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My book says the following:

If the body is acted upon by a system of forces, the resultant moment of the forces about point O can be determined by vector addition of the moment of each force. The resultant can be written symbolically

$$(M_R)_O = \Sigma (r \times F)$$

I don't understand why you can add the vectors. Shouldn't you just calculate the magnitude of each moment vector, if its counterclockwise make the magnitude positive, if its clockwise leave it positive and add the magnitudes?

For example if you have two moment vectors:

$$\vec{M_1} = (a,b,c)$$ (CLockwise) $$\vec{M_2} = (c,d,e)$$ (Anticlockwise)

So you know that there is a moment of magnitude $| \vec{M_1} |$ acting clockwise and a moment of magnitude $|\vec {M_2} |$ acting counterclockwise. So why can't you just do the following:

$$ x = \pm | \vec{M_1} | \pm |\vec {M_2} |$$ $$\text{If } x > 0, \text {the moment is counterclockwise, if its negative its clockwise} $$

Where the sign of the moments depend on whether they are clockwise or anticlockwise. Why is the above method invalid? The methods are obviously not the same since adding two vectors isn't equal to adding the magnitude unless the vectors are collinear.

Also, if you do add the vectors and get vector $M_3$, how do you know if $M_3$ is clockwise or counterclockwise?

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So why can't you just do the following:

$$ x = \pm| \vec{M_1} | \pm |\vec {M_2} |$$ $$\text{If } x > 0, \text {the moment is counterclockwise, if its negative its clockwise} $$

You notice that $x$ will never get negative using this method but will only gets bigger and bigger, right? I think you misunderstand that more negative means smaller in magnitude, whereas it is actually bigger in magnitude, just in opposite direction.

Moments are vector quantities because the result may be smaller when they are summed up. They can be "cancelled" out.

Also, if you do add the vectors and get vector $M_3$, how do you know if $M_3$ is clockwise or counterclockwise?

It depends on your definition on the directions. If you say "OK, I'll define clockwise as positive value", then all your clockwise moments must be positive, while anticlockwise must be negative. When you add the vectors and get the result $M_3$, just look at its sign. If it is positive then you know it is clockwise; if it is negative then you know it is anti-clockwise.

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  • $\begingroup$ Sorry, I made a typo when adding the magnitudes. See the revision, which can give you a negative x value. Also how can a vector be negative or positive? Say, $M_3 = (x,y,z)$. How can you tell if its positive or negative. $\endgroup$ – dfg Oct 6 '13 at 17:06
  • $\begingroup$ You cannot tell if it is a function. It has to be a number. $\endgroup$ – Neoh Oct 6 '13 at 17:12
  • $\begingroup$ For moment vector, its sense of "direction" is whether it is clockwise or anticlockwise. $\endgroup$ – Neoh Oct 6 '13 at 17:17
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Moments are positive or negative based on the right hand rule. Sign is arbitrary if you choose the direction for negative and positive yourself, as long as you define it from the beginning and don't change it while in the middle of a calculation.

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Basic understanding

What the moment vector tells you is where in space a force is applied. You can think of the magnitude of the moment is the minimum distance to the line of action of a force, and the individual components of the moment as the 3D location in space of said force.

It turns out that to account for the effects of two or more forces acting at different locations, all you have to do is add up the components of the moment vectors. $$ \vec{M} = \vec{M}_1 + \vec{M}_2 = \left( \vec{r}_1 \times \vec{F}_1 \right) + \left( \vec{r}_2 \times \vec{F}_2 \right) $$

If the result might be zero when the individual magnitudes $\| \vec{M}_1 \|$ and $\| \vec{M}_2 \|$ arent. This would be the situation when either

  1. Two equal and opposite forces act through the same location (line of action). $$ \vec{M} = \vec{r} \times \left( \vec{F} - \vec{F} \right) = 0$$
  2. Two equal forces act on opposing sides of the origin (measurement point). $$ \vec{M} = (\vec{r}) \times \vec{F} + (- \vec{r}) \times \vec{F} =0 $$
  3. Two parallel but different forces balance each other out. $$ \vec{M} = \vec{r}_1 \times \hat{e} F_1 + \vec{r}_2 \times \hat{e} F_2 = \hat{k} (d_1 F_1-d_2 F_2) =0 $$
  4. Two non-parallel and different forces act through the origin. $$ \vec{F} = \vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 = 0 $$

FYI - In general two skew non zero forces will always produce a non-zero resultant moment.

The deeper question of why is a little harder to explain.

In simpler terms, you can combine forces and moments into one thing with 6 components called a wrench (a type of screw) that represents the line in space where the force is acting through, as well as the magnitude of the force using something called Plücker line coordinates. Just as vectors convey magnitude and direction, all screws (like the wrench above) convey magnitude, direction, location and pitch.

So force & moments are screws that represent lines in space (by means of direction and location, like a ray) in Plücker coordinates. In addition, projective geometry tells us that geometrical objects such as points, lines and planes in space can be manipulated using linear algebra in ways that are geometrically meaningful.

One of these ways, is addition, where the addition of two Plücker line vectors (screws) results in a third line vector that is a linear combination of the two original ones. In other words, think of all the points that make up a line, and if the two lines share a point (are intersecting) then the linear combination is also going to contain this point.

For example, In projective geometry when a point $\boldsymbol{p}$ is on a line $\boldsymbol{\ell}$ then the following calculation is true $$ \boldsymbol{p} \cdot \boldsymbol{\ell} = 0 $$ where $\cdot$ is the dot product. The above is actually calculated with $\boldsymbol{p} ^\top \boldsymbol{\ell} = 0 $ with linear algebra.

So a point intersecting two lines $\boldsymbol{\ell}_1$ and $\boldsymbol{\ell}_2$ will also intersect the addition of the two lines via the distributive properties of linear algebra.

$$ \boldsymbol{p} \cdot \left( \boldsymbol{\ell}_1 + \boldsymbol{\ell}_2 \right) = \boldsymbol{p} \cdot \boldsymbol{\ell}_1 + \boldsymbol{p} \cdot \boldsymbol{\ell}_2 = 0 + 0 $$

If you have learned how to add force vectors you know the result is a linear combination of the two forces. The same applies to the lines of action of the forces. When combining wrenches, like the ones below, one adds up all the components one by one in the corresponding position in each object.

$$ \pmatrix{\vec{F} \\ \vec{M} } = \pmatrix{ \vec{F}_1 \\ \vec{M}_1 } + \pmatrix{ \vec{F}_2 \\ \vec{M}_2 } = \pmatrix{\vec{F}_1 + \vec{F}_2 \\ \vec{M}_1 + \vec{M}_2 } $$

You see that first three components of the above is the vector addition of forces (as you have learned in mechanics/physics) and the second part in the addition of moments as we deal with in the first part of this answer.

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