Lesson Explainer: Moment of a Force about a Point in 3D Mathematics

In this explainer, we will learn how to find the moment of vector forces acting on a body about a point in 3D.

A force, or a system of forces, acting on a body may have a turning effect on this body. This turning effect is described by the moment of the force (or of the system of forces). You may already be familiar with the moment of a force defined as a scalar given by the product of the magnitude of the force with the perpendicular distance between the line of action of the force and the point about which the moment is taken: ๐‘€=๐น๐‘‘.โŸ‚

In this explainer, we will learn that the moment of a force is properly defined as a vector.

The moment of the force depends on the point of the body about which it is taken. The direction of the moment vector gives the direction of rotation about that point that the force would generate, while its magnitude indicates the strength of the forceโ€™s turning effect.

Let us first see how the moment vector is defined.

Definition: Moment of a Force

The moment of force โƒ‘๐น acting on a body, taken about point ๐‘‚, is given by ๏ƒŸ๐‘€=โƒ‘๐‘Ÿร—โƒ‘๐น, where โƒ‘๐‘Ÿ is the position vector of ๐ด, the point of application of force โƒ‘๐น.

In this definition, we see that the coordinate system has been chosen such that its origin coincides with the point about which we take the moment. If we wanted to work out the moment of force โƒ‘๐น about point ๐‘ƒ that is not the origin, then we would simply replace โƒ‘๐‘Ÿ with ๏ƒŸ๐‘ƒ๐ด: ๏ƒŸ๐‘€=๏ƒŸ๐‘ƒ๐ดร—โƒ‘๐น.๏Œฏ

The letter ๐‘ƒ has been added as a subscript to ๏ƒŸ๐‘€ to indicate that the moment is taken about point ๐‘ƒ.

The properties of the cross product allow us to conclude first that ๏ƒŸ๐‘€ is a vector perpendicular to the plane defined by โƒ‘๐‘Ÿ and โƒ‘๐น. The direction of ๏ƒŸ๐‘€ is given by the right-hand rule. This rule is sometimes explained by referring to the rotation of a screw: the direction of the vector โƒ‘๐ดร—โƒ‘๐ต corresponds to the direction of the movement (up or down) of a bottle lid or a nut that one would turn in the same sense of rotation as when going from โƒ‘๐ด to โƒ‘๐ต, as illustrated in the following diagram.

This reference to rotation makes a lot of sense when talking about the moment of a force. Let us take the diagram used in the definition box above and imagine the movement of a rod fixed at ๐‘‚ and subjected to a single force โƒ‘๐น at ๐ด. It is worth noting that, for the sake of simplicity, the coordinate system is chosen here such that โƒ‘๐‘Ÿ and โƒ‘๐น are in the ๐‘ฅ๐‘ฆ-plane, so that ๏ƒŸ๐‘€ is parallel to the ๐‘ง-axis.

We see that the rod would turn clockwise until it is aligned with โƒ‘๐น, that is, when โƒ‘๐น has no turning effect anymore on the rod, which is indeed described by the fact that its moment is zero (which is consistent with the cross product of two collinear vectors being zero). This clockwise rotation corresponds to the rotation of vector โƒ‘๐‘Ÿ toward vector โƒ‘๐น.

Using the right-hand rule as explained above, it means that the moment of โƒ‘๐น about ๐‘‚ points down. In other words, its only component, which is along the ๐‘ง-axis, is negative since the positive ๐‘ง-axis points up. Remember that, in a three-dimensional Cartesian coordinate system, the right set of unit vectors ๏€ปโƒ‘๐‘–,โƒ‘๐‘—,โƒ‘๐‘˜๏‡ is such that โƒ‘๐‘˜=โƒ‘๐‘–ร—โƒ‘๐‘—.

The magnitude of the moment is given by โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–โ€–โ€–โƒ‘๐นโ€–โ€–|๐œƒ|,sin where ๐œƒ is the angle between โƒ‘๐‘Ÿ and โƒ‘๐น. We have used absolute value bars around sin๐œƒ here because if one uses directed angles, sin๐œƒ can be negative, as it is the case for the clockwise rotation discussed above, where ๐œƒ is then a negative angle. If one uses geometric angles, then 0โ‰ค๐œƒโ‰ค180โˆ˜, and thus sin๐œƒโ‰ฅ0.

Let us see now why this way of defining the magnitude of ๏ƒŸ๐‘€ is equivalent to what you may have learned before; namely, โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โ€–โ€–โƒ‘๐นโ€–โ€–๐‘‘โŸ‚, where ๐‘‘โŸ‚ is the perpendicular distance between the line of action of โƒ‘๐น and the point about which the moment is taken.

Taking our previous example of the rod, we can draw a circle with center ๐ด and radius โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–. We can think of it as a unit circle scaled by โ€–โ€–โƒ‘๐‘Ÿโ€–โ€– and rotated in such a way that the ๐‘ฅ-axis associated with the circle is in the same direction as โƒ‘๐น. The scalar projection of vector โƒ‘๐‘Ÿ on the ๐‘ฆ-axis associated with the unit circle is the segment ๐ด๐‘Œ, whose length is then โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–|๐œƒ|sin. Considering now the segment ๐‘‚๐ถ (observe that triangles ๐ด๐ต๐‘Œ and ๐‘‚๐ด๐ถ are congruent), we see that this length is also the perpendicular distance between the line of action of โƒ‘๐น and ๐‘‚.

To show that the perpendicular distance between the line of action of โƒ‘๐น and ๐‘‚ is given by โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–|๐œƒ|sin, we can also use the property that two supplementary angles have the same sine value ((180โˆ’๐›ผ)=๐›ผ)sinsinโˆ˜. As โˆ ๐‘‚๐ด๐ถ=180โˆ’๐œƒโˆ˜ (here ๐œƒ is a geometric angle), we have sinsinโˆ ๐‘‚๐ด๐ถ=๐œƒ. In triangle ๐‘‚๐ด๐ถ, we have ๐‘‚๐ถ=โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–โˆ ๐‘‚๐ด๐ถ=โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–๐œƒ.sinsin

This result means that, to work out the magnitude of the moment, we can take whatever angle between the line of action of โƒ‘๐น and the line โƒ–๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉ๏ƒฉโƒ—๐‘‚๐ด; the result will be the same.

As said above, the magnitude of the moment of a force indicates the strength of the forceโ€™s turning effect. We can link the equation โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โ€–โ€–โƒ‘๐นโ€–โ€–๐‘‘โŸ‚ with something we all have experienced in our daily life, namely, that the turning effect of a given force will increase if the force is applied further away from the pivot, the point about which the object turns. It is the working principle of a lever.

The method to calculate the cross product of two vectors from its components can be applied to calculate the moment of a force. With the force, โƒ‘๐น, and the position vector of the point of application of force โƒ‘๐น defined in the coordinate system ๏€ป๐‘‚,โƒ‘๐‘–,โƒ‘๐‘—,โƒ‘๐‘˜๏‡ as โƒ‘๐น=๐นโƒ‘๐‘–+๐นโƒ‘๐‘—+๐นโƒ‘๐‘˜๏—๏˜๏™ and โƒ‘๐‘Ÿ=๐‘Ÿโƒ‘๐‘–+๐‘Ÿโƒ‘๐‘—+๐‘Ÿโƒ‘๐‘˜๏—๏˜๏™, the moment of โƒ‘๐น about point ๐‘‚ is given by ๏ƒŸ๐‘€=โƒ‘๐‘Ÿร—โƒ‘๐น=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘Ÿ๐‘Ÿ๐‘Ÿ๐น๐น๐น|||||=๏€น๐‘Ÿ๐นโˆ’๐‘Ÿ๐น๏…โƒ‘๐‘–โˆ’(๐‘Ÿ๐นโˆ’๐‘Ÿ๐น)โƒ‘๐‘—+๏€น๐‘Ÿ๐นโˆ’๐‘Ÿ๐น๏…โƒ‘๐‘˜.๏—๏˜๏™๏—๏˜๏™๏˜๏™๏™๏˜๏—๏™๏™๏—๏—๏˜๏˜๏—

Let us look at an example where a three-dimensional moment vector is determined.

Example 1: Determining the Moment of a Force about a Point in Three Dimensions

A force of magnitude 6 N is acting on ๐ถ and is represented by a vector in a plane perpendicular to the ๐‘ฆ-axis as shown in the figure. Determine its moment vector about ๐ด in newton-centimeters.

Answer

To answer this question, we will first determine the components of vector ๏ƒ ๐ด๐ถ, which is the vector from the point about which the moment is taken and the point of application of the force, and of force โƒ‘๐น acting on ๐ถ.

From the figure, we find ๐ด(0,0,โˆ’16) and ๐ถ(0,16,8), with 1 cm being the unit length of the coordinate system. Hence, ๏ƒ ๐ด๐ถ=(0โˆ’0)โƒ‘๐‘–+(16โˆ’0)โƒ‘๐‘—+(8โˆ’(โˆ’16))โƒ‘๐‘˜๏ƒ ๐ด๐ถ=๏€ป16โƒ‘๐‘—+24โƒ‘๐‘˜๏‡.cm

The magnitude of force โƒ‘๐น is 6 N. Let us draw the direction of โƒ‘๐น in a plane perpendicular to the ๐‘ฆ-axis, that is, in a plane parallel to the ๐‘ฅ๐‘ง-plane. Beware of the orientation of the ๐‘ฅ-axis when working in three dimensions! We have represented here the plane by looking at it from left to right in the given diagram, that is, such that the unit vector โƒ‘๐‘— is pointing downward with respect to our screen.

โƒ‘๐น lies in a plane perpendicular to the ๐‘ฆ-axis; hence, ๐น=0๏˜. From the figure, we see that ๐น=โ€–โ€–โƒ‘๐นโ€–โ€–30=โ€–โ€–โƒ‘๐นโ€–โ€–2,๐น=โˆ’โ€–โ€–โƒ‘๐นโ€–โ€–30=โˆ’โˆš3โ€–โ€–โƒ‘๐นโ€–โ€–2.๏—โˆ˜๏™โˆ˜sincos

Since โ€–โ€–โƒ‘๐นโ€–โ€–=6, we find that โƒ‘๐น=๏€ป3โƒ‘๐‘–โˆ’3โˆš3โƒ‘๐‘˜๏‡.N

We can now work out the moment of โƒ‘๐น about ๐ด as ๏ƒŸ๐‘€=๏ƒ ๐ด๐ถร—โƒ‘๐น=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐ด๐ถ๐ด๐ถ๐ด๐ถ๐น๐น๐น|||||๏ƒŸ๐‘€=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜0162430โˆ’3โˆš3|||||๏ƒŸ๐‘€=๏€ป16โ‹…๏€ปโˆ’3โˆš3๏‡โˆ’0โ‹…24๏‡โƒ‘๐‘–โˆ’๏€ป0โ‹…๏€ปโˆ’3โˆš3๏‡โˆ’3โ‹…24๏‡โƒ‘๐‘—+(0โ‹…0โˆ’3โ‹…16)โƒ‘๐‘˜,๏ƒŸ๐‘€=๏€ปโˆ’48โˆš3โƒ‘๐‘–+72โƒ‘๐‘—โˆ’48โƒ‘๐‘˜๏‡โ‹….๏—๏˜๏™๏—๏˜๏™Ncm

Let us now look at an example where multiple forces act at a point to produce a moment.

Example 2: Finding the Resultant Moment Vector of Two Forces about the Origin in Three Dimensions

In the figure, if the forces โƒ‘๐น=โˆ’7โƒ‘๐‘–โˆ’โƒ‘๐‘—+3โƒ‘๐‘˜๏Šง and โƒ‘๐น=โˆ’7โƒ‘๐‘–+8โƒ‘๐‘—โˆ’6โƒ‘๐‘˜๏Šจ are acting on the point ๐ด, where โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ are measured in newtons, determine the moment vector of the resultant about the point ๐‘‚ in newton-centimetres.

Answer

Two forces act at ๐ด, โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ. Since the two forces act at the same point, the sum of their moments is equal to the moment of their resultant. For this reason, we are asked to find the moment of their resultant (i.e., their sum).

Let us start by finding the resultant, โƒ‘๐น, of โƒ‘๐น๏Šง and โƒ‘๐น๏Šจ: โƒ‘๐น=โƒ‘๐น+โƒ‘๐นโƒ‘๐น=โˆ’7โƒ‘๐‘–โˆ’โƒ‘๐‘—+3โƒ‘๐‘˜โˆ’7โƒ‘๐‘–+8โƒ‘๐‘—โˆ’6โƒ‘๐‘˜โƒ‘๐น=โˆ’14โƒ‘๐‘–+7โƒ‘๐‘—โˆ’3โƒ‘๐‘˜.๏Šง๏Šจ

We want to calculate the moment of โƒ‘๐น about point ๐‘‚, the origin, so we need to find the position vector of ๐ด, โƒ‘๐‘Ÿ. From the figure, we find that โƒ‘๐‘Ÿ=9โƒ‘๐‘–+12โƒ‘๐‘—+8โƒ‘๐‘˜.

The moment of โƒ‘๐น about point ๐‘‚ is given by ๏ƒŸ๐‘€=โƒ‘๐‘Ÿร—โƒ‘๐น=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘Ÿ๐‘Ÿ๐‘Ÿ๐น๐น๐น|||||๏ƒŸ๐‘€=||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜9128โˆ’147โˆ’3||||๏ƒŸ๐‘€=(12โ‹…(โˆ’3)โˆ’7โ‹…8)โƒ‘๐‘–โˆ’(9โ‹…(โˆ’3)โˆ’(โˆ’14)โ‹…8)โƒ‘๐‘—+(9โ‹…7โˆ’(โˆ’14)โ‹…12)โƒ‘๐‘˜๏ƒŸ๐‘€=๏€ปโˆ’92โƒ‘๐‘–โˆ’85โƒ‘๐‘—+231โƒ‘๐‘˜๏‡โ‹….๏—๏˜๏™๏—๏˜๏™Ncm

Let us now look at an example in which unknown components of a force vector are determined from the moment about a point due to the force.

Example 3: Finding the Unknown Components of a Force given Its Position Vector and the Moment Components about an Axis in Three Dimensions

If the force โƒ‘๐น=๐‘šโƒ‘๐‘–+๐‘›โƒ‘๐‘—โˆ’โƒ‘๐‘˜ is acting at a point whose position vector is โƒ‘๐‘Ÿ=14โƒ‘๐‘–โˆ’โƒ‘๐‘—+12โƒ‘๐‘˜ and the ๐‘ฅ- and ๐‘ฆ-components of the moment of the force โƒ‘๐น about the origin point are 73 and 242 units of moment, respectively, find the values of ๐‘š and ๐‘›.

Answer

As we are given here the ๐‘ฅ- and ๐‘ฆ-components of the moment of the force โƒ‘๐น about the origin, let us first work out the moment using the position and force vectors. It will give us the three components of the moment in terms of ๐‘š and ๐‘›: ๏ƒŸ๐‘€=โƒ‘๐‘Ÿร—โƒ‘๐น=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘Ÿ๐‘Ÿ๐‘Ÿ๐น๐น๐น|||||๏ƒŸ๐‘€=||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜14โˆ’112๐‘š๐‘›โˆ’1||||๏ƒŸ๐‘€=((โˆ’1)โ‹…(โˆ’1)โˆ’12๐‘›)โƒ‘๐‘–โˆ’(14โ‹…(โˆ’1)โˆ’12๐‘š)โƒ‘๐‘—+(14๐‘›+๐‘š)โƒ‘๐‘˜๏ƒŸ๐‘€=(1โˆ’12๐‘›)โƒ‘๐‘–+(14+12๐‘š)โƒ‘๐‘—+(14๐‘›+๐‘š)โƒ‘๐‘˜.๏—๏˜๏™๏—๏˜๏™

Equating the ๐‘ฅ- and ๐‘ฆ-components of ๏ƒŸ๐‘€ with the values given in the question gives

1โˆ’12๐‘›=73,14+12๐‘š=242.(1)(2)

Let us solve equation (1): โˆ’12๐‘›=72,๐‘›=72โˆ’12=โˆ’6.

And equation (2) gives 12๐‘š=228,๐‘š=22812=19.

The answer is ๐‘š=19 and ๐‘›=โˆ’6.

Let us now look at another example in which we are asked to determine unknown components of the position vector of the point at which the force acts using the moment of the force.

Example 4: Finding the Unknown Coordinates of a Point Lying on the Line of Action of a Force given Its Vector and the Moment of the force about a Point in Three Dimensions

The moment of the force โƒ‘๐น about the origin is ๏ƒŸ๐‘€๏Œฎ, where โƒ‘๐น=โƒ‘๐‘–โˆ’2โƒ‘๐‘—โˆ’โƒ‘๐‘˜ and ๏ƒŸ๐‘€=20โƒ‘๐‘–+27โƒ‘๐‘—โˆ’34โƒ‘๐‘˜๏Œฎ. Given that the force passes through a point whose ๐‘ฆ-coordinate is 4, find the ๐‘ฅ- and ๐‘ง-coordinates of the point.

Answer

The moment of a force acting at point ๐ด about the origin is given by the cross product of the position vector of ๐ด, โƒ‘๐‘Ÿ, and the force, โƒ‘๐น: ๏ƒŸ๐‘€=โƒ‘๐‘Ÿร—โƒ‘๐น.

In this question, we are not given the point of application of the force, but we are asked to find the ๐‘ฅ-coordinate of a point that lies on the line of action of the force.

We know, however, that โƒ‘๐‘Ÿร—โƒ‘๐น=โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–โ€–โ€–โƒ‘๐นโ€–โ€–|๐œƒ|โƒ‘๐‘›,sin where ๐œƒ is the angle between โƒ‘๐‘Ÿ and โƒ‘๐น and โƒ‘๐‘› is a unit vector perpendicular to the plane defined by โƒ‘๐‘Ÿ and โƒ‘๐น (provided โƒ‘๐‘Ÿ and โƒ‘๐น are noncollinear). As โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–|๐œƒ|sin is equal to the perpendicular distance between the line of action of โƒ‘๐น and the point about which the moment is taken (here the origin), ๐‘‘โŸ‚, we can write ๏ƒŸ๐‘€=โ€–โ€–โƒ‘๐นโ€–โ€–๐‘‘โƒ‘๐‘›.โŸ‚

We see that taking โƒ‘๐‘Ÿ as the position vector of any point on the line of action of โƒ‘๐น will give the same moment, ๏ƒŸ๐‘€.

Therefore, we can now calculate ๏ƒŸ๐‘€๏Œฎ by working out โƒ‘๐‘Ÿร—โƒ‘๐น with โƒ‘๐‘Ÿ being the position vector of the point on the line of action of โƒ‘๐น with coordinates (๐‘ฅ,4,๐‘ง): ๏ƒŸ๐‘€=โƒ‘๐‘Ÿร—โƒ‘๐น=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘Ÿ๐‘Ÿ๐‘Ÿ๐น๐น๐น|||||๏ƒŸ๐‘€=||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘ฅ4๐‘ง1โˆ’2โˆ’1||||๏ƒŸ๐‘€=((โˆ’1)โ‹…4โˆ’(โˆ’2)๐‘ง)โƒ‘๐‘–โˆ’((โˆ’1)๐‘ฅโˆ’1โ‹…๐‘ง)โƒ‘๐‘—+((โˆ’2)โ‹…๐‘ฅโˆ’1โ‹…4)โƒ‘๐‘˜๏ƒŸ๐‘€=(โˆ’4+2๐‘ง)โƒ‘๐‘–+(๐‘ฅ+๐‘ง)โƒ‘๐‘—+(โˆ’2๐‘ฅโˆ’4)โƒ‘๐‘˜.๏Œฎ๏—๏˜๏™๏—๏˜๏™๏Œฎ๏Œฎ๏Œฎ

We are told in the question that ๏ƒŸ๐‘€=20โƒ‘๐‘–+27โƒ‘๐‘—โˆ’34โƒ‘๐‘˜.๏Œฎ

Equating both expressions for ๏ƒŸ๐‘€๏Œฎ gives an equation for each component: โˆ’4+2๐‘ง=20,๐‘ฅ+๐‘ง=27,โˆ’2๐‘ฅโˆ’4=โˆ’34.

The first equation can be solved in this way: 2๐‘ง=24๐‘ง=12.

And the third equation gives the value of ๐‘ฅ: ๐‘ฅ=โˆ’34+4โˆ’2๐‘ฅ=15.

We can now check that these solutions verify the second equation to make sure our system of equations is consistent: ๐‘ฅ+๐‘ง=15+12=27.

The force passes through the point of coordinates (15,4,12).

Let us look at an example in which the perpendicular distance between the line of action of the force and a point is determined.

Example 5: Finding the Moment Vector of a Force Acting on a Given Point and the Length of the Perpendicular between the Origin Point and the Line of Action of the Force

Find the moment ๏ƒŸ๐‘€ of the force โƒ‘๐น about the origin point, given that โƒ‘๐น=โˆ’2โƒ‘๐‘–+โƒ‘๐‘—+โƒ‘๐‘˜ and is acting at a point ๐ด whose position vector is โƒ‘๐‘Ÿ=6โƒ‘๐‘–+6โƒ‘๐‘—โˆ’3โƒ‘๐‘˜ with respect to the origin point, then determine the length ๐ฟ of the perpendicular segment drawn from the origin point to the line of action of the force โƒ‘๐น.

Answer

We are asked to find first the moment of force โƒ‘๐น about the origin. It is given by ๏ƒŸ๐‘€=โƒ‘๐‘Ÿร—โƒ‘๐น=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘Ÿ๐‘Ÿ๐‘Ÿ๐น๐น๐น|||||๏ƒŸ๐‘€=||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜66โˆ’3โˆ’211||||๏ƒŸ๐‘€=(6โ‹…1โˆ’1โ‹…(โˆ’3))โƒ‘๐‘–โˆ’(6โ‹…1โˆ’(โˆ’2)โ‹…(โˆ’3))โƒ‘๐‘—+(6โ‹…1โˆ’(โˆ’2)โ‹…6)โƒ‘๐‘˜๏ƒŸ๐‘€=9โƒ‘๐‘–+18โƒ‘๐‘˜.๏—๏˜๏™๏—๏˜๏™

Then, we are asked to determine the length ๐ฟ of the perpendicular segment drawn from the origin point to the line of action of the force โƒ‘๐น. ๐ฟ is what is commonly referred to as the perpendicular distance between the origin and the line of action of โƒ‘๐น, denoted ๐‘‘โŸ‚.

Knowing that โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–โ€–โ€–โƒ‘๐นโ€–โ€–|๐œƒ|=โ€–โ€–โƒ‘๐นโ€–โ€–๐‘‘,sinโŸ‚ we see that, as ๐ฟ=๐‘‘โŸ‚, ๐ฟ is given by ๐ฟ=โ€–โ€–๏ƒŸ๐‘€โ€–โ€–โ€–โ€–โƒ‘๐นโ€–โ€–.

We have โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=๏„๐‘€+๐‘€+๐‘€โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โˆš9+18โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โˆš9+9โ‹…2โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โˆš9โ‹…5โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=9โˆš5๏Šจ๏—๏Šจ๏˜๏Šจ๏™๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ and โ€–โ€–โƒ‘๐นโ€–โ€–=๏„๐น+๐น+๐นโ€–โ€–โƒ‘๐นโ€–โ€–=โˆš2+1+1โ€–โ€–โƒ‘๐นโ€–โ€–=โˆš6.๏Šจ๏—๏Šจ๏˜๏Šจ๏™๏Šจ๏Šจ๏Šจ

Hence, ๐ฟ=โ€–โ€–๏ƒŸ๐‘€โ€–โ€–โ€–โ€–โƒ‘๐นโ€–โ€–๐ฟ=9โˆš5โˆš6๐ฟ=9โˆš306๐ฟ=3โˆš302.lengthunits

Let us now summarize what we have learned in this explainer.

Key Points

  • The moment of a force about a point describes the turning effect about this point of the force.
  • The moment of force โƒ‘๐น acting on a body, taken about point ๐‘‚, the origin of the coordinate system, is given by ๏ƒŸ๐‘€=โƒ‘๐‘Ÿร—โƒ‘๐น=|||||โƒ‘๐‘–โƒ‘๐‘—โƒ‘๐‘˜๐‘Ÿ๐‘Ÿ๐‘Ÿ๐น๐น๐น|||||,๏—๏˜๏™๏—๏˜๏™ where โƒ‘๐‘Ÿ is the position vector of ๐ด, the point of application of force โƒ‘๐น.
    When the moment of force โƒ‘๐น is taken about a point ๐‘ƒ that is not the origin, โƒ‘๐‘Ÿ in the above equation is replaced by ๏ƒŸ๐ด๐‘ƒ.
  • Given the properties of the cross product, the magnitude of the moment is given by โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–โ€–โ€–โƒ‘๐นโ€–โ€–|๐œƒ|,sin where ๐œƒ is the angle between โƒ‘๐‘Ÿ and โƒ‘๐น.
  • Since the perpendicular distance, ๐‘‘โŸ‚, between the line of action of force โƒ‘๐น and the point about which the moment is taken is equal to โ€–โ€–โƒ‘๐‘Ÿโ€–โ€–|๐œƒ|sin (or โ€–โ€–๏ƒŸ๐ด๐‘ƒโ€–โ€–|๐œƒ|sin), we have โ€–โ€–๏ƒŸ๐‘€โ€–โ€–=โ€–โ€–โƒ‘๐นโ€–โ€–๐‘‘;โŸ‚ that is, ๐‘‘=โ€–โ€–๏ƒŸ๐‘€โ€–โ€–โ€–โ€–โƒ‘๐นโ€–โ€–.โŸ‚

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.