When we are turning a wheel, from experience we would normally push with one hand and pull with the other.
Let’s analyze these two forces. 
Assume the two forces are of the same magnitude and opposite direction, therefore they can be known as F and negative F.
And they each has a perpendicular distance of half of d from the center of the wheel, point O.
We can write the force equilibrium equation that the resultant force acting on this wheel equals to F plus negative F, which is zero, 
which means that the forces cancel each other out, therefore have no translational effect on the wheel. 
These two forces are known as a couple.
However, we can also calculate the total moment caused by these two forces about the center of the wheel.
Since both forces are causing counterclockwise rotational effect about point O, therefore they both create positive moments about point O, 
and the total moment is positive F times d, with 
d being the perpendicular distance between the lines of action of these two forces. 
In fact, we can also try to summarize the total moment caused by these two same forces about another arbitrary point, point A, 
that is at a distance d prime from force F, as shown. 
Now this force
is creating counterclockwise rotational effect about point A, with a moment arm of d plus d prime,
while this force is creating a clockwise rotational effect about point A, with a moment arm of d prime, 
therefore,
the total moment can be calculated to be F times d plus d prime, minus F times d prime, which is still F times d, the same as 
the moment we calculated earlier about point O. 
the moment we calculated earlier about point O. 
Even if we want to summarize the total moment caused by these two forces about a point, say point B, that is not even in the current xy plane, 
we can use the vector formulation to get that the total moment equals to position vector r_1 
cross F plus position vector r_2 cross negative F, equals to 
position vector r_1 minus r_2 cross F. 
And after applying vector calculation, this still equals to a moment with the magnitude of F times d, pointing to the positive z direction.
And this is still the same moment calculated about point O and point A earlier. 
Therefore based on the previous three calculations we learned that the moment of 
And this is still the same moment calculated about point O and point A earlier. 
Therefore based on the previous three calculations we learned that the moment of 
a couple is a free vector because it does not depend on the reference point,
unlike the moment of a force which is calculated differently depending on what point or axis you choose to calculate this moment about.
The net external effect of a couple is that there is no net force, but only a net moment.
The magnitude is calculated as F times d, d being the perpendicular distance between the couple. 
The moment of a couple is also a vector, just like the other moment or force vectors. 
Therefore the calculation still follows the rules of vector calculation. 
In summary, to calculate the moment of a couple in scalar formulation, the moment equals to F times d. F is the magnitude of one force, 
and d is the perpendicular distance between the two forces. 
You need to determine if the moment is positive or negative based on if the rotational effect is counterclockwise or clockwise. 
In vector formulation, the moment equals to r cross F. F is one of the force vector, and r must be
the position vector that ends on this force vector.
As long as r is a position vector that ends on the force vector in your calculation, it can be
any position vector that's between these two force vectors.
Let's look at this example. There's a pair of forces applied on this member. The two forces both have the magnitude of 5 kN. 
They are of opposite direction, therefore they form a couple moment, and we need to determine the magnitude of this couple moment.
There are many ways to solve this problem.
We can always use the vector formulation. To do that we need to first set up our coordinate system and then represent one of these 
two force vectors in its Cartesian vector form. In this case I am choosing force vector F_1.
And then I need to draw a position vector that starts anywhere along the other force vector and ends anywhere along this chosen force vector 
F_1.
I have unlimited options but this is probably the most convenient way to draw this position vector, r.
And I need to also represent it in its Cartesian vector form.
And now I'm ready to apply the equation that the moment vector equals to the cross product of the position vector and the force vector, and 
the couple moment calculated is 12.5 k kilo-newton meter. This unit vector k indicates that the direction of this moment
is along the positive z direction which is perpendicular to the x-y plane, pointing outwards.
And the magnitude of the couple moment is simply 12.5 kilo-newton meter. it is positive and that indicates that the rotational effect is 
counterclockwise, within our x-y plane. And that's the answer to this problem.
But since this is a 2D problem it is easier to apply the scalar formulation instead.
For that we need to first know the magnitude of one of these two forces,
which is 5 kN, and we also need to know the perpendicular distance between these two parallel forces. In this case we can easily 
determine this distance from geometry to be 2.5 meter.
And then the couple moment is simply F times d, to be 12.5 kilo-newton meter. But when we apply scalar formulation we need to manually 
determine the direction of the rotational effect. In this case the rotational effect is
counterclockwise, therefore the moment is positive.
In some situations it is easier to apply the principle of moments instead to calculate the couple moment. To do that we need to first 
resolve the forces into their respective horizontal and vertical components.
As you can see now we have two pairs of forces since forming two couple moments shown in different colors.
The advantage of applying principle of moments is that now the moment arms are easy to determine.
So we simply need to calculate the two couple moments respectively,
and then add them together, and we will get to the same answer.
