Lesson Video: Resultant of Two Forces | Nagwa Lesson Video: Resultant of Two Forces | Nagwa

# Lesson Video: Resultant of Two Forces Mathematics

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

15:12

### Video Transcript

In this video, we will learn how to find the resultant of two forces acting on one point and how to find the direction of the resultant. We will begin by defining a force and exploring its properties.

Force is defined as the effect of one natural body upon another. Each force is described in terms of its magnitude, direction, point of action, and line of action. We represent a force using the notation vector π as shown.

Letβs now consider the properties of the magnitude, direction, point of action, and line of action of a force. The magnitude of a force is its size, which is measured in newtons. By using a directed line segment to represent the force π and drawing the line to a suitable scale, we can use the length of the line to denote the magnitude. This can be written as shown.

The direction of a force is the direction in which it acts. By once again using a directed line segment to represent the force π, we can use the direction of the arrow to show the direction of the force.

The point of action of a force is the point at which it is applied.

And finally, the line of action of a force is a geometric way to represent how the force is applied. It is drawn as a line through the point of action in the same direction as π. We can demonstrate this as follows.

The diagram shows the force π represented by the directed line segment π΄π΅. The magnitude of the force is determined by the magnitude of the line segment π΄π΅. The direction of the arrow corresponds to the direction of π. The point of action is π΄. And the line of action is indicated by extending π΄π΅ in the same direction, as shown by the dotted line.

Letβs now consider what happens when two forces act on a body together with their combined effect, which is known as the resultant. When two forces π sub one and π sub two 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 vector π is given by π is equal to π sub one plus π sub two. The vector quantity π, which is equal to π sub one plus π sub two, can be represented in two ways as shown. As π sub one, π sub two, and π 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.

If we let πΌ be the angle between forces π sub one and π sub two, π sub one be the angle between π and π sub one, and π sub two be the angle between π and π sub two as shown, then using our knowledge of corresponding and alternate angles, the law of sines gives us πΉ sub one over sin of π sub two is equal to πΉ sub two over sin of π sub one, which is equal to π over sin of 180 degrees minus πΌ, where πΉ sub one, πΉ sub two, and π are the magnitudes of vectors π sub one, π sub two, and π, respectively. We recall that sin of 180 degrees minus π₯ is equal to sin π₯. And as such, sin of 180 degrees minus πΌ is equal to sin πΌ. Our equation simplifies as shown.

Applying the law of cosines in our triangle, we have π squared is equal to πΉ sub one squared plus πΉ sub two squared minus two multiplied by πΉ sub one multiplied by πΉ sub two multiplied by cos of 180 degrees minus πΌ. cos of 180 degrees minus π₯ is equal to negative cos π₯. And as such, our equation simplifies as shown.

We now have two equations with many unknowns that we can simplify further. If we take the square root of both sides of the second equation and recalling that the magnitude of a vector is positive, we can obtain an explicit formula for π, the magnitude of vector π. It is also straightforward to derive an accompanying formula for the direction of vector π. We will state these results now.

Let vector π be the resultant of two forces vector π sub one and vector π sub two that act at a single point with an angle πΌ between them. Then, π is equal to the square root of πΉ sub one squared plus πΉ sub two squared plus two πΉ sub one πΉ sub two multiplied by cos πΌ. And tan π is equal to πΉ sub two sin πΌ divided by πΉ sub one plus πΉ sub two cos πΌ, where πΉ sub one, πΉ sub two, and π are the magnitudes of vectors π sub one, π sub two, and π, respectively. And π is the angle between vector π and vector π sub one.

We will now consider two examples, one where our two forces are perpendicular and one when they are not.

Two perpendicular forces of magnitudes 88 newtons and 44 newtons act at a point. Their resultant makes an angle π with the 88-newton force. Find the value of sin π.

In this question, it will be convenient to assume that one of the forces acts horizontally. We will let this be the 88-newton force and call it vector π sub one. We will therefore call the 44-newton force vector π sub two, and this will act vertically. Since the two perpendicular forces act at a point, we can represent this as shown. The resultant force, vector π, will act in the direction shown. And we are told that this makes an angle of π with the 88-newton force π sub one.

Since vectors have both direction and magnitude, we can create a triangle of forces. This is a right triangle, and we can use the properties of right triangles to find the value of sin π as required. As already mentioned, we know that the magnitude of vector π sub one is 88 newtons and the magnitude of π sub two is 44 newtons. Applying the Pythagorean theorem, we have π squared is equal to πΉ sub one squared plus πΉ sub two squared, where π, πΉ sub one, and πΉ sub two are the magnitudes of the corresponding vector forces. This means that π squared is equal to 88 squared plus 44 squared. The right-hand side of our equation simplifies to 9680. We can then square root both sides. And since π must be positive, we have π is equal to 44 root five. The magnitude of the resultant force is 44 root five newtons.

Next, we use our knowledge of right angle trigonometry, often referred to using the acronym SOH CAH TOA. The sine ratio tells us that sin π is equal to the opposite over the hypotenuse. This means that in this question, sin π is equal to the magnitude of π sub two over the magnitude of the resultant. Substituting in the values we know, this is equal to 44 over 44 root five. We can then divide the numerator and denominator by 44, giving us one over root five. Finally, we can rationalize the denominator by multiplying the numerator and denominator by root five, giving us sin π is equal to root five over five. This is the sine of the angle that the resultant makes with the 88-newton force.

Letβs now consider a second example where the angle between the two forces is not 90 degrees.

The angle between forces vector π sub one and vector π sub two is 112 degrees, and the measure of the angle between their resultant and vector π sub two is 56 degrees. If the magnitude of vector π sub one is 28 newtons, what is the magnitude of vector π sub two?

Letβs begin by sketching a diagram to model the situation. We are told that the angle between two forces π sub one and π sub two is 112 degrees. We are also told that the measure of the angle between the resultant force and π sub two is 56 degrees. We see that our two forces form a parallelogram, where the resultant is its diagonal. We can calculate the angle π between π sub one and the resultant by subtracting 56 degrees from 112 degrees. This is equal to 56 degrees. We can now add this angle and its alternate interior angle in our diagram as shown.

Next, we can apply the law of sines to either one of our triangles within the parallelogram. We have the magnitude of π sub one over sin of 56 degrees is equal to the magnitude of π sub two over the sin of 56 degrees. This means that the magnitude of π sub one is equal to the magnitude of π sub two. And since we are told in the question that the magnitude of π sub one is 28 newtons, then the magnitude of π sub two must also be 28 newtons.

We will now summarize the key points from this video. Force is defined as the effect of one body upon another. Each force is described in terms of its magnitude, direction, point of action, and line of action. We represent a force using the notation vector π. The resultant vector π of two forces vector π sub one and vector π sub two acting on a body at the same point is a single force that is given by vector π is equal to vector π sub one plus vector π sub two. The combined effect of vector π sub one and vector π sub two is the same as the effect of only vector π. Vectors π sub one, π sub two, and π 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 π sub one and π sub two and their resultant π gives πΉ sub one over sin of π sub two equals πΉ sub two over sin of π sub one, which is equal to π over sin πΌ, where πΉ sub one, πΉ sub two, and π are the magnitudes of vectors π sub one, π sub two, and π, respectively. πΌ is the angle between forces π sub one and π sub two. π sub one is the angle between the resultant π and π sub one. And π sub two is the angle between π and π sub two.

Applying the law of cosines in the same way under the same conditions gives us π squared is equal to πΉ sub one squared plus πΉ sub two squared plus two multiplied by πΉ sub one multiplied by πΉ sub two multiplied by cos πΌ. Square rooting both sides of this equation, we have π equals the square root of πΉ sub one squared plus πΉ sub two squared plus two multiplied by πΉ sub one multiplied by πΉ sub two multiplied by cos πΌ. And finally, we saw that if π is the angle between the resultant π and π sub one, then tan π is equal to πΉ sub two sin πΌ divided by πΉ sub one plus πΉ sub two cos πΌ.