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.
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 βπΉο§ and βπΉο¨ acting on a rod, perpendicularly to the length of the rod. Let πΉ be the magnitude of both forces. We have βπΉ=ββπΉ,ο§ο¨ and βββπΉββ=βββπΉββ=πΉ.ο§ο¨
Let us find the moments of forces βπΉο§ and βπΉο¨ about point π΄, the midpoint between the points of application of βπΉο§ and βπΉο¨. Recall that the moment of a force about a point is given by π=Β±πΉπ,β where πΉ is the magnitude of the force and πβ 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 π2 from π΄. The clockwise moment about π΄ of βπΉο§ is therefore given by π=βπΉπ2,ο§ and the moment about π΄ of βπΉο¨ is π=βπΉπ2.ο¨
The net moment of the couple is given by π=π+ππ=β2πΉπ2=βπΉπ.netnetο§ο¨
Let us now find the net moment of βπΉο§ and βπΉο¨ about π΅. Let π be the length of the rod. The moment about π΅ of βπΉο§ is negative because βπΉο§ produces a clockwise (negative) rotation about π΅. Hence, the moment is given by π=βπβ πΉ.ο§
By contrast, the moment of βπΉο¨ about π΅ is positive since βπΉο¨ produces a counterclockwise (positive) rotation about π΅. We find that π=(πβπ)β πΉ.ο¨
The net moment about π΅ of the couple is given by π=π+ππ=βππΉ+(πβπ)πΉπ=πΉ(βπ+πβπ)π=βπΉπ.netnetnetnetο§ο¨
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.
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 πΏ 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 ππsin. In the diagram on the left, where π is greater than 90β, we see that πΏ=ππβ²=π(180βπ)=ππsinsinsinβ, since sinsin(180βπ₯)=π₯β. Therefore, we see that the arm of the couple, the perpendicular distance between the forces, is always given by πΏ=ππ.sin
The moment of a couple is thus π=Β±πΉπΏ=Β±πΉππ.sin
We can also interpret πΉππsin as (πΉπ)πsin, 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.
The moment of a couple acting at π΄ and π΅ is given by π=Β±πΉπππ=Β±πΉπΏπ=Β±|πΉ|β π,sinβ where πΉ is the magnitude of both forces in the couple, π is the distance between π΄ and π΅, π is the angle between either force and the segment π΄π΅, πΏ is the arm of the couple, and πΉβ is the component of the force perpendicular to π΄π΅.
Let us look at an example about the moment of a couple.
If the norm of the moment of a couple is 750 Nβ 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 πΏ, the length of the moment arm. As the moment is given in newton-metres and the force is given in newtons, 750=50πΏπΏ=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.
The magnitude of the moment of a couple is given by |π|=πΉπΏ, where πΉ is the magnitude of either of the forces in the couple and πΏ 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.
In the figure below, πΉ=3ο§N and πΉο§ and πΉο¨ form a couple. Find the algebraic measure of the moment of that couple.
Answer
The forces in a couple must have equal magnitudes. If βπΉο§ has a magnitude of 3 N, then βπΉο¨ also has a magnitude of 3 N. The angle between βπΉο§ and βπΉο¨ and the line connecting the points that βπΉο§ and βπΉο¨ act from is 45β.
The rotation due to the couple is counterclockwise and, hence, positive. The magnitude of the moment is given by π=πΉπβ πsin to be π=3Γ45Γ7β2π=3Γβ22ο»7β2οπ=21β .sinNcm
The proper mathematical definition of the moment of a force is given by the cross product.
The moment of a force βπΉ about a point can be found using the cross product. The vector βπ 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. οπ=βπΓβπΉ
For a couple formed of two forces; βπΉο§ acting at point π΄ and βπΉο¨ acting at point π΅, the moment of the couple is given by, οπ=ο π΅π΄ΓβπΉ=ο π΄π΅ΓβπΉο§ο¨
Note that for the above equation for the moment of a couple, βπ has been substituted by the vector between the points of application of the forces. We can think of this as taking moments about point π΅ in the first case, and taking moments about point π΄ in the second case.
A useful way to evaluate the cross product, is by considering the determinant of 3Γ3 matrix. οπ=βπΓβπΉ=|||||βπβπβπππππΉπΉπΉ|||||οοοοοο
Although this method is mainly used when the vectors βπΉ and βπ exist in 3 dimensions, it can sometimes be useful for 2 dimensional systems. Let us look at one such example.
Given that two forces βπΉ=ββπ+2βπο§ and βπΉο¨ are acting at two points (2,2) and (β2,β2) 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 βπΉο¨, however it is not required to reach a solution. βπΉ+βπΉ=0βπΉ=ββπΉβπΉ=βπβ2βπο§ο¨ο¨ο§ο¨
The perpendicular distance between the lines of action of βπΉο§ and βπΉο¨ is the length of the line that is perpendicular to both. We can define this distance as π.
The moment of a couple can be found using the following formula where πΉ is the magnitude of either of the forces, π is the perpendicular distance between the lines of action of the forces, and the sign of the moment denotes the direction of rotation. π=Β±πΉπ
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 itβs perpendicular distance from the point about which a moment is being taken.
Let us imagine taking moments about the point π΅. Since force βπΉο§ acts at point π΄, we define the vector βπ as: βπ=ο π΅π΄.
By defining π as the positive acute angle between the line of action of βπΉο§ and βπ we can perform a useful simplification using right triangle geometry. Here we have expressed the perpendicular distance π in terms of the magnitude of βπ and the angle π. π=βββπββπsin
Note that this particular system, we are finding the distance π, 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 π 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: ββοπββ=βββπΉβββββπββπ=βββπΉββπ.ο§ο§sin
Rearranging this equation illustrates that if we are able to find the magnitudes of the moment οπ and the force βπΉο§ we will be able to find the perpendicular distance π. π=ββοπβββββπΉββο§
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 wonβt go into any detail here. οπ=βπΓβπΉ=ο»βββπΉβββββπββποβπο§sin
Recall that we are essentially taking a moment about point π΅. This means we will be considering the force βπΉο§ and the vector βπ defined from point π΅ to point π΄. βπ=ο π΅π΄=(2,2)β(β2,β2)=(4,4)
The standard method for the cross product in 3 dimensions involves finding the determinant of a 3Γ3 matrix, however since βπΉο§ and βπ are both 2 dimensional vectors, our calculations can be simplified. οπ=βπΓβπΉ=οΉπ,π,0ο ΓοΉπΉ,πΉ,0ο =οΉππΉβππΉο βπ=(4β 2β4β (β1))βπ=(8+4)βπ=12βπο§οοοοοοοο
We now find the magnitude of οπ and βπΉο§. ββοπββ=ββ12βπββ=12βββπΉββ=ββββπ+2βπββ=ο(β1)+2=β5ο§ο¨ο¨
And finally we have all of the components needed to find the distance π. π=ββοπβββββπΉββ=12β5=12β55ο§
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.
π΄π΅ is a horizontal light rod having a length of 60 cm, where two forces, each of magnitude 45 N, are acting vertically at π΄ and π΅ in two opposite directions. Two other forces, each of magnitude 120 N, are acting in two opposite directions at points πΆ and π· of the rod, where πΆπ·=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 πο§ formed by the forces that act at π΄ and π΅ is given by π=β(45Γ60)β .ο§Ncm
The couple πο¨ formed by the forces that act at πΆ and π· is given by π=β(120πΓ45)β .ο¨sinNcm
The question states that the couples are equivalent, so π=π.ο§ο¨
The angle π can be found by rearrangement: (45Γ60)=(120πΓ45)60120=ππ=30.sinsinβ
- 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 π=Β±πΉππsin, where πΉ is the magnitude of either of the forces in the couple, π is the length of the line connecting the points that the forces act from, and π is the angle between βπΉ and this line.
- The magnitude of the moment of a couple is given by |π|=πΉπΏ, where πΉ is the magnitude of either of the forces and πΏ is the arm of the couple.
- The moment of a couple can be found using the cross product. οπ=βπΓβπΉ
- For a couple formed of two forces; βπΉο§ acting at point π΄ and βπΉο¨ acting at point π΅: οπ=ο π΅π΄ΓβπΉ=ο π΄π΅ΓβπΉο§ο¨
- 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.