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 $P$.

(source: draw.to)

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?

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

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

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

  • $\begingroup$ Hm, where did the last expression come from? Also, how does this help me to determine the equation? $\endgroup$ – user40339 Feb 10 '14 at 23:31
  • $\begingroup$ $\vec{r} = (x,y,z)$ $\endgroup$ – ja72 Feb 11 '14 at 13:24
  • $\begingroup$ See updated answer. It all comes from screw theory. See physics.stackexchange.com/q/91181 $\endgroup$ – ja72 Feb 11 '14 at 14:35
  • $\begingroup$ And see physics.stackexchange.com/a/80547/392 $\endgroup$ – ja72 Feb 11 '14 at 18:22

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.