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In my textbook it states that:

In 3 dimensions however there is a fourth possibility. For example, consider two forces whose lines of action are skew (non-intersecting, non-parallel). Such a pair of forces can not be equivalent to a single force, couple or be in equilibrium, but are equivalent to a force and a couple whose plane does not include the force.

I understand the impossibility of equilibrium and it simplifying to a couple, but why can't it be equivalent to a single force acting through a specific point?

A diagram to illustrate would be greatly appreciated.

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    $\begingroup$ In order for both forces to act on a single point, their lines of action must intersect at a point (or in the case of a rigid body, if their lines of action are parallel). If their lines of action are skew, no such point exists. $\endgroup$ – probably_someone Jun 16 '18 at 5:55
  • $\begingroup$ :D you've simply restated the question. $\endgroup$ – Edward Garemo Jun 16 '18 at 9:32
  • $\begingroup$ The translation part could be simplified to one force, but the torque would be different than the real one. You need both contributions. $\endgroup$ – FGSUZ Jun 17 '18 at 22:56
  • $\begingroup$ What does your textbook say about the options for a pair of forces in 2 dimensions? $\endgroup$ – sammy gerbil Jun 18 '18 at 9:56
  • $\begingroup$ Which point would you choose? $\endgroup$ – ja72 Jun 18 '18 at 22:12
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Pick any point away from the line of action of a force and you need an equipollent moment to balance things out.

With two forces any point along one of the lines of action requires a moment for the other line. The question then becomes, is there a point in space where the two moments needed for the two forces cancel each other out.

The answer is yes, almost. What is going to be left is a component of the torque parallel to the combined line of action.

Here is the procedure mathematically.

  1. Two (non parallel) force vectors $\boldsymbol{F}_1$ and $\boldsymbol{F}_2$ that each passes through points $\boldsymbol{r}_1$ and $\boldsymbol{r}_2$ in space respectively.
  2. The combined loading is simply $$\boldsymbol{F} = \boldsymbol{F}_1 + \boldsymbol{F}_2 $$
  3. The combined moment about the origin is $$ \boldsymbol{M} = \boldsymbol{r}_1 \times \boldsymbol{F}_1 + \boldsymbol{r}_2 \times \boldsymbol{F}_2 $$
  4. The point closest to origin on the line of action of the combined loading is $$ \boldsymbol{r} = \frac{ \boldsymbol{F} \times \boldsymbol{M} }{ \| \boldsymbol{F} \|^2 } $$
  5. The "pitch", ratio of combined parallel moment to combined force magnitude, is $$ h = \frac{ \boldsymbol{F} \cdot \boldsymbol{M} } { \| \boldsymbol{F} \|^2 } $$
  6. The parallel moment at the point $\boldsymbol{r}$ is $$ \boldsymbol{M}_\parallel = h \boldsymbol{F} $$

Proof that the parallel torque at $\boldsymbol{r}$ has the same equipollent moment about the origin $\boldsymbol{M}$ as the combined force.

$$ \begin{aligned} \boldsymbol{M} &= \boldsymbol{r} \times \boldsymbol{F} + \boldsymbol{M}_\parallel \\ & = \left( \frac{ \boldsymbol{F} \times \boldsymbol{M} }{ \| \boldsymbol{F} \|^2 } \right) \times \boldsymbol{F}+\left( \frac{ \boldsymbol{F} \cdot \boldsymbol{M} } { \| \boldsymbol{F} \|^2 } \right) \boldsymbol{F} \\ & = \frac{ ( \boldsymbol{F}\cdot \boldsymbol{M}) \boldsymbol{F} - \boldsymbol{F} \times ( \boldsymbol{F}\times \boldsymbol{M}) }{\| \boldsymbol{F} \|^2 } \\ & = \frac{ (\boldsymbol{F} \cdot \boldsymbol{M})\boldsymbol{F} - \boldsymbol{F} ( \boldsymbol{F} \cdot \boldsymbol{M}) + \boldsymbol{M} ( \boldsymbol{F} \cdot \boldsymbol{F}) }{\| \boldsymbol{F} \|^2 } \\ & = \frac{ \boldsymbol{M} \| \boldsymbol{F} \|^2}{\| \boldsymbol{F}\|^2} \equiv \boldsymbol{M} \end{aligned} $$

Use vector triple product $a \times (b \times c) = b (a \cdot c) - c ( a \cdot b)$.

Summary - Any force and moment combo can be expressed as a force along a specific line and a parallel moment to the line. The direction of the line is parallel to the force,and the location of the line is found with step 4 above.

Related Answer link.

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There's two equations to be considered. First, is F = mA, and there, you can add forces (because it's only one mass, and only one real-world acceleration results). The second, though, is torque = Moment * d(omega)/dt, and there are in general three moments (around three axes) for a rigid body. Unless your system is constrained somehow, there are three equations to be solved there.

So, a two-forces-applied-skew poses the full problem, with four difference equations (three rotations, one velocity) to be solved. Two forces and two points of application gives you four data inputs, but one force and one point of application gives you only two.

So, it's not soluble. Many 'torque' treatments involve spinning mechanisms with only one axis, but those aren't fully three dimensional, only one moment of rotation is exercised; that special case IS soluble.

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One force could be considered equivalent to two other forces, if it would have the same effect on the surrounding world.

If two forces are not applied to the same point, they could be applied to two different non-overlapping objects. Obviously, it would be impossible to come up with one force that would have the same effect on these two objects.

In fact, in many cases, when the trajectories of the two objects do not overlap, one force would be able to have the same effect only on one of the two objects and have no effect on the other.

The same test could be used for two skewed forces acting on the same object, i.e., we can show that a pair of such forces will create moments that one force would not and, therefore the movement of the object or the stress in the object would be different and therefore, one force would not have the same effect on the world and, therefore, would not be equivalent to the two skewed forces.

For instance, with one force acting on an object, the movement of each point of the object, translational and rotational, would be restricted to one plane. With two skewed forces that would not be the case.

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