# Line of action force

The task is to reduce the two forces into a single net force and then find the equation of its line of action.

I chose to find the net force and momentum with respect to the origin. The magnitude of each force is $$P$$.

We can therefore write $$F_{net} = (P,0,0) + (0,0,-P) = (P,0,-P)$$. The distance vector from origin to $$F_1$$ is $$(0,b,c)$$ and from origin to $$F_2$$ is $$(a,0,c)$$. The net moment is then $$M_{net} = (0,b,c)\times (P,0,0) + (a,0,c)\times (0,0,-P) = (0, aP + cP, -bP ).$$

What I tried to do was to assume the attacking point of the net force to be $$(x,y,z)$$. This force must cause the net moment with respect to the origin. So I set up the cross product

$$(x,y,z) \times (P,0,-P) = (0, aP + cP, -bP).$$

but I was unable to determine $$x$$, $$y$$ and $$z$$. So what can we do?

• Do you mean "I chose to find the net force and moment with respect to the origin"? Moment and Momentum mean different things – Pranav Hosangadi Feb 11 '14 at 3:24
• Related post: physics.stackexchange.com/a/80547/392 – ja72 Feb 11 '14 at 18:22

## 1 Answer

If at point A with position $\vec{r}_A$ the sum of forces and moments is $\vec{F}$ and $\vec{M}_A$ then the force line of action has direction $$\vec{e} = \frac{\vec{F}}{|\vec{F}|}$$ and position closest to A as $$\vec{r} = \vec{r}_A + \frac{ \vec{F} \times \vec{M}_A}{|\vec{F}|^2}$$

where × is the vector cross product.

This comes from the net moment $\vec{M} = \sum_{i=1}^n \vec{r}_i \times \vec{F}_i$ and the net force $\vec{F} = \sum_{i=1}^n \vec{F}_i$. By definition the line of action is located where $\vec{M} = \vec{r} \times \vec{F}$ and by crossing with the net force both sides gives

\begin{aligned} \vec{F} \times \vec{M} & = \vec{F} \times ( \vec{r} \times \vec{F} ) \\ & = - \vec{F} \times ( \vec{F} \times \vec{r} ) \\ & = - \vec{F} (\vec{F}\cdot\vec{r}) + \vec{r} ( \vec{F}\cdot\vec{F} ) \end{aligned}

by using the vector triple product. Now since $\vec{r}$ is taken to be closest to the line of action, it means it points perpendicular to $\vec{F}$ and thus $\vec{r}\cdot\vec{F}=0$. So

$$\vec{F}\times\vec{M} = \vec{r} (\vec{F}\cdot\vec{F})$$ $$\boxed{ \vec{r} =\frac{ \vec{F}\times\vec{M} }{ \vec{F}\cdot\vec{F} } }$$

Here we have $$\vec{F} = (P,0,-P) \\ \vec{M} = (0,P (a+c),-P b)$$ and $$\vec{r} = ( \frac{a+c}{2}, \frac{b}{2}, \frac{a+c}{2} )$$

There is also a component of the net moment parallel to the net force. This is calculated with the pitch $$h = \frac{\vec{F}\cdot\vec{M}}{\vec{F}\cdot\vec{F}}$$ which in our case it is $h=\frac{b}{2}$. Together the net moment at the origin is

\begin{aligned} \vec{M} & = \vec{r} \times \vec{F} + h \vec{F} \\ & = ( \frac{a+c}{2}, \frac{b}{2}, \frac{a+c}{2} ) \times (P,0,-P) + \frac{b}{2} (P,0,-P) \\ & = (-P \frac{b}{2}, P (a+c), - P \frac{b}{2} ) + (P \frac{b}{2},0,P \frac{b}{2} ) \\ & = (0, P (a+c), -P b) \;\checkmark \end{aligned}

• Hm, where did the last expression come from? Also, how does this help me to determine the equation? – user40339 Feb 10 '14 at 23:31
• $\vec{r} = (x,y,z)$ – ja72 Feb 11 '14 at 13:24
• See updated answer. It all comes from screw theory. See physics.stackexchange.com/q/91181 – ja72 Feb 11 '14 at 14:35
• – ja72 Feb 11 '14 at 18:22