# Compute forces exerted on a pole in ground [closed]

The image below illustrates a pole which is in the ground with a depth of $L2$. Upon the pole a force $F1$ is exerted at height $L1$. As a result, the lever pushes with its bottom end against the ground to the right which causes the ground to exert a resistance force. $F1$ must overcome a certain force in order to get the pole's upper end to move in its direction; I wish to compute this force.

In the image we see multiple forces labelled Fr (only the black ones): as the pole pushes towards the right at every height 0 > L >= L2 (L = 0 on dotted area where air and ground meet), there is a resistance force Fr(L). For big L this force is small while for small L the force grows. I assume Fr(L) must grow in a linear fashion as it does with levers (F1 * L1 = F2 * L2).

Assume that the ground is made of only one material, with uniform density, disregarding any "dirty" effects, such as moisture, or temperature. The length of the pole, its radius (it is a cylinder), and its depth L2 are all constant. My computations are not required to simulate the exact real-world phenomena it must only appear to be reasonable in a 3d simulated world; as such, we may idealise as much as needed within reason.

For purposes of simplicity, we may compute a cumulative ground resistance force F2 replacing all the ground resistance forces as follows: F2 = $\int_0^{L2}$ Fr(L) dL.

Furthermore, let's view the pole as a lever with it pivot where air, ground, and pole meet. Then we have F1 * L1 = F2 * L2 when the lever is in balance. Now, I would like to find out at what force F1 the pole starts moving; the pole is moving when F1 * L1 > F2 * L2.

The problem would be solved if we find a means to compute Fr(L). If I made a conceptual mistake, please point it out, it might be I got the problem description wrong at some point.

Proposed solution:

My idea is Fr(L) = μ N(L) where μ is the coefficient of friction for the two surfaces and N(L) is the normal or perpendicular force pushing the two objects (ground and pole) together. The follows from the equation for dry friction (Wikipedia). The force N is dependent on both L and F1 (though we treat F1 as a constant in the formula not as a variable for simpler computations). My guess is that maybe N(L) = L/L2 * a * F1 as N must grow linearly with L; a is a scalar and thusly a * F1 the maximum resistance force at height L2. L/L2 then is a value between 0 and 1 scaling the force linearly. Does this idea seem plausible?

• The direction of forces in your diagram is not correct. You need to find a set of forces that, before the pole moves, result in zero net force AND zero net torque. This means that the forces under the soil actually need to change direction - near the surface they will oppose the force on the pole, deeper down they will be in the same direction. This is the only way I can think of that you can zero out both torque and net force. – Floris Sep 10 '16 at 12:52

[Not really an answer, but too long for a comment...]

Your situation is not something we would normally try to calculate because there are too many variables and too much variability. Everything depends on the details of the dirt. You can imagine that a post in sand would be much weaker than a post in clay. Many types of soil would behave nonlinearly; as the post is tipped compaction could increase the force of the soil on the post. The moisture content of the soil presumably changes all the time, and would have a huge effect on the forces. Etc. A physicist would be more likely to use your setup to measure Fr for a particular soil than to try to calculate it.

This situation isn't really the scenario to which dry friction is usually applied; that's more about force involving something sliding over a surface. I wouldn't be surprised if landscaping engineers have some rules of thumb to go by, but they'd really just be rules of thumb. For example, I know a standard rule of thumb for fences with 8' between posts is to bury 1/3 of the post in the ground — so if you want a 6' fence, you'd need a 9' post with 3' in the ground. Presumably, that's nicely related to the maximum wind forces on the fence panel. If you're really looking for practical advice like that, you'd be better off on the diy.stackexchange.

You also seem to be assuming that the post will pivot (stay in place) at the surface; in my experience, posts move most at the surface. If you wiggle a post out of the ground, it starts to give way at the surface before you get any movement at the bottom of the post. In practice, then, the pivot would be closer to the bottom. If you want the post to pivot at the top, you'd need to secure the top somehow. I've seen livestock fences where they bury a horizontal piece of 2x4 in the ground near the top to help ensure that there's no movement at the surface. As Floris points out, you need to balance both the torque and the total force; such a pivot would provide the balancing force to the right.

• There are a lot of details, such as moisture, playing a roll which I do not consider. My intent is to compute an approximate force under simplified circumstances: I wish to use this knowledge in a game simulation so the physics do not need to be exact so I am more looking at rules of thumb. I considered going out measuring out, too, and I guess this is what I'll end up doing. Thanks for the elaborate answer and redirection to diy.stackexchange. – univise Sep 10 '16 at 14:11
• Google also turns up this extensive and detailed document with references and lots of good formulas and typical values. – Mike Sep 10 '16 at 14:24