Lesson Explainer: Resolution of Forces Mathematics

In this explainer, we will learn how to solve problems about the resolution of a force into two directions.

Force is a vector quantity and so a force can be represented by an arrow in the direction of the force with a length proportional to the magnitude of the force.

The direction of a force can be expressed in terms of a coordinate system. Probably the most intuitive example of such a system is a two-dimensional system with perpendicular axes. It is a common convention to label these axes π‘₯ and 𝑦, as shown in the following figure.

The direction of a force may be parallel to one of the axes of the coordinate system. This is shown in the following figure for a force for which the point of action is the origin of the system.

When the line of action of a force is parallel to one axis of such a coordinate system, it is necessarily perpendicular to the other axis of the system.

The direction of force can however make an arbitrary angle with lines parallel to either of the axes of the system, as shown in the following figure.

The line of action of the force shown makes an angle πœƒ with the π‘₯-axis of the system and makes an angle πœ™ with the 𝑦-axis of the system.

A force acting in an arbitrary direction can be expressed in terms of two components. Each component is parallel to one of the axes of the system and perpendicular to the other axis. The directions of these components are, therefore, perpendicular to each other. The perpendicular components of a force and the force itself are shown in the following figure.

The magnitudes of the perpendicular components of a force can be determined from trigonometric rules for right triangles. Consider the following figure of a right triangle which has an angle πœƒ.

The ratios of the lengths of the opposite and adjacent sides to the length of the hypotenuse are determined by the following equations: sinoppositehypotenuseπœƒ= and cosadjacenthypotenuseπœƒ=.

A force, ⃑𝐹, and its perpendicular components form a right triangle, as shown in the following figure.

The arrows representing the force and its components are assumed to be in a coordinate system where angle πœƒ is the angle from the π‘₯-axis of the system.

The ratios of the magnitudes of the components of the force to the force are, therefore, given by sinπœƒ=𝐹𝐹; hence, 𝐹=πΉπœƒο˜sin and cosπœƒ=𝐹𝐹; hence, 𝐹=πΉπœƒ.cos

The perpendicular components of a force are often represented acting at the point at which the force acts. The following figure shows that representing the components of the force in this way is equivalent to representing the force as the sum of its perpendicular components.

Let us look at an example of resolving a force into perpendicular components.

Example 1: Resolving a Force into Two Perpendicular Components

Resolve a force of 81 N into two perpendicular components 𝐹 and 𝐹 as shown in the figure. Give your answer correct to two decimal places.

Answer

The component 𝐹 is given by 𝐹=81(54).cos

To two decimal places, 𝐹=47.61N.

The component 𝐹 is given by 𝐹=81(54).sin

To two decimal places, 𝐹=65.53N.

It is worth noting that the magnitudes of the components of a force are dependent on the coordinate system chosen to represent force vectors.

For example, a force that acts along the π‘₯-axis of a coordinate system has two nonzero components in another coordinate system, as shown in the figure for a coordinate system that is rotated by an angle πœƒ to the coordinate system in which ⃑𝐹 is parallel to the π‘₯-axis.

The lines of action of the components βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ are along the π‘₯- and 𝑦-axes of this rotated coordinate system.

Let us look at an example of resolving a force into perpendicular components relative to an arbitrary direction.

Example 2: Resolving a Particle’s Weight into the Directions Parallel and Perpendicular to the Plane

A body weighing 72 N is placed on a plane that is inclined at 45∘ to the horizontal. Resolve its weight into two components 𝐹 and 𝐹, where 𝐹 is the component in the direction of the plane and 𝐹 is the component normal to the plane.

Answer

The following figure shows the forces acting on the body, where π‘Š is the 72 N weight of the body.

The components of the weight parallel to and perpendicular to the plane are perpendicular to each other; hence, π‘Š=𝐹=π‘Š(45)=π‘Š(45)=72√2=36√2,οŒ₯sincos and π‘Š=𝐹=π‘Š(45)=π‘Š(45)=72√2=36√2.οŒ₯cossin

Let us now look at another such example.

Example 3: Resolving the Weight of a Body on an Inclined Plane

A particle weighing 69 N is placed on a plane inclined at an angle πœƒ to the horizontal, where tanπœƒ=43. Resolve the weight of the particle into two components, 𝐹 and 𝐹, where 𝐹 is parallel to a line of greatest slope and 𝐹 is perpendicular to 𝐹.

Answer

The following figure shows the force of the weight of the particle and its components parallel to and perpendicular to the plane. A section of the plane that corresponds to a right triangle is shown.

The question states that tan(πœƒ)=43.

From this, we can see that the sides of the triangle opposite to and adjacent to the angle πœƒ have lengths in the ratio 3∢4. In a right triangle with the sides having this ratio of lengths, the ratio of the side adjacent to πœƒ to the length of the hypotenuse is 3∢5, and the ratio of the side opposite πœƒ to the length of the hypotenuse is 4∢5.

From this we see that cossin(πœƒ)=35,(πœƒ)=45.

The component of the weight of the particle perpendicular to the plane, 𝐹, is rotated through an angle πœƒ from the direction of the weight of the particle, so 𝐹 has a magnitude given by 𝐹=69(πœƒ)=69ο€Ό35=41.4.cosN

Conversely, the component of the weight of the particle parallel to the plane, 𝐹, has a magnitude given by 𝐹=69(πœƒ)=69ο€Ό45=55.2.sinN

Thus far we have considered components of a force where the components are perpendicular to each other. We can consider a force as consisting of two nonperpendicular components. The following figure shows a force ⃑𝐹 and two nonperpendicular components βƒ‘πΉοŒΊ and βƒ‘πΉοŒ», where 𝐹=𝐹+𝐹.

For a force ⃑𝐹, there are infinite pairs of components that sum to ⃑𝐹. βƒ‘πΉοŒΊ and βƒ‘πΉοŒ» are only one example of such a pair of components.

For nonperpendicular components of a force, it is necessary to use different trigonometric rules to those of right triangles.

The components βƒ‘πΉοŒΊ and βƒ‘πΉοŒ» can be represented as acting from the same point as ⃑𝐹, as shown in the following figure.

Lines from the head of βƒ‘πΉοŒΊ to the head of ⃑𝐹 and from the head of βƒ‘πΉοŒ» to the head of ⃑𝐹 can be drawn. These lines complete a parallelogram, as shown in the following figure.

The line from the head of βƒ‘πΉοŒΊ to the head of ⃑𝐹 is parallel to βƒ‘πΉοŒ». The line from the head of βƒ‘πΉοŒ» to the head of ⃑𝐹 is parallel to βƒ‘πΉοŒΊ. The line of action of ⃑𝐹 is a line from which coangles can be defined for the parallelogram, as shown in the following figure.

The parallelogram consists of two similar triangles, for which the unknown angle, 𝛼, is given by 𝛼=180βˆ’(πœƒ+πœ™), as shown in the following figure.

Let us consider one of the triangles of the parallelogram, as shown in the following figure.

The lengths of the sides of a triangle 𝐴𝐡𝐢 are related to the angles of the triangle by the sine rule π‘Žπ΄=𝑏𝐡=𝑐𝐢,sinsinsin where π‘Ž, 𝑏, and 𝑐 are the lengths of the sides opposite to the angles 𝐴, 𝐡, and 𝐢.

For βƒ‘πΉοŒΊ, βƒ‘πΉοŒ», and ⃑𝐹, πΉπœ™=πΉπœƒ=𝐹𝛼.sinsinsin

Let us look at an example of resolving a force that has nonperpendicular components.

Example 4: Resolving a Force into Two Components Shown in a Diagram

A force of magnitude 41 N acts due south. It is resolved into two components as shown on the diagram. Find the magnitudes of βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨. Give your answer to two decimal places.

Answer

A parallelogram can be drawn with side lengths proportional to βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨. A straight line can be drawn from the vertex at which these sides meet to the opposite vertex of the parallelogram. This line has a length of 41 length units, as shown in the following figure.

If we consider one of the triangles of this parallelogram, it contains an angle πœƒ, as shown in the following figure.

The angle πœƒ is given by πœƒ=180βˆ’(60+45)=75.∘

The magnitudes of βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ can be determined by applying the sine rule in the triangle: 𝐹(45)=𝐹(60)=41(75).sinsinsin

We have, therefore, that 𝐹=41(45)(75).sinsin

To two decimal places, 𝐹=30.01N.

We also have that 𝐹=41(60)(75).sinsin

To two decimal places, 𝐹=36.76N.

Let us look at another such example.

Example 5: Resolving Forces in a Real-Life Context

The diagram shows a body of weight 69 N suspended by 2 light, inextensible strings, 𝐴𝐢 and 𝐡𝐢. Both strings make an angle of 37∘ with the horizontal. Resolve the weight of the body into two components in the direction of 𝐴𝐢 and in the direction of οƒͺ𝐡𝐢. Give your answers to the nearest newton.

Answer

The strings are at the same angle from the horizontal and so the components π‘ŠοŠ§ and π‘ŠοŠ¨ have equal magnitude. The line of action of the weight of the body is perpendicular to 𝐴𝐡.

The following figure shows two right triangles. For each triangle, the non-right angles are 37∘ and πœƒ.

The angle between each component and the weight is also πœƒ and so is given by πœƒ=90βˆ’37=53.∘

A parallelogram can be defined that has a vertex at 𝐢 and another vertex vertically below 𝐢 proportional to the weight, where the lengths of each side of the parallelogram are proportional to the magnitude of either component. A triangle of this parallelogram is shown in the following figure.

From this, we can see that 69(74)=π‘Š(53),π‘Š=69(53)(74).sinsinsinsin

To the nearest newton, π‘Š=57N.

Each component has a magnitude of 57 N.

Now let us summarize what we have learned in these examples.

Key Points

  • A two-dimensional force, ⃑𝐹, can be resolved into components, βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨. If these components are perpendicular to each other in a coordinate system used to represent the force, then these components have magnitudes given by 𝐹=πΉπœƒοŠ§cos and 𝐹=πΉπœƒ,sin where πœƒ is the angle between ⃑𝐹 and the π‘₯-axis of the coordinate.
  • A force can be resolved into components that are not perpendicular to each other. A parallelogram can be formed that has a diagonal of length proportional to the magnitude of the force and sides that have lengths proportional to the magnitudes of the components of the force that act at angles to the force proportional to the internal angles of the parallelogram. The lengths of the sides of the parallelogram can be determined using the sine rule π‘Žπ΄=𝑏𝐡=𝑐𝐢,sinsinsin where 𝐴𝐡𝐢 is a triangle of the parallelogram with side lengths π‘Ž, 𝑏, and 𝑐, where π‘Ž and 𝑏 are lengths of the sides of the triangle opposite the angles 𝐴, 𝐡, and 𝐢.

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