# Maximum acceleration without tipping over - torque balance analysis on an non inertial reference system

edit: This is not a homework question, but a question about whether choosing a different rotation axis can affect the torque balance analysis.

I am just using the following example to illustrate specifically how changing the rotation axis will give us different outcomes of torque balance analysis, which puzzled me.

A box of height $$h$$ and width $$w$$ and mass $$m$$ with two pedestals is placed on a truck. What is the maximum acceleration of the truck that this box won't be tipped over. Assume that the friction is large enough that the box won't slide.

1. Take the box as the object for free-body diagram analysis. It does not slide and nor tipping with respect to the truck, so I can decompose its moving status of the box by translation with the same acceleration of $$a$$ as the truck, and there is no rotation

1. for force analysis, we have:

$$N_1=mg$$

$$f_1=ma$$

The right pedestal has no $$N$$ and $$f$$ because we are calculating the maximum $$a$$ that is not tipping over

2. for torque balance, since we have no rotation, we should have $$\Sigma \tau=0$$, however, as $$N_1$$ and $$f_1$$ are all going through the rotation axis, the only $$\tau$$ will be caused by the weight $$mg$$ and $$\tau=mg\frac{w}{2}$$, which is contradicting with $$\Sigma \tau=0$$

What is wrong here?

On the other hand, i I move the rotation axis to the center of the mass of the box, I can get the correct answer: $$N_1\times\frac{w}{2} = f_1\times\frac{h}{2} \Rightarrow a=\frac{w}{h}g$$.

But in this problem, it is very clear that the rotation axis should be on the left pedestal, isn't it? What is wrong here?

• Notice that the truck is actually instantaneously "rotating" about the point in the floor that is not moving with the truck. The entire truck is moving past it from above, as if it were rotating. that is why the torque is not zero, The angular momentum is changing too, remember that point objects that move linearly have angular momentum in most reference frames (other than one in which the direction of motion passes through the origin). If you locate your origin in the center of mass there is no instantaneous rotation because there is mass moving above and below the origin,
– user338734
Aug 15, 2022 at 4:20
• @CarlosGauss, Thanks for your answer, but let's say the truck is"sliding" on the floor with acceleration of $a$, there should not have any rotation at all, isn't it?
– Y.G.
Aug 15, 2022 at 4:51
• It depends on what you actually mean by sliding. How would that change the answer? I didn't mean the rotation of the wheel, I meant the overall motion of the entire truck is like an instantaneous rotation. Imagine the truck were actually rotating clockwise around that point, how would it move? it would move tangentially to the left, the same way it moving when it translates
– user338734
Aug 15, 2022 at 5:00
• I edited the comment, I hope it is clearer now. Forget the wheels, imagine a block with kinetic friction
– user338734
Aug 15, 2022 at 5:03
• @CarlosGauss Thanks for the clarification. However, I am still confused. The truck (and the box) only have translational movement before tipping occurs, which means all particles in the system are having the same displacement (velocity, acceleration, etc) at any moment in time. If there is any rotation, the particles' displacement will vary with respect to the distance to the rotation axis because of the same angular speed. But that is not the case for this problem.
– Y.G.
Aug 15, 2022 at 5:38

The situation as described in an inertial reference frame has been discussed in the comments. For completness, we can also look at it in the accerelated reference frame. In this frame there is rotational and translational equilibrium. The real forces acting on the box is first gravity and the normal force. Then there is the friction force, $$f_1$$. Why is there a friction force when the box is standing still (as seen from the accelerated point of view)? It can only mean that there is a fourth force pushing the box backwards that we havent accounted for, and that the friction is counteracting. This force must be equal in magnitude and opposite to $$f_1$$. Furthermore, gravity imparts a torque about the left pedestal but there is no net torque on the box, so this fourth force must act through the CM and counteract the torque of gravity. Of course, this fourth force is just a pseudoforce caused by intertia.