# Understanding Torque for car rollover

1. May 29, 2014

### Maxo

1. The problem statement, all variables and given/known data

2. Relevant equations
$$\vec{\tau} = \vec{F}\vec{l}$$

3. The attempt at a solution
I understand this problem except for the part that involves Torque.
I would like some help in understanding how the Torque is calculated in this problem. Please look at Figure 9.13 c). I know the axis of rotation is perpendicular to the page, and that the rotation can be either clockwise or counterclockwise around this axis. I also see that the forces FN and fs are the forces which can cause this rotation. However I don't understand the general idea in figuring out how those dashed lines should be placed. Can someone please give me some suggestions on the general idea of how to analyze this kind of situation? I also don't know what the black dashed line in the picture actually represents?

Another question for this particular problem, I wonder how one would come up with the idea of looking at the car with only one side of the wheels in the ground? If one side of the cars wheels are already above ground, I would say the car already is going too fast. And the question was how fast it can go WITHOUT rolling over. I mean I wouldn't want to be in a car where one side of the wheels have already started to lift from the ground. But apparently one has to assume that in order to calculate this problem?

Last edited: May 29, 2014
2. May 29, 2014

### tms

$$\vec\tau = \vec r \times \vec F$$
with $\times$ being the vector cross product, $\vec F$ being the force, and $\vec r$ being the radius vector from the point about which the torque is being calculated to the (effective) point of application of the force. Remember that the vector cross product is not commutative.

3. May 29, 2014

### haruspex

The dashed lines only clarify what the distances refer to. They are not forces etc.
As tms says, you want the cross product of the displacement vector and the force vector. Since, in this case, there are only right angles involved, it's pretty easy. You can use the scalar form, Fd, where d is the distance from the axis to the line of action of the force.
For the critical point at which there is a danger of rolling, you can take the normal force on the tyres on one side to be zero. You don't need to suppose they have actually left the ground. Once they do leave the ground the torque imbalance gets worse, and the car will roll.

4. May 30, 2014

### Maxo

I know Torque is actually a cross product, in my book they use a simplified(?) version where the component of the length l that is perpendicular to both the force and to the rotatonal axis is measured, thence it becomes a normal product instead of cross product. This also means the direction of the Torque vector is lost, but that doesn't seem to matter so much here.

5. May 30, 2014

### haruspex

So is there some assistance you still need?