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.

14:54

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 recalling what we mean by a resultant force.

The resultant force is the single force that has the same effect as two or more forces acting together. In this video, we will only be dealing with two forces. Let’s begin by considering two forces 𝐹 sub one and 𝐹 sub two acting in the same direction. The resultant force 𝑅 will be equal to the sum of the forces. 𝑅 is equal to 𝐹 sub one plus 𝐹 sub two.

Let’s now consider what happens when the two forces 𝐹 sub one and 𝐹 sub two act in opposite directions. When two forces act in opposite directions, they produce a resultant force 𝑅 that is smaller than each individual force. To calculate this resultant, we subtract the magnitude of the smaller force from the magnitude of the larger force. In our diagram, if 𝐹 sub one is greater than 𝐹 sub two, then the resultant force 𝑅 will be equal to 𝐹 sub one minus 𝐹 sub two. The direction of the resultant force is in the same direction as the larger force. We can therefore conclude that if we have two forces acting parallel to one another, we can either add or subtract the forces to calculate the resultant.

Let’s now consider what happens when our two forces are acting in a perpendicular direction. When two forces are acting perpendicular to one another, we will use our knowledge of right triangles to help us calculate the resultant force. The resultant of the two forces 𝐹 sub one and 𝐹 sub two will act in the direction as shown. By creating a right triangle, we can calculate the magnitude of the resultant by using the Pythagorean theorem. 𝑅 squared is equal to 𝐹 sub one squared plus 𝐹 sub two squared. By square rooting both sides of this equation and using the fact that the magnitude of the resultant must be positive, 𝑅 is equal to the square root of 𝐹 sub one squared plus 𝐹 sub two squared.

We can also use our diagram to calculate the direction of the resultant. This is given as the angle between one of the forces and the resultant. Using our knowledge of right angle trigonometry, we know that the tan of any angle πœƒ is equal to the opposite over the adjacent. This means that the tan of our angle πœƒ is equal to 𝐹 sub two over 𝐹 sub one. The direction of the resultant force angle πœƒ is equal to the inverse tan of 𝐹 sub two over 𝐹 sub one.

Let’s now consider what happens if our two forces are neither parallel nor perpendicular. Let’s consider the two forces 𝐹 sub one and 𝐹 sub two acting at a point as shown. As any two equal vector forces will have the same magnitude and direction, we can redraw 𝐹 sub one and 𝐹 sub two nose to tail as shown. The resultant force 𝑅 will be the diagonal of the parallelogram. We can then use both the sine rule and cosine rule to help us calculate the magnitude of the resultant 𝑅 together with its direction the angle πœƒ.

We will now look at some questions involving perpendicular forces.

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

We are told that we have two forces of magnitude 35 newtons and 91 newtons acting at a particle. Whilst we do not know the angle between these two forces, we know that the resultant force is perpendicular to the 35-newton force. This can be demonstrated on the diagram as shown, where the magnitude of the resultant is 𝑅 newtons.

The Pythagorean theorem states that π‘Ž squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the length of the longest side of a right triangle. π‘Ž and 𝑏 are the lengths of the smaller sides. Substituting the values that we have for the magnitudes of the forces, we have 35 squared plus 𝑅 squared is equal to 91 squared. We can subtract 35 squared from both sides of this equation. 𝑅 squared is equal to 91 squared minus 35 squared. The right-hand side is equal to 7056. We can then square root both sides of this equation. As 𝑅 must be positive, 𝑅 is equal to 84 newtons. The magnitude of the resultant force is 84 newtons.

In our next question, we need to find the direction of the resultant of two perpendicular forces.

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

We are told that two perpendicular forces act at a point as shown. The resultant force 𝑅 makes an angle πœƒ with the 88-newton force. We are asked to calculate the value of sin πœƒ. And from our knowledge of right angle trigonometry, we know that sin of πœƒ is equal to the opposite over the hypotenuse. By creating a right triangle, the opposite is equal to 44, the adjacent 88, and the hypotenuse is the resultant force.

Using the Pythagorean theorem, we can calculate the magnitude of the resultant 𝑅. 𝑅 squared is equal to 44 squared plus 88 squared. 44 squared plus 88 squared is 9680. We can then square root both sides of our equation. And as 𝑅 must be positive, 𝑅 is equal to the square root of 9680. This simplifies to 44 root five. The magnitude of the resultant is equal to 44 root five newtons.

The sin of angle πœƒ is therefore equal to 44 over 44 root five. We can divide the numerator and denominator by 44. Multiplying the numerator and denominator of this new fraction by root five to rationalize the denominator gives us a value of sin πœƒ equal to root five over five.

In our next question, we need to calculate the magnitudes of two perpendicular forces.

Two perpendicular forces 𝐹 sub one and 𝐹 sub two act at a point. Their resultant 𝑅 has magnitude 188 newtons and makes an angle of 60 degrees with 𝐹 sub one. Find the magnitudes of 𝐹 sub one and 𝐹 sub two.

We can begin by drawing a diagram of the two perpendicular forces 𝐹 sub one and 𝐹 sub two. The resultant of these two forces has magnitude 188 newtons and makes an angle of 60 degrees with 𝐹 sub one. By creating a right triangle, we can use the trig ratios and the Pythagorean theorem to calculate 𝐹 sub one and 𝐹 sub two. The Pythagorean theorem states that π‘Ž squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the length of the longest side of a right triangle, known as the hypotenuse.

Substituting in our values from the diagram, we have 𝐹 sub one squared plus 𝐹 sub two squared is equal to 188 squared. The right-hand side simplifies to 35344. We will call this equation one. We also know that in a right triangle, the tan of angle πœƒ is equal to the opposite over the adjacent. Substituting in our values, we have the tan of 60 degrees is equal to 𝐹 sub two divided by 𝐹 sub one. The tan of 60 degrees is equal to root three.

We can then multiply both sides of this equation by 𝐹 sub one such that 𝐹 sub two is equal to root three 𝐹 sub one. Squaring both sides of this equation gives us three 𝐹 sub one squared is equal to 𝐹 sub two squared. We will call this equation two.

If we substitute equation two into equation one, we can replace 𝐹 sub two squared with three 𝐹 sub one squared. The left-hand side simplifies to four 𝐹 sub one squared, and this is equal to 35344. We can divide both sides of this equation by four so that 𝐹 sub one squared is equal to 8836. Square rooting both sides and as 𝐹 one must be positive, 𝐹 one is equal to 94. The force 𝐹 one is therefore equal to 94 newtons. We know that 𝐹 sub two is equal to root three multiplied by 𝐹 sub one. This means that 𝐹 sub two is equal to 94 root three newtons.

In our final question, we will need to calculate the resultant of two nonperpendicular forces.

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

Let’s begin by sketching the two forces with an angle of 112 degrees between them. We are also told in the question that the angle between the resultant force and 𝐹 sub two is 56 degrees. 112 minus 56 is equal to 56. Therefore, the angle between the resultant and 𝐹 sub one is also 56 degrees.

Using our knowledge of vector forces, we can create a parallelogram as shown. This can be split into two congruent triangles. Since 𝐹 sub one is equal to 28 newtons, we can use the sine rule to calculate 𝐹 sub two. The sine rule states that π‘Ž over sin 𝐴 is equal to 𝑏 over sin 𝐡, where the angles capital 𝐴 and capital 𝐡 are opposite the side lengths π‘Ž and 𝑏.

Substituting in our values, we have 28 over the sin of 56 degrees is equal to 𝐹 sub two over the sin of 56 degrees. Multiplying through by the sin of 56 degrees, we get 𝐹 sub two is equal to 28. The magnitude of the force 𝐹 sub two is equal to 28 newtons.

We notice that this is the same value as 𝐹 sub one, which leads us to a general rule. If the resultant force bisects two forces, then the two forces will have the same magnitude. In this question, as the angle between the resultant force and 𝐹 sub one was 56 degrees and the angle between the resultant and 𝐹 sub two was also 56 degrees, then 𝐹 sub two and 𝐹 sub one must have the same magnitudes.

We will now summarize the key points from this video. We can calculate the resultant of two forces acting on one point but in different directions by using firstly the Pythagorean theorem and trig ratios when the forces are perpendicular and secondly the sine and cosine rules when the forces are not perpendicular. The direction of the resultant force is given as the angle between the resultant and one of the forces.

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