Lesson Explainer: Moment of a Couple | Nagwa Lesson Explainer: Moment of a Couple | Nagwa

Lesson Explainer: Moment of a Couple Mathematics

In this explainer, we will learn how to calculate the moment of a couple of two forces and the resultant of two or more couples.

Let us first define a couple in mechanics.

Definition: Couple

A couple is a pair of forces, acting on the same body, that have parallel but noncoincident lines of action and that act in opposite directions and have equal magnitudes.

Although the sum of the forces is zero, there is a nonzero net moment (i.e., the sum of the moments of both forces) because the forces do not have the same line of action. Let us work out the net moment.

Consider the couple of forces 𝐹 and 𝐹 acting on a rod, perpendicularly to the length of the rod. Let 𝐹 be the magnitude of both forces. We have 𝐹=𝐹, and 𝐹=𝐹=𝐹.

Let us find the moments of forces 𝐹 and 𝐹 about point 𝐴, the midpoint between the points of application of 𝐹 and 𝐹. Recall that the moment of a force about a point is given by 𝑀=±𝐹𝑑, where 𝐹 is the magnitude of the force and 𝑑 is the perpendicular distance between the force and the point about which the moment is taken. If the force produces a clockwise rotation, the moment is negative. The moment is positive when the force produces a counterclockwise rotation.

Both forces produce clockwise (negative) rotation, and their points of application are both at a perpendicular distance of 𝑑2 from 𝐴. The clockwise moment about 𝐴 of 𝐹 is therefore given by 𝑀=𝐹𝑑2, and the moment about 𝐴 of 𝐹 is 𝑀=𝐹𝑑2.

The net moment of the couple is given by 𝑀=𝑀+𝑀𝑀=2𝐹𝑑2=𝐹𝑑.netnet

Let us now find the net moment of 𝐹 and 𝐹 about 𝐵. Let 𝑙 be the length of the rod. The moment about 𝐵 of 𝐹 is negative because 𝐹 produces a clockwise (negative) rotation about 𝐵. Hence, the moment is given by 𝑀=𝑙𝐹.

By contrast, the moment of 𝐹 about 𝐵 is positive since 𝐹 produces a counterclockwise (positive) rotation about 𝐵. We find that 𝑀=(𝑙𝑑)𝐹.

The net moment about 𝐵 of the couple is given by 𝑀=𝑀+𝑀𝑀=𝑙𝐹+(𝑙𝑑)𝐹𝑀=𝐹(𝑙+𝑙𝑑)𝑀=𝐹𝑑.netnetnetnet

We observe that the moment of the couple has the same value in both cases. It is general: the moment of a couple is the same about any point on the body that the couple acts on. It is worth noting the difference between points A and B. Point A is between the two points of application of the forces, and the forces produce a rotation in the same sense. Point B is outside the region between the two points of application of both forces of the couple, and the moments of the two forces in the couple have opposite signs as they produce rotation in opposite senses.

Property: Moment of a Couple

The moment of a couple is independent of the point about which moments of the couple are taken.

The forces in a couple do not necessarily act perpendicularly to the line connecting the points that they act from. The following figure shows three examples of couples where the forces in a couple do not act perpendicularly to the line connecting the points that they act from.

In this case, the perpendicular distance, also called the arm of the couple, denoted by 𝐿 in the above figure, is not the distance between the two points of application of the forces. We see that for the two diagrams on the right, the arm of the couple is given by 𝑑𝜃sin. In the diagram on the left, where 𝜃 is greater than 90, we see that 𝐿=𝑑𝜃=𝑑(180𝜃)=𝑑𝜃sinsinsin, since sinsin(180𝑥)=𝑥. Therefore, we see that the arm of the couple, the perpendicular distance between the forces, is always given by 𝐿=𝑑𝜃.sin

The moment of a couple is thus 𝑀=±𝐹𝐿=±𝐹𝑑𝜃.sin

We can also interpret 𝐹𝑑𝜃sin as (𝐹𝜃)𝑑sin, that is, the product of the absolute value of the component of the force perpendicular to the rod and the distance between the two points of application of the forces.

Property: The Moment of a Couple

The moment of a couple acting at 𝐴 and 𝐵 is given by 𝑀=±𝐹𝑑𝜃𝑀=±𝐹𝐿𝑀=±|𝐹|𝑑,sin where 𝐹 is the magnitude of both forces in the couple, 𝑑 is the distance between 𝐴 and 𝐵, 𝜃 is the angle between either force and the segment 𝐴𝐵, 𝐿 is the arm of the couple, and 𝐹 is the component of the force perpendicular to 𝐴𝐵.

Let us look at an example about the moment of a couple.

Example 1: Determining the Arm of the Couple Equivalent to a System of Two Forces

If the norm of the moment of a couple is 750 N⋅m, and the magnitude of one of its two forces is 50 N, determine the length of the moment arm.

Answer

The norm of the moment is the magnitude of the moment. The magnitude of the moment of the couple is the product of the magnitude of either of the forces in the couple and 𝐿, the length of the moment arm. As the moment is given in newton-metres and the force is given in newtons, 750=50𝐿𝐿=75050=15.m

With the previous example, we see that the magnitude of the moment is simply given by the product of the magnitude of one of the forces and the arm of the couple.

Property: Magnitude of the Moment of a Couple

The magnitude of the moment of a couple is given by |𝑀|=𝐹𝐿, where 𝐹 is the magnitude of either of the forces in the couple and 𝐿 is the arm of the couple.

Let us look at an example involving the moment of a couple where the forces are not perpendicular to the line connecting their points of application.

Example 2: Finding the Magnitude of a Couple of Two Inclined Forces Acting on Two Ends of a Line

In the figure below, 𝐹=3N and 𝐹 and 𝐹 form a couple. Find the algebraic measure of the moment of that couple.

Answer

The forces in a couple must have equal magnitudes. If 𝐹 has a magnitude of 3 N, then 𝐹 also has a magnitude of 3 N. The angle between 𝐹 and 𝐹 and the line connecting the points that 𝐹 and 𝐹 act from is 45.

The rotation due to the couple is counterclockwise and, hence, positive. The magnitude of the moment is given by 𝑀=𝐹𝜃𝑑sin to be 𝑀=3×45×72𝑀=3×2272𝑀=21.sinNcm

The proper mathematical definition of the moment of a force is given by the cross product.

Definition: Moment of a Couple using the Cross Product

The moment of a force 𝐹 about a point can be found using the cross product. The vector 𝑟 represents the position vector from the point about which a moment is being taken to any point on the line of action of the force. 𝑀=𝑟×𝐹

For a couple formed of two forces; 𝐹 acting at point 𝐴 and 𝐹 acting at point 𝐵, the moment of the couple is given by, 𝑀=𝐵𝐴×𝐹=𝐴𝐵×𝐹

Note that for the above equation for the moment of a couple, 𝑟 has been substituted by the vector between the points of application of the forces. We can think of this as taking moments about point 𝐵 in the first case, and taking moments about point 𝐴 in the second case.

A useful way to evaluate the cross product, is by considering the determinant of 3×3 matrix. 𝑀=𝑟×𝐹=|||||𝑖𝑗𝑘𝑟𝑟𝑟𝐹𝐹𝐹|||||

Although this method is mainly used when the vectors 𝐹 and 𝑟 exist in 3 dimensions, it can sometimes be useful for 2 dimensional systems. Let us look at one such example.

Example 3: Finding the Perpendicular Distance between Two Force Vectors Forming a Couple

Given that two forces 𝐹=𝑖+2𝑗 and 𝐹 are acting at two points (2,2) and (2,2) respectively to form a couple, find the perpendicular distance between the two forces.

Answer

Because the forces form a couple, they must sum to 0. For completeness we can use this to find 𝐹, however it is not required to reach a solution. 𝐹+𝐹=0𝐹=𝐹𝐹=𝑖2𝑗

The perpendicular distance between the lines of action of 𝐹 and 𝐹 is the length of the line that is perpendicular to both. We can define this distance as 𝑑.

The moment of a couple can be found using the following formula where 𝐹 is the magnitude of either of the forces, 𝑑 is the perpendicular distance between the lines of action of the forces, and the sign of the moment denotes the direction of rotation. 𝑀=±𝐹𝑑

To gain a better understanding of our system we can switch to vector notation and remind ourselves of where this formula comes from. The moment of a force is equal to the magnitude of the force multiplied by it’s perpendicular distance from the point about which a moment is being taken.

Let us imagine taking moments about the point 𝐵. Since force 𝐹 acts at point 𝐴, we define the vector 𝑟 as: 𝑟=𝐵𝐴.

By defining 𝜃 as the positive acute angle between the line of action of 𝐹 and 𝑟 we can perform a useful simplification using right triangle geometry. Here we have expressed the perpendicular distance 𝑑 in terms of the magnitude of 𝑟 and the angle 𝜃. 𝑑=𝑟𝜃sin

Note that this particular system, we are finding the distance 𝑑, which is a non-negative scalar. This means we are not concerned with the sign in our equation, and hence the direction of rotation can be ignored.

It is for this reason that we are able to define 𝜃 as the positive acute angle, in essence, discarding any negative solutions. In doing so we have simplified our system to consider only the magnitude of the moment, which can be expressed as: 𝑀=𝐹𝑟𝜃=𝐹𝑑.sin

Rearranging this equation illustrates that if we are able to find the magnitudes of the moment 𝑀 and the force 𝐹 we will be able to find the perpendicular distance 𝑑. 𝑑=𝑀𝐹

To proceed we recall that the moment of a couple can be found using the cross product. As an added bonus, if we recall the definition of the cross product, we confirm our previous logic, although we won’t go into any detail here. 𝑀=𝑟×𝐹=𝐹𝑟𝜃𝑛sin

Recall that we are essentially taking a moment about point 𝐵. This means we will be considering the force 𝐹 and the vector 𝑟 defined from point 𝐵 to point 𝐴. 𝑟=𝐵𝐴=(2,2)(2,2)=(4,4)

The standard method for the cross product in 3 dimensions involves finding the determinant of a 3×3 matrix, however since 𝐹 and 𝑟 are both 2 dimensional vectors, our calculations can be simplified. 𝑀=𝑟×𝐹=𝑟,𝑟,0×𝐹,𝐹,0=𝑟𝐹𝑟𝐹𝑘=(424(1))𝑘=(8+4)𝑘=12𝑘

We now find the magnitude of 𝑀 and 𝐹. 𝑀=12𝑘=12𝐹=𝑖+2𝑗=(1)+2=5

And finally we have all of the components needed to find the distance 𝑑. 𝑑=𝑀𝐹=125=1255

This is the perpendicular distance between the lines of action of the forces in the couple.

Multiple couples can act on a body simultaneously. When multiple couples act on a body, the resultant moment due to the couples is the sum of the moments due to the couples. Let us look at an example involving multiple couples.

Example 4: Analysis of a System of Four Forces Acting on a Horizontal Rod Equivalent to a Couple

𝐴𝐵 is a horizontal light rod having a length of 60 cm, where two forces, each of magnitude 45 N, are acting vertically at 𝐴 and 𝐵 in two opposite directions. Two other forces, each of magnitude 120 N, are acting in two opposite directions at points 𝐶 and 𝐷 of the rod, where 𝐶𝐷=45cm. If they form a couple equivalent to the couple formed by the first two forces, find the measure of the angle of inclination that the second two forces make with the rod.

Answer

The couple 𝑐 formed by the forces that act at 𝐴 and 𝐵 is given by 𝑐=(45×60).Ncm

The couple 𝑐 formed by the forces that act at 𝐶 and 𝐷 is given by 𝑐=(120𝜃×45).sinNcm

The question states that the couples are equivalent, so 𝑐=𝑐.

The angle 𝜃 can be found by rearrangement: (45×60)=(120𝜃×45)60120=𝜃𝜃=30.sinsin

Key Points

  • A couple is a pair of forces that have parallel and distinct lines of action and equal magnitudes but opposite directions.
  • The moment due to a couple is given by 𝑀=±𝐹𝑑𝜃sin, where 𝐹 is the magnitude of either of the forces in the couple, 𝑑 is the length of the line connecting the points that the forces act from, and 𝜃 is the angle between 𝐹 and this line.
  • The magnitude of the moment of a couple is given by |𝑀|=𝐹𝐿, where 𝐹 is the magnitude of either of the forces and 𝐿 is the arm of the couple.
  • The moment of a couple can be found using the cross product. 𝑀=𝑟×𝐹
  • For a couple formed of two forces; 𝐹 acting at point 𝐴 and 𝐹 acting at point 𝐵: 𝑀=𝐵𝐴×𝐹=𝐴𝐵×𝐹
  • Multiple couples can act on a body simultaneously. When multiple couples act on a body, the resultant moment due to the couples is the sum of the moments due to the couples.

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