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

In this explainer, we will learn how to find the moment of a planar system of forces acting on a body about a point as a vector.

We know that a force, or a system of forces, can have a rotational effect on a body, which is described by the moment of the force, or the system of forces, about a point. We recall that in planar motion, the moment 𝑀 of force ⃑𝐹 about a point is defined to be a scalar whose magnitude is given by |𝑀|=‖‖⃑𝐹‖‖𝑑,βŸ‚ where π‘‘βŸ‚ is the perpendicular distance between the point and the line of action for force ⃑𝐹. We can then determine the sign of the moment by considering whether the rotational effect is clockwise or counterclockwise. By convention, we define the moment with the counterclockwise effect to be positive, which means the moment with the clockwise rotational effect is defined to be negative.

While this definition works well for planar motion, it is insufficient when we consider the motion with a 3-dimensional space because the notion of clockwise or counterclockwise notation does not hold here. Hence, we would like to extend the definition of the moment to 3D motion from the scalar moment defined for planar motion. In order to preserve the notion of the orientation of a rotation, we define a moment to be a vector as follows.

Definition: Moment of a Force

The moment of a force ⃑𝐹 acting on a body, taken about point 𝑂, is given by οƒŸπ‘€=βƒ‘π‘ŸΓ—βƒ‘πΉ, where π‘Ÿ is the position vector of 𝐴, the point of application of force ⃑𝐹.

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 ⃑𝐹 about point 𝑃 that is not the origin, then we would simply replace βƒ‘π‘Ÿ with οƒŸπ‘ƒπ΄: οƒŸπ‘€=οƒŸπ‘ƒπ΄Γ—βƒ‘πΉ.

The letter 𝑃 has been added as a subscript to οƒŸπ‘€ to indicate that the moment is taken about point 𝑃.

In our first example, we will use this formula to compute the vector moment of a force on a plane about a point.

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

If the force ⃑𝐹=βˆ’5⃑𝑖+π‘šβƒ‘π‘— is acting at the point 𝐴(7,3), determine the moment of ⃑𝐹 about the point 𝐡(7,βˆ’2).

Answer

In this example, we need to find the moment of a planar force about a point. Recall that the vector moment of force ⃑𝐹 acting at point 𝐴 about point 𝐡 is given by οƒŸπ‘€=𝐡𝐴×⃑𝐹.

Let us begin by finding the vector 𝐡𝐴: 𝐡𝐴=(7,3,0)βˆ’(7,βˆ’2,0)=(0,5,0).

We can write ⃑𝐹 as ⃑𝐹=βˆ’5⃑𝑖+π‘šβƒ‘π‘—+0βƒ‘π‘˜=(βˆ’5,π‘š,0).

Taking the cross product, 𝐡𝐴×⃑𝐹=||||βƒ‘π‘–βƒ‘π‘—βƒ‘π‘˜050βˆ’5π‘š0||||=(5Γ—0βˆ’0Γ—π‘š)βƒ‘π‘–βˆ’(0Γ—0βˆ’0Γ—(βˆ’5))⃑𝑗+(0Γ—π‘šβˆ’5Γ—(βˆ’5))βƒ‘π‘˜=25βƒ‘π‘˜.

We note that the unknown constant π‘š in the force ⃑𝐹 cancelled out when we computed the cross product. Hence, the moment of ⃑𝐹 about point 𝐡 is 25βƒ‘π‘˜.

In the previous example, we computed the vector moment of a planar force about a point using the formula οƒŸπ‘€=βƒ‘π‘ŸΓ—βƒ‘πΉ.

We can see that the resulting vector of the cross product only contained a βƒ‘π‘˜ component, and the ⃑𝑖 and ⃑𝑗 components vanished. This is not surprising if we consider the geometric property of a cross product. Recall that a vector resulting from the cross product of two vectors must be perpendicular to the two vectors. Since οƒŸπ‘€ is defined to be the cross product of vectors βƒ‘π‘Ÿ and ⃑𝐹, it must be perpendicular to both vectors. We know that βƒ‘π‘Ÿ and ⃑𝐹 both lie on the π‘₯𝑦-plane, so οƒŸπ‘€ should be perpendicular to the π‘₯𝑦-plane. A vector that is perpendicular to the π‘₯𝑦-plane should be parallel to the unit vector βƒ‘π‘˜ in the 3-dimensional coordinate system. This means βƒ‘π‘ŸΓ—βƒ‘πΉ=π‘βƒ‘π‘˜ for some scalar 𝑐. Since this is always the case, we can simplify the computation of this cross product by using the 2D cross product.

Definition: 2D Cross Product

Given two 2D vectors (π‘Ž,𝑏) and (𝑐,𝑑), the 2D cross product is defined by (π‘Ž,𝑏)Γ—(𝑐,𝑑)=(π‘Žπ‘‘βˆ’π‘π‘)βƒ‘π‘˜.

As we can see, the 2D cross product is quicker to compute. We will use this formula to compute the cross product between 2D vectors for the remainder of this explainer.

Next, let us discuss the magnitude of the moment, which is equal to the magnitude of the cross product: β€–β€–οƒŸπ‘€β€–β€–=β€–β€–βƒ‘π‘ŸΓ—βƒ‘πΉβ€–β€–.

Recall that the cross product between two vectors gives the area of the parallelogram whose two adjacent sides are formed by the two vectors. Let us observe this using the following diagram.

In the diagram above, the area of the highlighted region represents the magnitude of the cross product βƒ‘π‘ŸΓ—βƒ‘πΉ and hence the magnitude of the momentum οƒŸπ‘€. We can also find the area of this parallelogram geometrically by using the geometric formula lengthofthebaseperpendicularheightΓ—.

In the diagram, the base of this parallelogram is formed by vector ⃑𝐹, and the height is the perpendicular distance from the origin to the line of action of ⃑𝐹, which is denoted π‘‘βŸ‚.

This leads to the following formula for the magnitude of the vector moment for a 2D force about a point.

Property: Magnitudes of the Vector Moment of a Force

The magnitude of the vector moment of a planar force ⃑𝐹 about a point is given by β€–β€–οƒŸπ‘€β€–β€–=‖‖⃑𝐹‖‖𝑑,βŸ‚ where π‘‘βŸ‚ is the perpendicular distance between the point and the line of action of force ⃑𝐹.

We can see that the magnitude of the vector moment given above is equal to the magnitude of the scalar moment. Hence, the magnitude of the vector moment is consistent with that of the scalar moment for planar motion.

When we rearrange this equation, we obtain a useful formula for computing the perpendicular distance between a point and a line of action of a force.

Formula: Perpendicular Distance between a Point and a Line of Action

Let οƒŸπ‘€ be the vector moment of a force, or a system of forces, on a plane about a point. Then, the perpendicular distance between the point and the line of action of the force is given by 𝑑=β€–β€–οƒŸπ‘€β€–β€–β€–β€–βƒ‘πΉβ€–β€–.βŸ‚

In the next example, we will compute the moment of a planar force about a point and then use this formula to find the perpendicular distance between the point and the line of action of the force.

Example 2: Finding the Moment Vector of a Force Acting at a Point and the Perpendicular between the Moment and the Line of Action of the Force

Given that force ⃑𝐹=4βƒ‘π‘–βˆ’3⃑𝑗 acts through the point 𝐴(3,6), determine the moment οƒŸπ‘€ about the origin 𝑂 of the force ⃑𝐹. Also, calculate the perpendicular distance 𝐿 between 𝑂 and the line of action of the force.

Answer

In this example, we need to first find the moment οƒŸπ‘€ about 𝑂 of the force ⃑𝐹 and then calculate the perpendicular distance between 𝑂 and the line of action of ⃑𝐹. Let us begin by finding the moment. Recall that the vector moment of force ⃑𝐹 acting at point 𝐴 about the origin 𝑂 is given by οƒŸπ‘€=𝑂𝐴×⃑𝐹.

We are given the coordinates of 𝐴, which means that 𝑂𝐴 is the position vector given by 𝑂𝐴=(3,6).

We can write ⃑𝐹 in component form as ⃑𝐹=4βƒ‘π‘–βˆ’3⃑𝑗=(4,βˆ’3).

Now, we are ready to compute the cross product 𝑂𝐴×⃑𝐹. Recall that the cross product of 2D vectors is defined by (π‘Ž,𝑏)Γ—(𝑐,𝑑)=(π‘Žπ‘‘βˆ’π‘π‘)βƒ‘π‘˜.

Applying this formula, we obtain 𝑂𝐴×⃑𝐹=(3,6)Γ—(4,βˆ’3)=(3Γ—(βˆ’3)βˆ’6Γ—4)βƒ‘π‘˜=βˆ’33βƒ‘π‘˜.

Hence, the moment of ⃑𝐹 about the origin is βˆ’33βƒ‘π‘˜.

Next, let us find the perpendicular distance between the origin and the line of action for ⃑𝐹. Recall that the magnitude of the vector moment of a planar force ⃑𝐹 about a point is given by β€–β€–οƒŸπ‘€β€–β€–=‖‖⃑𝐹‖‖𝐿, where 𝐿 is the perpendicular distance between the point and the line of action for force ⃑𝐹. We can rearrange this equation to write 𝐿=β€–β€–οƒŸπ‘€β€–β€–β€–β€–βƒ‘πΉβ€–β€–.

Since we know οƒŸπ‘€=βˆ’33βƒ‘π‘˜, we can obtain β€–β€–οƒŸπ‘€β€–β€–=33. Let us find ‖‖⃑𝐹‖‖: ‖‖⃑𝐹‖‖=4+(βˆ’3)=√25=5.

Substituting these values into the formula for 𝐿, we obtain 𝐿=335=6.6.

Hence, οƒŸπ‘€=βˆ’33βƒ‘π‘˜,𝐿=6.6.lengthunits

We have noted that the moment of a force about a point results in a vector that is parallel to the unit vector βƒ‘π‘˜. In other words, there is some scalar 𝑐 such that οƒŸπ‘€=π‘βƒ‘π‘˜.

Additionally, we observed that the magnitude of the moment is equal to the magnitude of the scalar moment |𝑀|. This means that either 𝑐=𝑀 or 𝑐=βˆ’π‘€. To determine which one is true, we need to examine whether or not the sign of 𝑐 matches the sign of the scalar moment 𝑀.

The properties of the cross product allow us to conclude first that οƒŸπ‘€ is a vector perpendicular to the plane defined by βƒ‘π‘Ÿ and ⃑𝐹. The direction of οƒŸπ‘€ 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 ⃑𝐴×⃑𝐡 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 ⃑𝐴 to ⃑𝐡, as illustrated in the following diagram.

Remember that we have οƒŸπ‘€=βƒ‘π‘ŸΓ—βƒ‘πΉ=π‘βƒ‘π‘˜.

If 𝑐>0, the moment vector would be coming out of the plane (up), which corresponds to counterclockwise rotation according to the figure above. If 𝑐<0, the moment vector would go into the plane (down), which indicates clockwise rotation. Recall that for a scalar moment 𝑀, the counterclockwise orientation corresponds to the positive sign, while the clockwise rotation leads to the negative sign. This tells us that the sign of the scalar momentum 𝑀 agrees with the sign of the scalar 𝑐. Hence, we have shown that 𝑐=𝑀.

Property: 2D Vector Moment of a Force

Let 𝑀 and οƒŸπ‘€ be the scalar and vector moments of a force, or a system of forces, on a plane about a point. Then, οƒŸπ‘€=π‘€βƒ‘π‘˜.

This property firmly establishes why this vector moment is a reasonable extension of the scalar moment for a planar force. Furthermore, the vector moment can be generalized to represent a moment of a general 3D force about a point since it is obtained using the cross product.

We can deduce several useful observations from this property. Firstly, we know that the scalar moment does not depend on the location of the point the force is acting on, as long as the point lies on the same line of action of the force. This is because the scalar moment is obtained by only using the magnitude of the force ‖‖⃑𝐹‖‖ and the perpendicular distance π‘‘βŸ‚. This means that the vector moment is also independent of the location of the point at which the force is acting. We can understand this better when we compare the magnitude of the momentum when we move this point along the line of action.

We can see that the areas of both parallelograms are equal since the length of the base ‖‖⃑𝐹‖‖ and the height π‘‘βŸ‚ are the same for both parallelograms. This tells us that the magnitude of the momentum for these two systems is the same. Furthermore, we can see that both system would cause the clockwise rotation about the origin, meaning that the sign of the moment would be the same for both systems. Hence, the vector moment is the same for these two systems. This leads to the following useful property.

Property: Vector Moment of a Force

The vector moment οƒŸπ‘€ of a force about a point is independent of the point at which the force acts, as long as the point lies in the same line of action.

In the next example, we will find the vector moment of a planar force about a point when the initial point 𝐴 is not given.

Example 3: Finding the Moment of a Force Vector Acting at a Point

End 𝐴 of 𝐴𝐡 is at (βˆ’6,7) and 𝐴𝐡 has midpoint 𝐷(βˆ’7,1). If the line of action of the force ⃑𝐹=βˆ’2βƒ‘π‘–βˆ’6⃑𝑗 bisects 𝐴𝐡, determine the moment of ⃑𝐹 about point 𝐡.

Answer

In this example, we need to find the moment of a planar force about a point. Recall that the vector moment of force ⃑𝐹 acting at point 𝑃 about point 𝑂 is given by οƒŸπ‘€=οƒŸπ‘‚π‘ƒΓ—βƒ‘πΉ.

While we are not given the point at which the force is acting, we are given that the line of action of force ⃑𝐹 bisects 𝐴𝐡. This means that the line of action passes through the midpoint 𝐷 of 𝐴𝐡. Recall that the vector moment οƒŸπ‘€ of a force about a point is independent of the initial point, as long as the point lies in the same line of action. Hence, we can compute the moment by considering the initial point to be at 𝐷(βˆ’7,1). This means that the moment of ⃑𝐹 about 𝐡 is given by οƒŸπ‘€=𝐡𝐷×⃑𝐹.

Let us begin by finding vector 𝐡𝐷. Since 𝐷 is the midpoint of 𝐴, we know that ‖‖𝐴𝐷‖‖=‖‖𝐡𝐷‖‖.

Also, these vectors have the opposite direction, which means 𝐡𝐷=βˆ’οƒ π΄π·.

We can find 𝐴𝐷 by using the coordinates of points 𝐴 and 𝐷: 𝐴𝐷=(βˆ’7,1)βˆ’(βˆ’6,7)=(βˆ’1,βˆ’6).

Hence, 𝐡𝐷=βˆ’(βˆ’1,βˆ’6)=(1,6).

Now, we are ready to compute the cross product 𝐡𝐷×⃑𝐹. Recall that the cross product of 2D vectors is defined by (π‘Ž,𝑏)Γ—(𝑐,𝑑)=(π‘Žπ‘‘βˆ’π‘π‘)βƒ‘π‘˜.

Applying this formula, we obtain 𝐡𝐷×⃑𝐹=(1,6)Γ—(βˆ’2,βˆ’6)=(1Γ—(βˆ’6)βˆ’6Γ—(βˆ’2))βƒ‘π‘˜=6βƒ‘π‘˜.

Hence, the moment of ⃑𝐹 about point 𝐡 is 6βƒ‘π‘˜.

In the next example, we will find the moment of a system of planar forces acting at a single point about another point by first finding the resultant of the forces.

Example 4: Calculating the Moment of Three Forces Acting on a Single Point about a Given Point and the Distance between the Points

Given that ⃑𝐹=βˆ’2⃑𝑖+2βƒ‘π‘—οŠ§, ⃑𝐹=βˆ’3βƒ‘π‘–βˆ’βƒ‘π‘—οŠ¨, and ⃑𝐹=βƒ‘π‘–βˆ’4βƒ‘π‘—οŠ© are acting at the point 𝐴(2,3), determine the moment βƒ‘π‘š of the resultant of the forces about the point 𝐡(βˆ’2,βˆ’1), and calculate the length of the perpendicular line 𝐿 joining the point 𝐡 to the resultant’s line of action.

Answer

In this example, we are given a system of planar forces acting at the same point. Let us begin by finding the resultant of the forces. Recall that the resultant of a system of forces acting at the same point is the sum of all force vectors in the system. Hence, the resultant ⃑𝐹 is given by ⃑𝐹=⃑𝐹+⃑𝐹+⃑𝐹=ο€Ίβˆ’2⃑𝑖+2⃑𝑗+ο€Ίβˆ’3βƒ‘π‘–βˆ’βƒ‘π‘—ο†+ο€Ίβƒ‘π‘–βˆ’4⃑𝑗=ο€Ίβˆ’2βƒ‘π‘–βˆ’3⃑𝑖+⃑𝑖+ο€Ί2βƒ‘π‘—βˆ’βƒ‘π‘—βˆ’4⃑𝑗=βˆ’4βƒ‘π‘–βˆ’3⃑𝑗.

This tells us that the resultant of the forces is ⃑𝐹=βˆ’4βƒ‘π‘–βˆ’3⃑𝑗. Next, let us find the moment βƒ‘π‘š of the resultant about point 𝐡(βˆ’2,βˆ’1). Recall that the vector moment of force ⃑𝐹 acting at point 𝐴 about point 𝐡 is given by οƒŸπ‘€=𝐡𝐴×⃑𝐹.

Using the coordinates of 𝐴 and 𝐡, we can find 𝐡𝐴=(2,3)βˆ’(βˆ’2,βˆ’1)=(4,4).

Now, we are ready to compute the cross product 𝐡𝐴×⃑𝐹. Recall that the cross product of 2D vectors is defined by (π‘Ž,𝑏)Γ—(𝑐,𝑑)=(π‘Žπ‘‘βˆ’π‘π‘)βƒ‘π‘˜.

This leads to 𝐡𝐴×⃑𝐹=(4,4)Γ—(βˆ’4,βˆ’3)=(4Γ—(βˆ’3)βˆ’4Γ—(βˆ’4))βƒ‘π‘˜=4βƒ‘π‘˜.

Hence, the moment of the resultant of forces about point 𝐡 is 4βƒ‘π‘˜.

Next, let us find the length of the perpendicular line 𝐿 joining point 𝐡 to the resultant’s line of action. This length 𝐿 is also known as the perpendicular distance between point 𝐡 and the resultant’s line of action. To compute this length, we recall that the magnitude of the vector moment of a planar force ⃑𝐹 about a point is given by β€–β€–οƒŸπ‘€β€–β€–=‖‖⃑𝐹‖‖𝐿, where 𝐿 is the perpendicular distance between the point and the line of action for ⃑𝐹. We can rearrange this equation to write 𝐿=β€–β€–οƒŸπ‘€β€–β€–β€–β€–βƒ‘πΉβ€–β€–.

Since we know οƒŸπ‘€=4βƒ‘π‘˜, we can obtain β€–β€–οƒŸπ‘€β€–β€–=4. Let us find ‖‖⃑𝐹‖‖: ‖‖⃑𝐹‖‖=(βˆ’4)+(βˆ’3)=√25=5.

Substituting these values into the formula for 𝐿, we obtain 𝐿=45=0.8.

Hence, οƒŸπ‘€=4βƒ‘π‘˜,𝐿=0.8.lengthunits

In the previous example, we found the moment of a system of planar forces acting at the same point about another point. We can note that the process of finding the moment for the system of forces is the same as that for one force, if the forces are acting at the same point.

Let us now consider the problem of finding the moment of a system of planar forces where the forces are not acting at the same point.

Definition: The Moment of a System of Planar Forces

Consider the system of forces βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, …, and βƒ‘πΉοŠ acting at 𝐴, 𝐴, …, and 𝐴 respectively. To find the moment of this system of forces about point 𝑂, we need to find the moments οƒŸπ‘€οŠ§, οƒŸπ‘€οŠ¨, …, and οƒŸπ‘€οŠ of forces βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, …, and βƒ‘πΉοŠ about point 𝑂. Then, the moment of the system οƒŸπ‘€ about point 𝑂 is given by οƒŸπ‘€=οƒŸπ‘€+οƒŸπ‘€+β‹―+οƒŸπ‘€.

This definition tells us that the moment of a system of forces is equal to the sum of individual moments of each force in the system about the same point.

In our final example, we will find the unknown constants in forces in a system acting at different points when we are given the moment of the system of forces about two different points.

Example 5: Finding Unknown Components of Two Forces given the Sum of Their Moments about Two Points

⃑𝐹=π‘šβƒ‘π‘–+βƒ‘π‘—οŠ§ and ⃑𝐹=π‘›βƒ‘π‘–βˆ’5βƒ‘π‘—οŠ¨, where βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ are two forces acting at the points 𝐴(3,1) and 𝐡(βˆ’1,βˆ’1) respectively. The sum of moments about the point of origin equals zero. The sum of the moments about the point 𝐢(1,2) also equals zero. Determine the values of π‘š and 𝑛.

Answer

In this example, we need to find the unknown constants π‘š and 𝑛 in the forces βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ when we are given that the sum of the moments of the two forces about the origin and also about the point 𝐢 is zero. We can find the unknown constants by identifying a pair of simultaneous equations involving π‘š and 𝑛. We will obtain the first equation by computing the sum of moments of βƒ‘πΉοŠ§ and βƒ‘πΉοŠ¨ about the origin and setting them equal to zero.

Recall that the vector moment of force ⃑𝐹 acting at point 𝑃 about point 𝑄 is given by οƒŸπ‘€=βƒ‘π‘ŸΓ—βƒ‘πΉ, where βƒ‘π‘Ÿ is the vector from point 𝑄 to point 𝑃. Let us first find the moment of βƒ‘πΉοŠ§ about the origin. Since βƒ‘πΉοŠ§ acts at point 𝐴, we can write βƒ‘π‘Ÿ=οƒŸπ‘‚π‘ƒ=(3,1).

We can write βƒ‘πΉοŠ§ in component form as ⃑𝐹=π‘šβƒ‘π‘–+⃑𝑗=(π‘š,1).

Now, we are ready to compute the cross product οƒŸπ‘‚π‘ƒΓ—βƒ‘πΉ. Recall that the cross product of 2D vectors is defined by (π‘Ž,𝑏)Γ—(𝑐,𝑑)=(π‘Žπ‘‘βˆ’π‘π‘)βƒ‘π‘˜.

This leads to βƒ‘π‘ŸΓ—βƒ‘πΉ=(3,1)Γ—(π‘š,1)=(3Γ—1βˆ’1Γ—π‘š)βƒ‘π‘˜=(3βˆ’π‘š)βƒ‘π‘˜.

Next, let us find the moment of βƒ‘πΉοŠ¨ about the origin. Since βƒ‘πΉοŠ¨ acts at point 𝐡, we can write βƒ‘π‘Ÿ=οƒŸπ‘‚π΅=(βˆ’1,βˆ’1).

We can write βƒ‘πΉοŠ¨ in component form as ⃑𝐹=π‘›βƒ‘π‘–βˆ’5⃑𝑗=(𝑛,βˆ’5).

Taking the cross product, βƒ‘π‘ŸΓ—βƒ‘πΉ=(βˆ’1,βˆ’1)Γ—(𝑛,βˆ’5)=(βˆ’1Γ—(βˆ’5)βˆ’(βˆ’1)×𝑛)βƒ‘π‘˜=(5+𝑛)βƒ‘π‘˜.

Then, the sum of these two moments about the origin is (3βˆ’π‘š)βƒ‘π‘˜+(5+𝑛)βƒ‘π‘˜=(8βˆ’π‘š+𝑛)βƒ‘π‘˜.

Since we are given that the sum of these moments should equal zero, we obtain

8βˆ’π‘š+𝑛=0.(1)

This gives us one equation involving π‘š and 𝑛. We can repeat this computation for the moment about point 𝐢 to obtain another equation, but we can also find the second equation by using properties of the moments. Let us find the moment of βƒ‘πΉοŠ§ about point 𝐢: βƒ‘π‘Ÿ=𝐢𝐴=(3,1)βˆ’(1,2)=(2,βˆ’1).

Taking the cross product, βƒ‘π‘ŸΓ—βƒ‘πΉ=(2,βˆ’1)Γ—(π‘š,1)=(2Γ—1βˆ’(βˆ’1)Γ—π‘š)βƒ‘π‘˜=(2+π‘š)βƒ‘π‘˜.

Next, for the moment of βƒ‘πΉοŠ¨ about 𝐢, βƒ‘π‘Ÿ=οƒŸπΆπ΅=(βˆ’1,βˆ’1)βˆ’(1,2)=(βˆ’2,βˆ’3).

Taking the cross product, βƒ‘π‘ŸΓ—βƒ‘πΉ=(βˆ’2,βˆ’3)Γ—(𝑛,βˆ’5)=(βˆ’2Γ—(βˆ’5)βˆ’(βˆ’3)×𝑛)βƒ‘π‘˜=(10+3𝑛)βƒ‘π‘˜.

Summing these two moments about 𝐢, (2+π‘š)βƒ‘π‘˜+(10+3𝑛)βƒ‘π‘˜=(12+π‘š+3𝑛)βƒ‘π‘˜.

Since we are given that the sum of these moments should equal zero, we obtain

12+π‘š+3𝑛=0.(2)

Now that we have obtained two equations for π‘š and 𝑛, let us write equations (1) and (2) here: 8βˆ’π‘š+𝑛=0,12+π‘š+3𝑛=0.

We can sum the two equations to eliminate π‘š. This leads to 20+4𝑛=0.

Rearranging this equation so 𝑛 is the subject gives us 𝑛=βˆ’5. We can substitute this value into equation (1) to write 8βˆ’π‘šβˆ’5=0.

Rearranging this equation so π‘š is the subject leads to π‘š=3. Hence, we have π‘š=3,𝑛=βˆ’5.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • The vector moment of force ⃑𝐹 acting at point 𝐴 about point 𝑂 is given by οƒŸπ‘€=βƒ‘π‘ŸΓ—βƒ‘πΉ, where βƒ‘π‘Ÿ is the vector from point 𝑂 to point 𝐴.
  • The magnitude of the vector moment of a planar force ⃑𝐹 about a point is given by β€–β€–οƒŸπ‘€β€–β€–=‖‖⃑𝐹‖‖𝑑,βŸ‚ where π‘‘βŸ‚ is the perpendicular distance between the point and the line of action of force ⃑𝐹.
  • The vector moment οƒŸπ‘€ of a force about a point is independent of the initial point, as long as the point lies in the same line of action.
  • Let 𝑀 and οƒŸπ‘€ be the scalar and vector moments of a force, or a system of forces, on a plane about a point. Then, οƒŸπ‘€=π‘€βƒ‘π‘˜.
  • The computation of the cross product βƒ‘π‘ŸΓ—βƒ‘πΉ to compute the moment οƒŸπ‘€ of a planar force about a point can be simplified by using the 2D cross product, which is defined by (π‘Ž,𝑏)Γ—(𝑐,𝑑)=(π‘Žπ‘‘βˆ’π‘π‘)βƒ‘π‘˜.
  • Consider the system of forces βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, …, and βƒ‘πΉοŠ acting at 𝐴, 𝐴, …, and 𝐴 respectively. To find the moment of this system of forces about point 𝑂, we need to find the moments οƒŸπ‘€οŠ§, οƒŸπ‘€οŠ¨, …, and οƒŸπ‘€οŠ of forces βƒ‘πΉοŠ§, βƒ‘πΉοŠ¨, …, and βƒ‘πΉοŠ about point 𝑂. Then, the moment of system οƒŸπ‘€ about point 𝑂 is given by οƒŸπ‘€=οƒŸπ‘€+οƒŸπ‘€+β‹―+οƒŸπ‘€.

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