Physics on a stake in the ground

In summary, the conversation discusses the amount of force needed to remove a three inch piton driven into rock or concrete. The factors involved include the depth of the concrete, the force of the rock pressing against the sides of the piton, and the coefficient of friction between steel and concrete. It is suggested to experimentally hang weights on a piton to determine the force needed, as other factors such as the surface area of the spike may not greatly impact the result.
  • #1
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i'm trying to figure out how much force it would take to remove a three inch piton driven into rock or concrete. For example, assuming the concrete in a 50 square foot parking lot was more than three inches deep, and you drove a 3 inch piton into the center of it, how many pounds of force would it take to pull it out? The only factor i haven't been able to find is how much force the rock, which you forced apart while driving in the piton, is pressing against the sides of the piton or stake. if anyone can tell me or lead me to anywhere or formula that would be helpful, thanks in advance. I've found that the coefficient of friction of steel on concrete to be .5 (on one surface of the piton) so probably closer to .8 figuring the other sides... I'm really just looking for a rough answer:tongue2:
 
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  • #2
Why not just do it experimentally hanging weights on a piton in a vertical position (just not at great altitude).
 
  • #3
If you multiply the force pressing on the sides by the friction coefficient 0.5 then you will get a rough but decent approximation to the force needed to pull out the spike. I know it sounds too simplistic, but not even the area of the surface of the spike will matter.

To get any better than this approximation you will have to go experimental. Also, you might be able to calibrate your coefficient of static friction.
 

1. What is "Physics on a stake in the ground"?

"Physics on a stake in the ground" refers to a simplified model used to understand the behavior of physical objects attached to a rigid stake in the ground. This model is often used in introductory physics courses to demonstrate concepts such as forces, moments, and equilibrium.

2. How does "Physics on a stake in the ground" work?

The model assumes that the stake is completely rigid and that the object is attached to the stake with a single point of contact. This allows for the simplification of forces and moments acting on the object, making it easier to analyze and understand its behavior.

3. What are the limitations of "Physics on a stake in the ground"?

This model is a simplified representation of real-world scenarios and does not take into account factors such as friction, air resistance, and the flexibility of the stake. It is most useful for analyzing objects in static equilibrium and may not accurately predict the behavior of objects in motion.

4. How is "Physics on a stake in the ground" used in real-world applications?

The concept of a stake in the ground is commonly applied in engineering and architecture to understand the stability and structural integrity of buildings and structures. It can also be used in analyzing the forces acting on objects such as flagpoles, streetlights, and power lines.

5. Can "Physics on a stake in the ground" be applied to non-physical concepts?

While the model is primarily used to understand physical forces, it can also be applied to other concepts such as decision-making and problem-solving. The stake can represent a fixed goal or constraint, and the forces acting on the object can represent different factors or considerations that must be taken into account.

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