Torque - Why does a block topple and net torque when net force is non-zero?

Question

1. Suppose there's an inclined plane with friction $\mu$ (possibly $\mu = 0$), and there's a rigid cuboid block on it. Now while it's physically intuitive to me that the block will topple over only when it's very tall, but in the below picture after some gory calculation, it looks like the net torque around $P$ is nonzero?

2. Also, through where the normal force acts? Somewhere I read that it acts through the center of mass, but in a video, I was watching regarding toppling it somehow magically shifted to the perpendicular to the plane at point P (bottom right of the block)!

3. Suppose I have a rigid body with a net force acting on it $\mathbf{\vec{F}}$. Now the net acceleration of the centre of mass would be $\mathbf{\vec{a_{cm}}} = \frac{\mathbf{\vec{F}}}{M}$. Now when $\mathbf{\vec{F}} = 0$, it's easy to see if the body will spin or not - we should just calculate $\tau_x, \tau_y, \tau_z$ through any axes parallel to the $x,y,z$ axis respectively and check if they equal to zero or not. But when $\mathbf{\vec{F}} \neq 0$, then how we know if a body will spin or not? Through which axis we need to calculate the torque? (It may happen, say, for some axis parallel to the $x$ axis the torque is nonzero though for some other axis parallel to the $x$ axis !)

Answer 1

Let's start with $\theta=0$. As the block rests on a horizontal plane, the net plane reaction force does indeed act through the center of the bottom face. On a microscopic level, the block deforms the surface underneath equally, so each subforce $df$ acting on a each subarea $dA$ (pressure) is the same, resulting in a net reaction torque as if the net reaction force was acting through the center.

Now let's incline the plane slightly ($theta>0$). At the very first moment, we have ruined the torque balance (the torque of gravity has increased while the torque of reaction force has not). The block starts to rotate clockwise. As it has rotated very-very-very tiny angle, it dents the plane in the front more and release the deformation in the rear. So pressure becomes not the same across the bottom face: it is larger closer to the point P. Macroscopically, it results in the net reaction force shifting its application point towards point P, and the torque balance is restored.

As we increase $\theta$, the application point is shifting closer and closer to P and at some $\theta$ it has nowhere to go, as it reaches P. That's why when your video is actually discussing the toppling moment, it has the reaction force acting through P.

The toppling takes place when focusing all the reaction force in the end point of the face (point P) isn't enough to restore the torque balance.

P.S. But if you in some way provide the application point with the means to go further, you can prevent the toppling. That's how you can perform Michael Jackson moves on skis.

Answer 2

Good questions all. I'll answer them out of order.

• Question 2: The normal force can act anywhere it needs to along the surface of contact. More precisely, it can act everywhere along the surface of contact: the single vector you've drawn is really the result of skillions of tiny vectors all along the contact surface. We can define a "center of force" of all these infinitesimal force vectors, similar to the idea of a "center of mass" for a bunch of masses. We can then mentally replace all of these infinitesimal force vectors with a single normal force vector whose magnitude is the sum of all the infinitesimal vectors, and which acts at the "center of force".
  
  From this argument, though, it should be evident that the "center of force" can be anywhere along the contact surface; the only restriction is that it can't lie outside of the surface of contact. The distribution of normal force along the surface can't actually be determined from the laws of statics acting on rigid bodies, and so you can't really calculate it a priori. Effectively, this is an indeterminate system if you don't use the fact that the net torque is zero, and you'd have to resort to elasticity and deformation techniques.

• Question 1: From above, we know that the position of the "center of force" for the normal force can shift. This means that if we know that the block is static, we can use this fact to figure out where the "center of force" must be: it's the location that is necessary to balance out the gravitational torque about $P$. (Or more generally, the normal force acts at the point that is necessary to make the net torque about any point vanish.)
  
  To illustrate this, imagine slowly increasing $\theta$ from 0. When the angle $\theta$ is near zero, there is a relatively large positive (CCW) torque about $P$ due to gravity, and so the normal force must act relatively far away from $P$ to balance this torque out. In fact, if you do the gory math, you'll find that it does indeed act through the CM. However, if we increase $\theta$, then the lever arm of the gravitational force about $P$ will decrease; and this means that the gravitational torque will decrease. The torque about $P$ from the normal force must therefore decrease as well. But the normal force is constrained by the fact that the block isn't accelerating; and so the only way for the normal torque to decrease is for its lever arm to decrease. In the limit where the block is about to topple over, the gravitational force acts directly through $P$, with zero torque; and so the normal force must also exert zero torque, and therefore its "center of force" must be at $P$ as well.
  
  Finally, we see from this sort of argument that a tall block will be more likely to topple than a short block. A block will fall over once its CM moves past the point $P$ horizontally: at that point, the gravity will exert a negative (CW) torque, and there is no possible source of positive torque to counteract it. So it will start pivoting in a clockwise direction about $P$. A bit of geometry should convince you that it takes a smaller angle for this to happen for a taller block than for a short block.

• Question 3: Even if a body is accelerating, it is still the case that the torque measured about the CM is the rate of change of angular momentum about the CM: $$ \vec{\tau}_{CM} = \frac{d \vec{L}_{CM}}{dt}. $$ In particular, if the torque about the CM vanishes, the body's rate of rotation will not change. This isn't an entirely trivial theorem, but it is in fact true; consult an intermediate-level text on classical mechanics for a proof.

Answer 3

after some gory calculation, it looks like the net torque around P is nonzero?

If the block is static and sitting on the ramp, then the net torque you calculate about any point should be zero.

You didn't say if the block were sitting or sliding, but the sliding case is different. If it's sliding, then it's probably accelerating, and your axis of choice ($P$) is no longer at rest in any inertial frame. Since it's now in a non-inertial frame, fictitious forces appear in the frame in the opposite direction of the acceleration. If you add this fictitious force (acting through the center of mass), then you'll find the net torque is once again zero (or the block is tumbling).

I think this answers your third question as well. All the points are equal, but if your considered axis is the center of mass, you can ignore the acceleration (because the fictitious force acts as though it passes through that point, so torque there is also zero).