œƒ๏Šง is the angle between โƒ‘๐‘… and โƒ‘๐น๏Šง, and ๐œƒ๏Šจ is the angle between โƒ‘๐‘… and โƒ‘๐น๏Šจ.
  • Applying the law of cosines in the triangle formed by two forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ and their resultant, โƒ‘๐‘…, gives ๐‘…=๐น+๐น+2๐น๐น๐›ผ,๏Šจ๏Šง๏Šจ๏Šจ๏Šจ๏Šง๏Šจcos where ๐น๏Šง, ๐น๏Šจ, and ๐‘… are the magnitudes of โƒ‘๐น๏Šง, โƒ‘๐น๏Šจ, and โƒ‘๐‘…, respectively, ๐›ผ is the angle between forces โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, ๐œƒ๏Šง is the angle between โƒ‘๐‘… and โƒ‘๐น๏Šง, and ๐œƒ๏Šจ is the angle between โƒ‘๐‘… and โƒ‘๐น๏Šจ.

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