Suppose we have two objects $A$ and $B$ that are attached to each other. Suppose further there is a force $\vec{F}_{A}$ acting vertically on $A$ at $\vec{r}_{A}$ and there is a force $\vec{F}_{B}$ acting vertically acting on $B$ at $\vec{r}_{B}$. My main question is, can we combine these two forces into an equivalent resultant force $\vec{F}_{R}$ acting on the combined $A+B$ object at some position $\vec{r}_{R}$?
This post is relevant, but the replies there honestly don't resolve my specific quandary. I am dealing with a specific problem that I will outline in detail below.
Suppose we have two objects $A$ and $B$ as shown below that are attached to one another.
We assume both objects have a uniform depth $d$ in the $z$-direction. Now suppose one sub-object experiences a force $\vec{F}_{A} = F_{A}\vec{e}_{y}$ acting at $\vec{r}_{A}$, and the other sub-object experiences a force $\vec{F}_{B} = F_{B}\vec{e}_{y}$ acting at $\vec{r}_{B}$, and these forces are possibly different (maybe for concreteness one part is magnetic and the other is not). This is shown below with point $O$ being the origin of coordinates.
Assuming $A$ and $B$ are attached to each other, I want to find the resultant force $\vec{F}_{R}$ acting on $\vec{r}_{R}$, as shown below.
To do this, I assume that $\vec{F}_{R}$ alone and the pair of forces $\vec{F}_{A}$, $\vec{F}_{B}$ both result in the same change in momentum and the same change in angular momentum:
\begin{align*}
\vec{F}_{R} = \frac{d\vec{p}}{dt} = \vec{F}_{A} + \vec{F}_{B} \qquad\text{ and }\qquad \vec{\tau}_{R} = \frac{d\vec{L}}{dt} = \vec{\tau}_{A} + \vec{\tau}_{B}.
\end{align*}
The first equation easily gives us
$$ \vec{F}_{R} = (F_{A} + F_{B})\vec{e}_{y}. $$
The second equation gives
\begin{align*}
\vec{r}_{R}\times\vec{F}_{R} &= \vec{r}_{A}\times\vec{F}_{A} + \vec{r}_{B}\times\vec{F}_{B} \\
(r_{R, x}\cdot F_{R, y} – r_{R, y}\cdot 0) &= (r_{A, x}\cdot F_{A} – r_{A, y}\cdot 0) + (r_{B, x}\cdot F_{B} – r_{A, y}\cdot 0) \\
r_{R, x}F_{R, y} &= r_{A, x}F_{A} + r_{B, x}F_{B} \\
r_{R, x}(F_{A} + F_{B}) &= r_{A, x}F_{A} + r_{B, x}F_{B}.
\end{align*}
Now $F_{A}$, $F_{B}$, $r_{A, x}$, and $r_{B, x}$ are all known, so then we can solve for $r_{R, x}$ and obtain
$$ r_{R, x} = \frac{r_{A, x}F_{A} + r_{B, x}F_{B}}{F_{A} + F_{B}}. $$
Now the last part is that I need to find $r_{R, y}$, but thinking about torque (or moment of forces) simply does not give us any info about $r_{R, y}$. Is the problem here underdetermined? If not, how can we deduce $r_{R, y}$?
Best Answer
You can do this if you're considering a rigid body, otherwise you can't replace a set of forces with an equivalent one, since the points of application of the forces matter.
For rigid bodies, you can find the resultant of the forces (vector sum of the forces acting on the rigid body) and the line of action of the resultant of the forces: the line of action has the direction of the resultant of the forces, and is the set of infinite number of points where you can apply the resultant. So, when you look for the position of the point of application of the resultant force, the problem os underdetermined and you get an infinite number of solutions, corresponding to all the points of line of action.