Can net torque $\sum_i\mathbf r_i\times\mathbf F_i$ be expessed as $\mathbf r\times$ (net force) for some $\mathbf r$?

Question

Let $\mathbf F_i$ be forces each of which is applied on $\mathbf r_i$ of a rigid body. Then is there a position vector $\mathbf r$ that satisfies

$$\displaystyle\sum_i\mathbf r_i\times\mathbf F_i =\mathbf r\times\displaystyle\sum_i\mathbf F_i~ ? \tag{1}$$

Well, what I get is by letting $\mathbf r_i=(r_{i,x},r_{i,y},r_{i,z})$, $\mathbf F_i=(F_{i,x},F_{i,y},F_{i,z})$, $\mathbf r=(r_x,r_y,r_z)$ is

$$\begin{bmatrix} 0 & \sum F_{i,z} & -\sum F_{i,y}\\ -\sum F_{i,z} & 0 & \sum F_{i,x}\\\sum F_{i,y} & -\sum F_{i,x} & 0 \end{bmatrix} \begin{bmatrix} r_x\\ r_y\\r_z \end{bmatrix}= \begin{bmatrix} \sum (r_{i,y}F_{i,z}-r_{i,z}F_{i,y})\\\sum (r_{i,z}F_{i,x}-r_{i,x}F_{i,z})\\\sum (r_{i,x}F_{i,y}-r_{i,y}F_{i,x}) \end{bmatrix}, \tag{2}$$

and $$\begin{vmatrix} 0 & \sum F_{i,z} & -\sum F_{i,y}\\ -\sum F_{i,z} & 0 & \sum F_{i,x}\\\sum F_{i,y} & -\sum F_{i,x} & 0 \end{vmatrix}=0. \tag{3}$$

Since the matrix is singular, the system might not have a unique solution.

So is it the case that generally such $\mathbf r$ may not be unique (or even nonexistant)? If so what is the criterion for uniqueness of $\mathbf r$?

Answer 1

Yes. The solution is:

$$ \bf{r} = \dfrac{\left( \sum {\bf F}_i\right) \times \left( \sum ({\bf r}_i \times {\bf F}_i) \right)} {\| \sum {\bf F}_i \|^2} =\dfrac{{\bf F} \times {\bf \tau}}{{\bf F}\cdot{\bf F}}$$

Then you can show that

$$ {\bf r}\times \left(\sum {\bf F}_i \right)= \sum ({\bf r}_i \times {\bf F}_i) = {\bf }\tau$$

Use ${\bf F} =\sum {\bf F}_i$ and ${\bf \tau} = \sum {\bf r}_i \times {\bf F}_i $ and the vector triple product $\vec{a}\times(\vec{b}\times\vec{c}) = \vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})$

$$\begin{aligned} {\bf r}\times {\bf F} &= \left(\dfrac{{\bf F} \times {\bf \tau}}{{\bf F}\cdot{\bf F}}\right)\times {\bf F} \\ & = \dfrac{\left({\bf F} \times {\bf \tau}\right)\times {\bf F}}{{\bf F}\cdot{\bf F}} \\ & =- \dfrac{{\bf F}\times \left({\bf F} \times {\bf \tau}\right)}{{\bf F}\cdot{\bf F}} \\ & = - \dfrac{{\bf F} ({\bf F}\cdot{\bf \tau})-{\bf \tau} ( {\bf F}\cdot{\bf F})}{{\bf F}\cdot{\bf F}} \\ & = {\bf \tau} \end{aligned} $$ Since ${\bf F}\cdot{\bf \tau}=0$

See more details in this answer https://physics.stackexchange.com/a/70445/392. Each loading defines a screw axis, and screws can be combined linearly (addition of two screws is a screw, and a scalar times a screw is a screw). For the resultant screw you can extract its direction $\vec{e} = \frac{\bf F}{\| {\bf F}\|}$, its position $\vec{r} = \frac{{\bf F}\times{\bf \tau}}{{\bf F}\cdot{\bf F}}$ and its pitch $h=\frac{{\bf F}\cdot{\bf \tau}}{{\bf F}\cdot{\bf F}}$.

Answer 2

Let me introduce the notation

$$\sum F_{i,x} = F_x, \ \ \ \sum F_{i,y} = F_y, \ \ \ \sum F_{i,z} = F_z, \tag{i}$$

Since the determinant is zero, there may be indeed, no solution of the system. But if the system of equations has a solution, recall that the body doesn't rotate around a point, but around an axis. So, your $\vec r$ is bound to be on an direction perpendicular to the resultant force $\vec F$ and on the rotation axis. A direction in the space may be described by a simple system of equations, e.g.

$$\begin{cases} {y = a x + b} \\ {z = a' x + b'} \end{cases}. \tag{ii}$$

Now, for simplicity let's introduce one more notation

$$\begin{cases} {\sum (r_{i,y}F_{i,z}-r_{i,z}F_{i,y}) = \tau_x} \\ {\sum (r_{i,z}F_{i,x}-r_{i,x}F_{i,z}) = \tau_y} \\ {\sum (r_{i,x}F_{i,y}-r_{i,y}F_{i,x}) = \tau_z} \end{cases}, \tag{iii}$$

From your system $(2)$ and using the notations $\text {(i)}$ and $\text {(iii)}$ one obtains three equations, two of which of the form $\text {(ii)}$.

Well, for this system of equations to have a solution, between the constants $\tau_x, \tau_y, \tau_z$ has to be a linear and homogeneous relation. From now on the job is yours. Find the relation.