0
$\begingroup$

I was told that

If net external moment of some forces is zero about a point, then the net external force passes through the point.

I know it's not true in general, what was the special condition imposed which I forgot to include to make it true? Also, is there anything close to the statement we can say?

Some of my friends guess that if the additional condition that the net external force on the body is zero would be also imposed the statement might become true, while some say that the forces were meant to be co-planar. I do not understand why that would be true. It will be a great help if someone could provide some insight into what exactly is happening here.

Please ask for clarifications if something is not clear.

$\endgroup$
2
  • $\begingroup$ how can you say that its not true in general? $\endgroup$
    – crabNebula
    Sep 5 '20 at 6:50
  • $\begingroup$ What is your specific question? $\endgroup$
    – Bill N
    Sep 6 '20 at 0:56
0
$\begingroup$

If net external moment of some forces is zero about a point, then the net external force passes through the point

It means that when n number of forces are acting on a body,and if about certain point the net torque is zero (say P), then you find the net force on the body and apply it at that point, P (in free body diagram) and you will get the same dynamics of the body as earlier.

Take a slab in space (because its easy to visualize). Now apply coplanar forces(say in xy plane). The forces should be opposite and of different magnitude (for simplicity). you can find a point about which the body will not rotate along z axis(P) as we are applying the force in XY plane .And also calculate the torque about other points on the body(say S).

since the forces are of different magnitude, calculate the net force on body and in the free body diagram, apply the net force at that point where the torque is zero(P). calculate the torque about other points(S) and you will find that you will get the same torque (about S) as calculated earlier(about S) when the force was not translated.

For translation of body we always take force about COM , so translational dynamics is same in both cases, and rotation is also same in both cases(as same torque). So you will get the same dynamics.

you change the Plane from xy- to any other , the law will hold for any plane. you can apply forces in arbitrary direction, resolve those forces in x, y, z and individually find the dynamics along 3 direction, and they are independent and same (about S and P). so overall dynamics is same.Therefore the rule applies for forces in any arbitrary direction

$\endgroup$
0
$\begingroup$

Two systems of forces are dynamically equivalent if they have the same sum $\bf R$ (net force) and the same moment $\bf M$ about a given pole $O$. It's worth noting that $\bf M\cdot R$ is an invariant independent of the choice of the pole $O$, because the transformation rule of the moment is $\bf M' = \bf M + (O-O')\times \bf R $.

If $\bf M\cdot R = 0$ and $\bf R \neq 0$ the system is equivalent to a single force $\bf R$ applied at $O$.

if $\bf M \neq 0$ and $\bf R = 0$ the system is equivalent to a torque (pair of forces independent of the pole).

if $\bf M=0$ and $\bf R = 0$ the system is balanced.

If $\bf M \cdot R \neq 0$: the system of forces is reducible to a single force and a torque.

That's all.

$\endgroup$
1
  • $\begingroup$ Lets take a Single force ,isn't torque dot force is zero $\endgroup$
    – crabNebula
    Sep 5 '20 at 11:53
0
$\begingroup$

The moment due to a force $\boldsymbol{F}$ acting through some location $\boldsymbol{r}_1$ in space is $$\boldsymbol{M}_0 = \boldsymbol{r}_1 \times \boldsymbol{F} \tag{1}$$

Now transfer the moment from the origin 0 to point 1 where the force applies

$$ \boldsymbol{M}_1 = \boldsymbol{M}_0 + (- \boldsymbol{r}_1) \times \boldsymbol{F} = \boldsymbol{r}_1 \times \boldsymbol{F} - \boldsymbol{r}_1 \times \boldsymbol{F} = \boldsymbol{0} $$

So it is true that

The moment caused by a force is zero if the force goes through the point where the moment is measured. The assumption here is that the net force is non-zero $\boldsymbol{F} \neq 0$.

Now consider the more general case of a moment $\boldsymbol{M}_0$ that is due to an offset force $\boldsymbol{F}$ at $\boldsymbol{r}_1$, just as before and a parallel moment $\boldsymbol{M}_1$ about 1. This means that $\boldsymbol{M}_1$ and $\boldsymbol{F}$ act along the same direction, such that $\boldsymbol{M}_1 = h \boldsymbol{F}$ where the scalar $h$ (in distance units) is the "pitch". Then the moment at 0 is

$$ \boldsymbol{M}_0 = h\,\boldsymbol{F} + \boldsymbol{r}_1 \times \boldsymbol{F} \tag{2}$$

The above vector spans $\mathbb{R}^3$ since the cross product $\times$ eliminates all components parallel to the force, but the pitch $h$ includes them. So all possible directions are accounted for.

From $\boldsymbol{M}_0$ and $\boldsymbol{F}$ you can recover where the force is acting $\boldsymbol{r}_1$ and the pitch $h$ (and hence the parallel moment).

$$\begin{aligned} \boldsymbol{r}_1 &= \frac{ \boldsymbol{F} \times \boldsymbol{M}_0}{\| \boldsymbol{F} \|^2} & h & = \frac{ \boldsymbol{F} \cdot \boldsymbol{M}_0}{\| \boldsymbol{F} \|^2} \end{aligned} \tag{3} $$

Where $\cdot$ is the dot product, and $\times$ the cross product.

To prove the above, use the vector triple product identity $a \times (b \times c) = b (a \cdot c) - c ( a \cdot b)$ and the self-dot $a \cdot a = \| a \|^2$.

Use (3) into (2)

$$ \begin{aligned}\boldsymbol{M}_{0} & =\tfrac{\left(\boldsymbol{F}\cdot\boldsymbol{M}_{0}\right)\boldsymbol{F}}{\|\boldsymbol{F}\|^{2}}+\tfrac{\left(\boldsymbol{F}\times\boldsymbol{M}_{0}\right)\times\boldsymbol{F}}{\|\boldsymbol{F}\|^{2}}\\ & =\tfrac{\boldsymbol{F}\left(\boldsymbol{F}\cdot\boldsymbol{M}_{0}\right)-\boldsymbol{F}\times\left(\boldsymbol{F}\times\boldsymbol{M}_{0}\right)}{\|\boldsymbol{F}\|^{2}}\\ & =\tfrac{\boldsymbol{F}\left(\boldsymbol{F}\cdot\boldsymbol{M}_{0}\right)-\left(\boldsymbol{F}\left(\boldsymbol{F}\cdot\boldsymbol{M}_{0}\right)-\boldsymbol{M}_{0}\left(\boldsymbol{F}\cdot\boldsymbol{F}\right)\right)}{\|\boldsymbol{F}\|^{2}}\\ & =\tfrac{\boldsymbol{M}_{0}\left(\boldsymbol{F}\cdot\boldsymbol{F}\right)}{\|\boldsymbol{F}\|^{2}}=\tfrac{\boldsymbol{M}_{0}\|\boldsymbol{F}\|^{2}}{\|\boldsymbol{F}\|^{2}}=\boldsymbol{M}_{0}\;\checkmark \end{aligned} $$

Which makes the following statement true as well, (also when the net force is non-zero $\boldsymbol{F} \neq 0$)

A rigid body has a non-zero net force $\boldsymbol{F}$ applied and a net moment $\boldsymbol{M}$ applied on some point in space. The moment can be transform to any other point in space, but when transformed along the points of a specific line in space (called the line of action of the force) the resulting moment is minimal and the components are parallel to the line of action. The location of the line is given by (3) and the direction of the line is given by the direction of the force $\boldsymbol{F}$.

To be precise, the point given by (3) is the closest point on the line to the place of summation of moment $\boldsymbol{M}_0$.

See Also

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.