Force sensors output when inclined

Question

So here is the context:

I am working on a motorcycle simulator. It consists of a motorcycle frame mounted on different actuators (see picture on the left).

In particular, the two actuators at the front serve the purpose of rendering the motorcycle roll angle. There is a force sensor on both sides between the motorcycle frame and the actuator (see picture on the right). The force sensors are tensile/compressive force transducers. These are are 1-axis only force sensors.

Imagine a rider on the simulator.

So here is the behavior that I observe in terms of sensor output :

• at rest, i.e. when the roll angle is $0°$, both sensor outputs are equal (let's call it $V_0$).
• when the roll angle is $>0°$ (the motorcycle and the rider are inclined towards the left), the output of the sensor on the left is $V_l > V_0$ while the output of the sensor on the right is $V_r < V_0$.
• when the roll angle is $<0°$ (the motorcycle and the rider are inclined towards the right), the output of the sensor on the left is $V_l < V_0$ while the output of the sensor on the right is $V_r > V_0$.

I would like to model the expected output of both of the sensors as a function of the roll angle. However, I have thus far been unsuccessful.

My hypothesis is that at rest the sensors both measure half of the weight of the frame + the rider and when inclined they measure some kind of function of this weight and of the roll angle. But when I draw the free body diagram I always find an equal force applied on both sides, which is not coherent with the actual results.

Edit: This is not a homework question, but let me show my work anyway.

At rest, the forces involved are:

• $W$, the weight of the frame + the rider (which is along the $y$-axis);
• $R_{l_x}$ and $R_{l_y}$ the reaction of the left support (resp. along $x$ and $y$-axis);
• $R_{r_x}$ and $R_{r_y}$ the reaction of the right support (resp. along $x$ and $y$-axis).

So the equilibrium equations are:

1. $R_{l_x} - R_{r_x} = 0$
2. $R_{l_y} - W + R_{r_y} = 0$
3. $-W \times L/2 + R_{r_y} \times L = 0$ (where $L$ is the distance between the two sensors)

Because the sensors are only sensible to traction/compression along their y-axis, I am not interested in equation (1). Equations (2) and (3) result in $R_{r_y} = W/2$ and $R_{l_y} = R_{r_y} = W/2$. According to this, it makes sense that at rest the sensors output the same voltage $V_0$ (which corresponds to them being subjected to a compressive force of magnitude $W/2$).

What I am having trouble doing is using the same logic in the inclined case.

Could somebody please help me figure this out?

Answer 1

This is me trying again with the help of @Chuck's answer.

So in the general case, the torque equilibrium equation about A is $\vec{r}_{A \rightarrow COM} \times \vec{W} + \vec{r}_{A \rightarrow B} \times \vec{F}_{B_{reaction}} = \vec{0}$

Let's write the Cartesian components of each of these vectors:

• $ \vec{r}_{A \rightarrow COM} = \begin{pmatrix}|\vec{r}_{A \rightarrow COM}| \cdot \sin(\theta_W) \\ |\vec{r}_{A \rightarrow COM}| \cdot \cos(\theta_W) \\ 0\end{pmatrix}$ with $\theta_W$ is the angle between $\vec{r}_{A \rightarrow COM}$ and $\vec{W}$. Also, we have $\theta_W=\theta_{W_0} - \alpha$ where $\theta_{W_0}$ is the angle between $\vec{r}_{A \rightarrow COM}$ and $\vec{W}$ when the motorcycle is not inclined (left case in Chuck's graph) and $\alpha$ is the roll angle of the motorcycle (in the right case in Chuck's graph, $\alpha < 0°$). This will be useful later;
• $\vec{W} = \begin{pmatrix}0 \\ -W \\ 0\end{pmatrix}$;
• $ \vec{r}_{A \rightarrow B} = \begin{pmatrix}L \cdot \cos(\alpha) \\ -L \cdot \sin(\alpha) \\ 0\end{pmatrix}$;
• $\vec{F}_{B_{reaction}} = \begin{pmatrix}F_{B_{reaction}} \sin(\alpha) \\ F_{B_{reaction}} \cos(\alpha) \\ 0\end{pmatrix}$.

Now if we take the norm of the torque equilibrium equation and substitute everything, we have: $|\vec{r}_{A \rightarrow COM}| \cdot \sin(\theta_{W_0} - \alpha) \cdot (-W) + L \cdot F_{B_{reaction}} \cdot (\cos^2(\alpha) + \sin^2(\alpha)) = 0$

$\theta_{W_0}$ is such that $\sin(\theta_{W_0}) = \frac{L/2}{|\vec{r}_{A \rightarrow COM}|}$ and $\cos(\theta_{W_0}) = \frac{h}{|\vec{r}_{A \rightarrow COM}|}$ where $h$ is the height of the center of mass. With a few more developments and if we do the same for the torque about B, this is the result:

• $F_{B_{reaction}} = \frac{W \cdot [\frac{L}{2} \cdot \cos(\alpha) - h \cdot \sin(\alpha)]} {L}$
• $F_{A_{reaction}} = \frac{W \cdot [\frac{L}{2} \cdot \cos(\alpha) + h \cdot \sin(\alpha)]} {L}$

Answer 2

The thing you're missing from your equations of motions are the torques. Typically in these kinds of problems we assume the actuator ends are "pinned" connections so you can't transmit a torque through that mounting point. This lets you assume that the actuator force must cancel out any moments generated in your system.

Consider the following graphic:

The center of mass is the red circle. The left case has the equal torques about A and B. Remember that the equation for torque is $r \times F$. The vector $r$ is the vector from the pivot to the center of mass, and $F$ is the force vector, which is $mg$ and points down.

The equations you're missing are the ones that reconcile the applied and reactionary torques. If you consider just the torque about the pinned joint A, and your test platform is stationary, then the sum of the applied torque due to the center of mass and the reactionary torque applied at B must be zero:

$$ \left(r_{A\rightarrow COM} \times mg \right) + \left( r_{A\rightarrow B} \times F_{B_{reaction}} \right) = 0 \\ $$

Then you have to consider the torque about B and do it all over again. In the left scenario you wind up with the same counter torques and you get the load sharing - each actuator supports half the weight.

In the right scenario, in my graphic, the center of mass is nearly over B. In this case you'll get a very large torque about A that B's reaction needs to support. Because the center of mass is almost directly over B, there's not very much torque about B, and so A has virtually no reactionary force.

If you were to extend this more, if the center of mass is directly over B then there is NO torque about B and A doesn't support any weight at all. If you had the following scenario:

If the distance from A to the center of mass is twice the distance from A to B then you wind up with B in compression at double the weight and A's reaction is in tension equal to the weight - this becomes a lever and fulcrum problem.