Lesson Explainer: Moment of a Force about a Point in 2D: Scalar Mathematics

In this explainer, we will learn how to find the sum of moments of a group of forces acting on a body about a point in 2D.

A nonzero net force acting on a rigid body produces a linear acceleration of the body in the direction that the net force acts in, which results in the translation of the center of mass of the body in that direction.

A force acting on a body may also result in the angular acceleration of the body about a point, producing rotation of the body. The magnitude of the angular acceleration of the body due to the action of the force is proportional to the moment of the force about the point.

The following figure shows a thin rod suspended vertically at a point š‘ƒ. A force āƒ‘š¹ acts on the rod horizontally. The moment of the force acts to rotate the rod about š‘ƒ.

The following figure shows that the rod rotates clockwise due to the moment of āƒ‘š¹ about š‘ƒ.

Suppose that the line of action of āƒ‘š¹ changes so that the line passes through š‘ƒ, as shown in the following figure.

The rod does not rotate about š‘ƒ due to the moment of āƒ‘š¹. For āƒ‘š¹ to have a nonzero moment about š‘ƒ, there must be a nonzero distance between š‘ƒ and the line of action of āƒ‘š¹.

When the line connecting a point š‘ƒ and the point where a force āƒ‘š¹ acts and the line of action of the force āƒ‘š¹ are perpendicular, the magnitude of the moment of the force about point š‘ƒ is the product of the magnitude of āƒ‘š¹ and the distance š‘‘ between the point and the line of action of the force. This can be expressed as š‘€=š¹š‘‘.

Using newtons as the unit of force and metres as the unit of distance, the unit of the moment of a force is the newton-metre (Nā‹…m).

The unit of newton-metre appears to be the same unit as that of the work done by a force, but the distance in metres has a very different meaning for the moment of a force than it does for the work done by a force. For the work done by a force, š‘‘ is the distance a body moves along the line of the force that acts on the body to move it. For the moment of a force, š‘‘ is the distance between the line of the force and the point about which a moment is produced.

Let us now look at an example of the calculation of a moment about a point.

Example 1: Finding the Magnitude of the Moment of a Force about a Point

If a force of magnitude 498 N is 8 cm away from a point š“, find the norm of the moment of the force about point š“, giving your answer in newton-metres.

Answer

The question tells us that there is a force of magnitude 498 N acting at a point 8 cm from a point š“. The question does not specify that the 8 cm distance and the line of action of the force are perpendicular, but this can be assumed if nothing indicates otherwise. The question asks for the norm of the force, meaning its magnitude.

The moment can be calculated using the formula š‘€=š¹š‘‘.

The answer is to be given in newton-metres. The magnitude of the force is 498 N.

The value of š‘‘ is not 8, as the distance is not 8 m but 8 cm. The 8 cm must be converted to a value in metres and, therefore, š‘‘=0.08m.

The values of the force are is now used in the formula to give the solution: š‘€=498Ɨ0.08=39.84ā‹….Nm

A moment of a force can produce rotation of a body either clockwise or counterclockwise, as the following figure shows.

The force āƒ‘š¹ļŠ§ produces a clockwise moment š‘€ļŠ§ about š‘ƒ, and the force āƒ‘š¹ļŠØ produces a counterclockwise moment š‘€ļŠØ about š‘ƒ. The sum of š‘€ļŠ§ and š‘€ļŠØ is š‘€net, where š‘€=š‘€āˆ’š‘€,netļŠØļŠ§ as counterclockwise moments are taken as positive.

Let us now look at an example of the calculation of a moment about a point in which the direction of rotation due to the moment is considered and the distance from the point that the moment is about to the point that the force acts at is not directly given.

Example 2: The Moment of a Force about a Point in Two Dimensions

Determine the moment of the force of magnitude 11 N about point š‘‚, giving your answer in Nā‹…m.

Answer

For this question to be answered, we must assume that the 11 N force is the only force acting on the object. As no other values are given, this is a reasonable assumption.

The question appears fairly complicated, so it is useful to remember that only two values are required to determine the moment of the force about š‘‚. One of the required values is the magnitude of the force, which is stated to be 11 N. The other value required is the distance from š‘‚ to the line of action of the force. This distance is shown in the following figure as š‘‘.

The moment produced by the force acting on the body is equivalent to a force that acts at a point at a horizontal distance š‘‘ and zero vertical distance from š‘‚. The force could act on the body at any point on the body along the line of action of the force to produce the same moment.

It is possible to determine š‘‘ by constructing a right triangle with a vertical side connecting the point on the body 36 cm horizontally distant from š‘‚ to a point vertically below that, which intersects a line perpendicular to the line of action of the force that passes through the point where the force acts, as shown in the following figure.

The side of the triangle formed opposite angle šœƒ is š·, and determining š· allows š‘‘ to be determined, using the equation š‘‘=0.36āˆ’(0.26āˆ’š·).

Note that the distances of 36 cm and 26 cm have been converted to 0.36 m and 0.26 m, respectively, so that the answer can be given in newton-metres.

The value of š· can be found if angle šœƒ is known, as šœƒ is the angle of a right triangle opposite š·, where the hypotenuse of š· has a length of 29 cm. The 29 cm is also converted to 0.29 m so that the answer can be given in newton-metres.

The following equation is used to determine š·: sinšœƒ=š·0.29.

Because the angle of 60āˆ˜ shown in the figure is part of a right angle consisting of the 60āˆ˜ angle and angle šœƒ, it follows that šœƒ=(90āˆ’60)=30.āˆ˜āˆ˜āˆ˜

As sin(30)=12āˆ˜, 12=š·0.29.

Making š· the subject gives š·=0.292=0.145.

It was previously established that š‘‘=0.36āˆ’0.26+š·, so š‘‘ is found from š‘‘=0.36āˆ’0.26+0.145=0.245.m

As š‘€=š¹š‘‘,š‘€=11Ɨ0.245=2.695ā‹….Nm

This is not the complete solution, however, as the rotation due to š‘€ is either clockwise or counterclockwise. The following figure shows that the force must move the object downward, which results in clockwise rotation for the body.

Because clockwise moments are taken as negative, the moment is given by š‘€=āˆ’2.695ā‹….Nm

Let us now look at an example in which the net moment due to multiple forces is determined.

Example 3: Finding the Sum of the Moments of the Forces Acting on a Rod

š“šµ is a rod of length 114 cm and negligible weight. Forces of magnitudes 83 N, 225 N, 163 N, and 136 N are acting on the rod as shown in the following figure. š¶ and š· are the points of trisection of š“šµ, and point š‘‚ is the midpoint of the rod. Find the algebraic sum of the moments of these forces about point š‘‚.

Answer

The length of the rod is 114 cm, and the lengths of šµš·, š·š¶, and š¶š“ are all given by š‘‘=1143=38.cm

The length of šµš‘‚ is given by šµš‘‚=1142=57,cm which equals the length of š“š‘‚.

The length of š·š‘‚ is given by š·š‘‚=38ļ€¼12ļˆ=19,cm which equals the length of š¶š‘‚.

The forces acting at š“ and at š· produce counterclockwise moments about š‘‚, and the forces acting at šµ and at š¶ produce clockwise moments about š‘‚. The net moment about š‘‚ is, therefore, given by š‘€=(83Ɨ57)+(163Ɨ19)āˆ’(136Ɨ57)āˆ’(225Ɨ19)=āˆ’4199ā‹….netNcm

For a force to produce a moment, the force must have a nonzero component acting perpendicularly to the line connecting the point around which a moment is produced and the point at which the force acts. Consider the following figure.

If šœƒ is zero, then the line of force āƒ‘š¹ must pass through š‘ƒ and so it would produce zero moment about š‘ƒ. If šœƒ=90āˆ˜, then āƒ‘š¹ produces its maximum moment about š‘ƒ. The calculation of the moment of a force must therefore include the angle at which the force acts.

Definition:

The moment of a force about a point š‘ƒ is the distance š‘‘ from š‘ƒ to the point where the force acts, multiplied by the component of the force perpendicular to the direction of the line intersecting š‘ƒ and the point where the force acts. This can be written as š‘€=š¹š‘‘šœƒ,sin where š¹ is the magnitude of the force and šœƒ is the angle between the direction of the force and the direction of the line intersecting š‘ƒ and the point where the force acts.

Now let us now look at an example where the angles at which forces act must be considered.

Example 4: Finding the Magnitude of the Sum of the Moments of Three Forces Acting along an Equilateral Triangle

Three forces, measured in newtons, are acting along the sides of an equilateral triangle š“šµš¶ as shown in the figure. Given that the triangle has a side length of 7 cm, determine the algebraic sum of the moments of the forces about the midpoint of š“šµ rounded to two decimal places.

Answer

The question asks for the moments about the midpoint of š“šµ, which is shown as the point š‘ƒ in the following figure. The length of š“šµ is 7 cm. Because š‘ƒ is at the midpoint of š“šµ, lengthlengthcmš“š‘ƒ=šµš‘ƒ=3.5.

The line of action of the 300 N force is along š“šµ and so it passes through š‘ƒ producing zero moment about š‘ƒ and can be ignored.

Because triangle š“šµš¶ is equilateral, all its internal angles are 60āˆ˜. The following figure shows the nonzero moments about š‘ƒ.

Triangle š‘ƒšµš¶ is a right triangle with an angle of 60āˆ˜ for which the length of the side adjacent to the angle is 3.5 cm and the length of the side opposite the angle is š‘ƒš¶. We have, therefore, that tan(60)=š‘ƒš¶3.5;āˆ˜ hence, š‘ƒš¶=3.5āˆš3.cm

The following figure shows the magnitudes of the forces acting, the angles at which they act, and their distances from š‘ƒ.

The 150 N force and the 100 N force both act clockwise, so both are negative.

Using the formula š‘€=š¹š‘‘šœƒ,sin the moment about š‘ƒ due to the 150 N force is given by š‘€=āˆ’150Ɨ3.5āˆš3(30)ā‹…š‘€=āˆ’525āˆš32ā‹….ļŠ§āˆ˜ļŠ§sinNcmNcm

The moment about š‘ƒ due to the 100 N force is given by š‘€=āˆ’100Ɨ3.5(60)ā‹…,š‘€=āˆ’175āˆš3ā‹….ļŠØāˆ˜ļŠØsinNcmNcm

The sum of the moments about š‘ƒ is š‘€net, which is the sum of š‘€ļŠ§ and š‘€ļŠØ. The question asks for this value to two decimal places: š‘€=āˆ’ļ€æ525āˆš32+175āˆš3ļ‹=āˆ’875āˆš32āˆ’757.77ā‹….netNcm

Let us now look at another such example.

Example 5: Locating a Point on a Rectangle given the Sum of Moments of Forces about It

š“šµš¶š· is a rectangle, where š“šµ=6cm and šµš¶=8cm, and forces of magnitudes 24, 30, 8, and 30 newtons are acting along ļƒ«šµš“, ļƒŖšµš¶, ļƒ«š¶š·, and ļƒ«š¶š“, respectively. If point šøāˆˆšµš¶, where the sum of the moments of the forces about šø is 53 Nā‹…cm in the direction of š“šµš¶š·, determine the length of šµšø.

Answer

The forces acting on the rectangle and the points at which they act are shown in the following figure.

The force at š“ that acts along š¶š“ can be resolved into perpendicular components. Resolving this force requires knowing the angle from š“šµ to the line of action of the force, which is shown as šœƒ in the following figure and is equal to the angle made by the line of action of the force with š¶š·.

š“š·š¶ is a right triangle. The length of š“š¶ is given by š“š¶=āˆš8+6=āˆš100=10.ļŠØļŠØcm

We have, therefore, that sincosšœƒ=810=45,šœƒ=610=35.

The component of the 30-newton force along š·š“ is š¹=30ļ€¼45ļˆ=24,ļŒ£ļŒ N and the component of the 30-newton force along šµš“ is š¹=30ļ€¼35ļˆ=18.ļŒ”ļŒ N

š¹ļŒ”ļŒ  and the 24-newton force acting at š“ both act along the same line, so the net force acting at š“ is š¹=24+18=42.ļŒ N

The following figure shows the components of the forces acting perpendicular and parallel to šµš¶, which intersects a point šø about which the net moment is known.

The moment about šø due to each component is the distance from the line of action of the component to šø, perpendicular to the direction of the component.

The following figure shows the magnitudes and directions of the moments about šø due to each force, and the net moment about šø.

The net moment about šø can be equated to the sum of the moments about šø due to the components: 144+64āˆ’8š‘¦āˆ’42š‘¦=53.

We can rearrange this expression to make š‘¦ the subject as follows: āˆ’50š‘¦=53āˆ’64āˆ’144=āˆ’155š‘¦=āˆ’155āˆ’50=3.1.cm

Let us summarize what we have learned in these examples.

Key Points

  • The moment of a force āƒ‘š¹ about a point š‘ƒ is the distance š‘‘ from š‘ƒ to the point where the force acts, multiplied by the component of the force perpendicular to the direction of the line intersecting š‘ƒ and the point where the force acts. This can be written as š‘€=š¹š‘‘šœƒ,sin where š¹ is the magnitude of the force and šœƒ is the angle between the direction of the force and the direction of the line intersecting š‘ƒ and the point where the force acts.
  • The net moment due to a set of moments about a point is the sum of the clockwise and counterclockwise moments about the point, where counterclockwise moments are positive.

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