Lesson Explainer: Resolution of Forces | Nagwa Lesson Explainer: Resolution of Forces | Nagwa

ู†ู‚ุงุท ุงู„ู…ูƒุงูุขุช

ุงูƒุณุจ ู†ู‚ุงุทู‹ุง ู…ู† ุฎู„ุงู„ ุชูุงุนู„ูƒ ู…ุน ุงู„ู…ูุฏุฑุณ ุฃุซู†ุงุก ุงู„ุญุตุฉ ูˆุฅุฌุงุจุชูƒ ุนู† ุงู„ุฃุณุฆู„ุฉุŒ ูˆุงุณุชุฎุฏู…ู‡ุง ู„ุงุญู‚ู‹ุง ู„ู„ุญุตูˆู„ ุนู„ู‰ ุฎุตู… ุนู†ุฏ ุดุฑุงุก ุญุตุต ุฃุฎุฑู‰.

Lesson Explainer: Resolution of Forces Mathematics • Second Year of Secondary School

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 ๐‘ฅ and ๐‘ฆ, 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 ๐œƒ with the ๐‘ฅ-axis of the system and makes an angle ๐œ™ with the ๐‘ฆ-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 ๐œƒ.

The ratios of the lengths of the opposite and adjacent sides to the length of the hypotenuse are determined by the following equations: sinoppositehypotenuse๐œƒ= and cosadjacenthypotenuse๐œƒ=.

A force, โƒ‘๐น, 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 ๐œƒ is the angle from the ๐‘ฅ-axis of the system.

The ratios of the magnitudes of the components of the force to the force are, therefore, given by sin๐œƒ=๐น๐น;๏˜ hence, ๐น=๐น๐œƒ๏˜sin and cos๐œƒ=๐น๐น;๏— hence, ๐น=๐น๐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 ๐น๏Šง and ๐น๏Šจ as shown in the figure. Give your answer correct to two decimal places.

Answer

The component ๐น๏Šง is given by ๐น=81(54).๏Šงcos

To two decimal places, ๐น=47.61๏ŠงN.

The component ๐น๏Šจ is given by ๐น=81(54).๏Šจsin

To two decimal places, ๐น=65.53๏ŠจN.

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 ๐‘ฅ-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 ๐œƒ to the coordinate system in which โƒ‘๐น is parallel to the ๐‘ฅ-axis.

The lines of action of the components โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ are along the ๐‘ฅ- and ๐‘ฆ-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 Particleโ€™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 45โˆ˜ to the horizontal. Resolve its weight into two components ๐น๏Šง and ๐น๏Šจ, where ๐น๏Šง is the component in the direction of the plane and ๐น๏Šจ is the component normal to the plane.

Answer

The following figure shows the forces acting on the body, where ๐‘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, ๐‘Š=๐น=๐‘Š(45)=๐‘Š(45)=72โˆš2=36โˆš2,๏Œฅ๏Šง๏Ž sincos and ๐‘Š=๐น=๐‘Š(45)=๐‘Š(45)=72โˆš2=36โˆš2.๏Œฅ๏Šจ๏Žกcossin

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 ๐œƒ to the horizontal, where tan๐œƒ=43. Resolve the weight of the particle into two components, โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ, where โƒ‘๐น๏Šง is parallel to a line of greatest slope and โƒ‘๐น๏Šจ is perpendicular to โƒ‘๐น๏Šง.

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(๐œƒ)=43.

From this, we can see that the sides of the triangle opposite to and adjacent to the angle ๐œƒ have lengths in the ratio 4โˆถ3. In a right triangle with the sides having this ratio of lengths, the ratio of the side adjacent to ๐œƒ to the length of the hypotenuse is 3โˆถ5, and the ratio of the side opposite ๐œƒ to the length of the hypotenuse is 4โˆถ5.

From this we see that cossin(๐œƒ)=35,(๐œƒ)=45.

The component of the weight of the particle perpendicular to the plane, โƒ‘๐น๏Šจ, is rotated through an angle ๐œƒ from the direction of the weight of the particle, so โƒ‘๐น๏Šจ has a magnitude given by โƒ‘๐น=69(๐œƒ)=69๏€ผ35๏ˆ=41.4.๏ŠจcosN

Conversely, the component of the weight of the particle parallel to the plane, โƒ‘๐น๏Šง, has a magnitude given by โƒ‘๐น=69(๐œƒ)=69๏€ผ45๏ˆ=55.2.๏ŠงsinN

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 โƒ‘๐น and two nonperpendicular components โƒ‘๐น๏Œบ and โƒ‘๐น๏Œป, where โƒ‘๐น=โƒ‘๐น+โƒ‘๐น.๏Œบ๏Œป

For a force โƒ‘๐น, there are infinite pairs of components that sum to โƒ‘๐น. โƒ‘๐น๏Œบ and โƒ‘๐น๏Œป 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 โƒ‘๐น๏Œบ and โƒ‘๐น๏Œป can be represented as acting from the same point as โƒ‘๐น, as shown in the following figure.

Lines from the head of โƒ‘๐น๏Œบ to the head of โƒ‘๐น and from the head of โƒ‘๐น๏Œป to the head of โƒ‘๐น can be drawn. These lines complete a parallelogram, as shown in the following figure.

The line from the head of โƒ‘๐น๏Œบ to the head of โƒ‘๐น is parallel to โƒ‘๐น๏Œป. The line from the head of โƒ‘๐น๏Œป to the head of โƒ‘๐น is parallel to โƒ‘๐น๏Œบ. The line of action of โƒ‘๐น 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, ๐›ผ, is given by ๐›ผ=180โˆ’(๐œƒ+๐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 ๐ด๐ต๐ถ are related to the angles of the triangle by the sine rule ๐‘Ž๐ด=๐‘๐ต=๐‘๐ถ,sinsinsin where ๐‘Ž, ๐‘, and ๐‘ are the lengths of the sides opposite to the angles ๐ด, ๐ต, and ๐ถ.

For โƒ‘๐น๏Œบ, โƒ‘๐น๏Œป, and โƒ‘๐น, ๐น๐œ™=๐น๐œƒ=๐น๐›ผ.๏Œบ๏Œปsinsinsin

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 โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ. Give your answer to two decimal places.

Answer

A parallelogram can be drawn with side lengths proportional to โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ. 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 ๐œƒ, as shown in the following figure.

The angle ๐œƒ is given by ๐œƒ=180โˆ’(60+45)=75.โˆ˜

The magnitudes of โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ can be determined by applying the sine rule in the triangle: ๐น(45)=๐น(60)=41(75).๏Šง๏Šจsinsinsin

We have, therefore, that ๐น=41(45)(75).๏Šงsinsin

To two decimal places, ๐น=30.01๏ŠงN.

We also have that ๐น=41(60)(75).๏Šจsinsin

To two decimal places, ๐น=36.76๏ŠจN.

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, ๐ด๐ถ and ๐ต๐ถ. Both strings make an angle of 37โˆ˜ with the horizontal. Resolve the weight of the body into two components in the direction of ๏ƒซ๐ด๐ถ and in the direction of ๏ƒช๐ต๐ถ. Give your answers to the nearest newton.

Answer

The strings are at the same angle from the horizontal and so the components ๐‘Š๏Šง and ๐‘Š๏Šจ have equal magnitude. The line of action of the weight of the body is perpendicular to ๐ด๐ต.

The following figure shows two right triangles. For each triangle, the non-right angles are 37โˆ˜ and ๐œƒ.

The angle between each component and the weight is also ๐œƒ and so is given by ๐œƒ=90โˆ’37=53.โˆ˜

A parallelogram can be defined that has a vertex at ๐ถ and another vertex vertically below ๐ถ 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)=๐‘Š(53),๐‘Š=69(53)(74).sinsinsinsin๏Šง๏Šง

To the nearest newton, ๐‘Š=57๏ŠงN.

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, โƒ‘๐น, can be resolved into components, โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ. If these components are perpendicular to each other in a coordinate system used to represent the force, then these components have magnitudes given by ๐น=๐น๐œƒ๏Šงcos and ๐น=๐น๐œƒ,๏Šจsin where ๐œƒ is the angle between โƒ‘๐น and the ๐‘ฅ-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 ๐‘Ž๐ด=๐‘๐ต=๐‘๐ถ,sinsinsin where ๐ด๐ต๐ถ is a triangle of the parallelogram with side lengths ๐‘Ž, ๐‘, and ๐‘, where ๐‘Ž, ๐‘, and ๐‘ are lengths of the sides of the triangle opposite the angles ๐ด, ๐ต, and ๐ถ.

ุงู†ุถู… ุฅู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ

ุดุงุฑูƒ ููŠ ุงู„ุญุตุต ุงู„ู…ุจุงุดุฑุฉ ุนู„ู‰ ู†ุฌูˆู‰ ูƒู„ุงุณูŠุฒ ูˆุญู‚ู‚ ุงู„ุชู…ูŠุฒ ุงู„ุฏุฑุงุณูŠ ุจุฅุฑุดุงุฏ ูˆุชูˆุฌูŠู‡ ู…ู† ู…ุฏุฑุณ ุฎุจูŠุฑ!

  • ุญุตุต ุชูุงุนู„ูŠุฉ
  • ุฏุฑุฏุดุฉ ูˆุฑุณุงุฆู„
  • ุฃุณุฆู„ุฉ ุงู…ุชุญุงู†ุงุช ูˆุงู‚ุนูŠุฉ

ุชุณุชุฎุฏู… «ู†ุฌูˆู‰» ู…ู„ูุงุช ุชุนุฑูŠู ุงู„ุงุฑุชุจุงุท ู„ุถู…ุงู† ุญุตูˆู„ูƒ ุนู„ู‰ ุฃูุถู„ ุชุฌุฑุจุฉ ุนู„ู‰ ู…ูˆู‚ุนู†ุง. ุงุนุฑู ุงู„ู…ุฒูŠุฏ ุนู† ุณูŠุงุณุฉ ุงู„ุฎุตูˆุตูŠุฉ