Moment of a couple
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Hi,
It is unusual that the problem states the condition "if the group is equivalent to a couple...", since we can add the forces F₁, F₂, and F₃, as given, and readily see that the resultant is 0, and thus the only possible result is a couple resultant. In addition, every system of forces is equivalent to a force-couple system, even if the force, or the couple, or both, are 0. That is, the condition was not an additional assumption that it was necessary to state in order to start working on the problem.
In either case, in order to resolve the resultant couple of a planar system of forces, it can be helpful to first find the point of intersection of the lines of action of 2 or more forces. Computing the moment of each force about such a point would be simpler than choosing any other point, due to the fact that the moment of a force about a point on its line of action is 0. By choosing a point on the line of action of at least two forces, we maximize that simplification.
The equations for the lines of action are readily found by using the Point-Slope form of the equation of a line.
For example, consider the line of action of force F₁. It will necessarily have a slope of ⅓, since that the rise of the vector is 1 and the run is 3. It passes through the point of action (1, 2), so an equation of the line is y - 2 = ⅓(x - 1).
Similarly, the other two lines of action are y + 3 = -³⁄₂(x - 2) and y + 1 = -⅖(x - 3).
Via algebra, we can find that the lines of action of F₁ and F₃ intersect at the point (-2, 1), so we may more easily compute the moment of the system about that point (see https://www.desmos.com/geometry/2bnkbxb6cj for an illustration of the forces, their points of application, and their lines of action).
Since (-2, 1) lies on the lines of action of forces F₁ and F₃, the moment of each of these forces about (-2, 1) is 0.
We thus only have to compute the moment of F₂ about (-2, 1). We recall that the moment of a force about a point is r × F, where r is the vector from the point to the point of application of F, and × is the cross product.
In the case of a 2-dimensional coordinate system, as in this problem statement, it is common to use the (pseudo)scalar cross product, which is the z-component of the ordinary cross product, since all of the other components are 0.
The vector from (-2, 1) to (2, -3) is ⟨4, -4⟩ (this is the light-blue vector in the linked diagram).
It follows that the (pseudo)scalar moment of F₂ about (-2, 1) is the z-component of ⟨4, -4, 0⟩ × ⟨2, -3, 0⟩, which is 4∙(-3) - 2∙(-4) = -4 units of moment, which is the moment of the couple resultant of the system of forces. Being negative, this corresponds to a clockwise moment of magnitude 4 units of moment (recall that the units of moment are units of force times units of length. Since neither unit is given in the problem, we cannot assign a more specific unit of moment to the given magnitude).
It is unusual that the problem states the condition "if the group is equivalent to a couple...", since we can add the forces F₁, F₂, and F₃, as given, and readily see that the resultant is 0, and thus the only possible result is a couple resultant. In addition, every system of forces is equivalent to a force-couple system, even if the force, or the couple, or both, are 0. That is, the condition was not an additional assumption that it was necessary to state in order to start working on the problem.
In either case, in order to resolve the resultant couple of a planar system of forces, it can be helpful to first find the point of intersection of the lines of action of 2 or more forces. Computing the moment of each force about such a point would be simpler than choosing any other point, due to the fact that the moment of a force about a point on its line of action is 0. By choosing a point on the line of action of at least two forces, we maximize that simplification.
The equations for the lines of action are readily found by using the Point-Slope form of the equation of a line.
For example, consider the line of action of force F₁. It will necessarily have a slope of ⅓, since that the rise of the vector is 1 and the run is 3. It passes through the point of action (1, 2), so an equation of the line is y - 2 = ⅓(x - 1).
Similarly, the other two lines of action are y + 3 = -³⁄₂(x - 2) and y + 1 = -⅖(x - 3).
Via algebra, we can find that the lines of action of F₁ and F₃ intersect at the point (-2, 1), so we may more easily compute the moment of the system about that point (see https://www.desmos.com/geometry/2bnkbxb6cj for an illustration of the forces, their points of application, and their lines of action).
Since (-2, 1) lies on the lines of action of forces F₁ and F₃, the moment of each of these forces about (-2, 1) is 0.
We thus only have to compute the moment of F₂ about (-2, 1). We recall that the moment of a force about a point is r × F, where r is the vector from the point to the point of application of F, and × is the cross product.
In the case of a 2-dimensional coordinate system, as in this problem statement, it is common to use the (pseudo)scalar cross product, which is the z-component of the ordinary cross product, since all of the other components are 0.
The vector from (-2, 1) to (2, -3) is ⟨4, -4⟩ (this is the light-blue vector in the linked diagram).
It follows that the (pseudo)scalar moment of F₂ about (-2, 1) is the z-component of ⟨4, -4, 0⟩ × ⟨2, -3, 0⟩, which is 4∙(-3) - 2∙(-4) = -4 units of moment, which is the moment of the couple resultant of the system of forces. Being negative, this corresponds to a clockwise moment of magnitude 4 units of moment (recall that the units of moment are units of force times units of length. Since neither unit is given in the problem, we cannot assign a more specific unit of moment to the given magnitude).
Hi,
F₁ and F₃ do not intersect; (-2, 1) is the intersection point of the lines of action of F₁ and F₃.
In the previous comment, we derived the fact that the line of action of F₁ has the equation y - 2 = ⅓(x - 1), and the line of action of F₃ has the equation y + 1 = -⅖(x - 3).
To find the point of intersection of two lines, we can find the x-coordinate at which the two lines have the same y-coordinate (thus concluding that both lines share the coordinate pair (x, y)).
To facilitate this idea, we can first solve both equations for y in terms of x.
For the first equation, we may add 2 to both sides of the equation to get y = ⅓(x - 1) + 2, which simplifies to y = (⅓)x + ⁵⁄₃.
For the second equation, we may subtract 1 from both sides to get y = -⅖(x - 3) - 1, which simplifies to y = -(⅖)x + ⅕.
Thus, both lines pass through the same y-coordinate when (⅓)x + ⁵⁄₃ = -(⅖)x + ⅕.
Multiplying both sides of the equation by 15 in order to rid ourselves of fraction arithmetic, we get 5x + 25 = -6x + 3.
Adding 6x to both sides of the equation, then subtracting 25 from both sides of the equation, we get 11x = -22.
Dividing both sides of the equation by 11 reveals the value of x: x = -2.
Since both lines pass through this point, we may use either line's equation to find the associated y-coordinate: y = ⅓(-2 - 1) + 2 simplifies to y = 1.
We may prove that an arbitrary system of forces (acting on an ideal rigid body) is equivalent to a force-couple system via induction on the number of forces.
Consider a system of one force applied to a point on a rigid body.
If the center of mass of the body lies on the line of action of the force, then the resulting motion is equivalent to the motion produced by that force, plus a couple of moment 0 (in many texts, this is how the concept of a center of mass is derived, so this is a bit of a tautology).
Let us now prove a lemma about forces for which this case is not true:
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Lemma 1: If the center of mass does not lie on the line of action of the force, then the resulting motion is equivalent to the motion produced by a force of the same magnitude on the center of the body, plus a couple of non-zero moment, equal to the moment of the original force about the center of mass.
Proof of Lemma 1:
Let F be the force and let O be the center of mass. Construct a line L through O parallel to F and introduce a couple of forces of the same magnitude as F at O on L. Since this pair of forces contributes a net force of 0, this system is mechanically equivalent to the original single-force system.
Label one of the new forces F₁ and the other F₂.
Since this pair is a couple, one of the forces acts in the opposite direction to F.
Suppose that force is F₁. Then F and F₁ form a couple.
Since F₁ acts on O, the moment of this couple is equivalent to the moment of F about O.
Thus, we have an equivalent system consisting of a force of equal magnitude to F, F₂, plus a couple whose moment is equivalent to the moment of F about O, as was to be shown. ∎
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This ends the argument for a system of one force.
We may now assume that the original statement holds for a system of k forces on a rigid body, as we know it holds for at least one case: k = 1.
To continue the inductive process, suppose we have a system of k+1 forces on a rigid body. We know we can reduce a system of k forces to a force-couple system.
We may then reduce the (k+1)th force to a force-couple system about O using the procedure outlined above, and add the resultant to the resultant of the k forces, and the couple-resultant to the couple-resultant of the k forces to produce the necessary single force-couple system.
We may also note that we can compute the couple about any point in the plane of the couple, as the moment of a couple is the same at every point in the plane of the couple.
F₁ and F₃ do not intersect; (-2, 1) is the intersection point of the lines of action of F₁ and F₃.
In the previous comment, we derived the fact that the line of action of F₁ has the equation y - 2 = ⅓(x - 1), and the line of action of F₃ has the equation y + 1 = -⅖(x - 3).
To find the point of intersection of two lines, we can find the x-coordinate at which the two lines have the same y-coordinate (thus concluding that both lines share the coordinate pair (x, y)).
To facilitate this idea, we can first solve both equations for y in terms of x.
For the first equation, we may add 2 to both sides of the equation to get y = ⅓(x - 1) + 2, which simplifies to y = (⅓)x + ⁵⁄₃.
For the second equation, we may subtract 1 from both sides to get y = -⅖(x - 3) - 1, which simplifies to y = -(⅖)x + ⅕.
Thus, both lines pass through the same y-coordinate when (⅓)x + ⁵⁄₃ = -(⅖)x + ⅕.
Multiplying both sides of the equation by 15 in order to rid ourselves of fraction arithmetic, we get 5x + 25 = -6x + 3.
Adding 6x to both sides of the equation, then subtracting 25 from both sides of the equation, we get 11x = -22.
Dividing both sides of the equation by 11 reveals the value of x: x = -2.
Since both lines pass through this point, we may use either line's equation to find the associated y-coordinate: y = ⅓(-2 - 1) + 2 simplifies to y = 1.
We may prove that an arbitrary system of forces (acting on an ideal rigid body) is equivalent to a force-couple system via induction on the number of forces.
Consider a system of one force applied to a point on a rigid body.
If the center of mass of the body lies on the line of action of the force, then the resulting motion is equivalent to the motion produced by that force, plus a couple of moment 0 (in many texts, this is how the concept of a center of mass is derived, so this is a bit of a tautology).
Let us now prove a lemma about forces for which this case is not true:
----------------------------------
Lemma 1: If the center of mass does not lie on the line of action of the force, then the resulting motion is equivalent to the motion produced by a force of the same magnitude on the center of the body, plus a couple of non-zero moment, equal to the moment of the original force about the center of mass.
Proof of Lemma 1:
Let F be the force and let O be the center of mass. Construct a line L through O parallel to F and introduce a couple of forces of the same magnitude as F at O on L. Since this pair of forces contributes a net force of 0, this system is mechanically equivalent to the original single-force system.
Label one of the new forces F₁ and the other F₂.
Since this pair is a couple, one of the forces acts in the opposite direction to F.
Suppose that force is F₁. Then F and F₁ form a couple.
Since F₁ acts on O, the moment of this couple is equivalent to the moment of F about O.
Thus, we have an equivalent system consisting of a force of equal magnitude to F, F₂, plus a couple whose moment is equivalent to the moment of F about O, as was to be shown. ∎
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This ends the argument for a system of one force.
We may now assume that the original statement holds for a system of k forces on a rigid body, as we know it holds for at least one case: k = 1.
To continue the inductive process, suppose we have a system of k+1 forces on a rigid body. We know we can reduce a system of k forces to a force-couple system.
We may then reduce the (k+1)th force to a force-couple system about O using the procedure outlined above, and add the resultant to the resultant of the k forces, and the couple-resultant to the couple-resultant of the k forces to produce the necessary single force-couple system.
We may also note that we can compute the couple about any point in the plane of the couple, as the moment of a couple is the same at every point in the plane of the couple.
