Wind gust duration to overturn a block

[hillbilly63]

TL;DR Summary

How long does a force of H have to be applied to a block to push it beyond it's tipping point?

I'm familiar with statics and calculating the force that will start to tip a block (lets call it H). I'm also familiar with when a block will get to the point of tipping over, rather than sitting back down if the load is removed.

What I'm not familiar with is how to calculate the duration for which H would need to be applied to get the block past the point of tipping over.

I'm assuming an energy based approach is the way to, i.e. the centroid has to be lifted up a certain height in order to get to the tipping point, which can be equated to an increase in potential energy, but I'm not really sure where to go from there.

The image link shows the nomenclature I'm using, as well as force application (H is the total force applied to the vertical face of the block):

Simplifications/Thoughts:
- Assuming block won't slide
- In my application, the force is due to wind. For now, I'm happy to ignore the fact that the area that presents itself to the wind varies as the block tips, as potentially does the drag coefficient.
- Maybe the angle of force should remain perpendicular to the face (force H is due to pressure difference on front/rear surfaces and can only act perpendicular to the surfaces).

 

[erobz]

Probably $$\sum \mathbf {\tau} = \frac{d}{dt} \left( I \mathbf {\omega} \right)$$ is going to be helpful. Can you try to come up with some ideas about how to apply it? Probably energy too is useful, as you say $H$ is responsible for lifting the center of mass vertically above the corner of the box. I'm sure there is more than one way to skin the cat.

P.S. Is it just me or does

\mathbf{...}

not seem like its "bolding" the symbols $\tau, \omega$?

Probably should move this to physics homework. It will get better coverage there anyhow.

 

[berkeman]

Probably should move this to physics homework.

It doesn't seem schoolwork-like, so it's probably okay here for now.

 

[Lnewqban]

TL;DR Summary: How long does a force of H have to be applied to a block to push it beyond it's tipping point?

... the centroid has to be lifted up a certain height in order to get to the tipping point, which can be equated to an increase in potential energy, but I'm not really sure where to go from there.

That force H induces a moment about the bottom edge of the block that rotates the center of mass.
The weight induces another moment in the opposite direction.

The net moment has to fight the rotational inertia of the block during the completion of the geometrical angle needed to relocate the center of mass directly above the pivot or bottom edge.

The time you look for is based on the magnitude of the achievable angular acceleration.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#rlin

 

[tech99]

It seems to me that when the block reaches maximum height with the lowest possible wind speed, it will be in a position of unstable equilibrium. We cannot quantify the increment of additional energy required to start it to tip further; neither can be quantify the duration. In principle, the duration could be very long.

 

[erobz]

It seems to me that when the block reaches maximum height with the lowest possible wind speed, it will be in a position of unstable equilibrium. We cannot quantify the increment of additional energy required to start it to tip further; neither can be quantify the duration. In principle, the duration could be very long.

Seems like a great point. Trouble in paradise. I was imagining it having some non-zero angular velocity at the equilibrium position, but that doesn't have to be true.

EDIT:
But wait, if $H$ is a constant force its never going to be in equilibrium at that point because the weight is providing no counter torque there?