Calculating the force of a kick

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The discussion centers on calculating the force of a kick to determine if it could rupture a testicle, with an average kick speed of 30 m/s and a leg mass of 11.5 kg. The initial calculation yields a force of 345 Newtons, approximately 77 lbs, but participants note that the problem is not fully solvable without additional information, such as the distance the foot travels during impact. Instead, they suggest calculating kinetic energy to better understand the dynamics involved. Additionally, it is clarified that pounds (a force unit) and psi (a pressure unit) are not interchangeable; psi depends on the area over which the force is applied. The conversation emphasizes the complexity of accurately determining the force involved in a kick.
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Homework Statement


I'm writing a paper about wearing jockstraps in sports, and I'm trying to calculate whether the force of an average kick could rupture a testicle (Yeah, it hurts just to even think about). I would like to get the end result in lbs, so I'm using the metric system in the equation and then converting Newtons to lbs. I would just like to confirm that I'm calculating this correctly. As a side note, can this be called psi as well? Like if the end result is 77lbs, would that also mean 77 psi?

Variables:
Average kick speed: 30 m/s (I'm aware that this is the velocity of a kick, but I'm guessing that the average kick will not exceed 30 m/s coming from rest).
Average mass of a leg: 11.5 kgs.
Testicular rupture occurs at ~110lbs of force

Homework Equations


F = MA

The Attempt at a Solution



F = (30)(11.5)
F = 345 Newtons, or ~77lbs.
 
Last edited:
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Troll alert.
 
AJ Bentley said:
Troll alert.

... Why?
 
I don't know that I would say "troll" but the problem you give simply is not solvable. You could use the information you have to find the kinetic energy in the kick and then set "force times distance" equal to that energy but you have two unknowns. You would also need to calculate the distance the foot moves from the time of impact to stopping. That would require information such as the resistance to the kick.
 
HallsofIvy said:
I don't know that I would say "troll" but the problem you give simply is not solvable. You could use the information you have to find the kinetic energy in the kick and then set "force times distance" equal to that energy but you have two unknowns. You would also need to calculate the distance the foot moves from the time of impact to stopping. That would require information such as the resistance to the kick.

Ah alright that makes sense, thank you. I will probably just settle with finding the kinetic energy and go from there, thanks for your help.
 
As a side note, pounds and psi are not the same. A pound is a force unit and psi is shorthand for pounds per square inch, which is a pressure or stess unit. 77 pounds acting on 1 sq in is 77 psi pressure; 77 pounds acting on 2 sq in is 38.5 psi, etc.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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