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Assume all other variables the same (same object, same surface area, etc).

Someone on Quora claimed the clavicle breaks with 8 pounds of force. When I said it doesn’t break when resting an 8-pound weight they claimed it would be different if it were "an abrupt application of force". I said 8 pounds is already a measure of instantaneous force, so that doesn't make sense. They said I am ridiculous.

https://www.quora.com/How-is-it-possible-for-an-axe-kick-to-be-powerful/answer/Keith-Dale-1/comment/36607882

https://www.quora.com/unanswered/Is-abruptly-applying-X-pounds-of-force-more-likely-break-a-bone-than-gradually-applying-X-pounds-of-force

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  • $\begingroup$ Maybe it has to do more with impulse than force. $\endgroup$
    – user122423
    Jun 5, 2017 at 18:25
  • $\begingroup$ Impulse = F*T. Force is the derivative of impulse. If you cushion a blow with pillows, the impulse will be the same, and the time will be lengthened, so the force is lower. $\endgroup$
    – pete
    Jun 5, 2017 at 18:35
  • $\begingroup$ Relevant video: youtube.com/watch?v=edvpnfvmEYU $\endgroup$
    – Mark H
    Jun 18, 2017 at 7:34
  • $\begingroup$ This is exactly the fallacy I was trying to correct in the first place. If axe kick 1 and axe kick 2 both move a truck forward by the same amount (same momentum change aka impulse), and axe kick 1 is delivered more suddenly than axe kick 2, then axe kick 1 has MORE FORCE, PERIOD, and any measuring device would show this; it isn't because it's a "more abrupt application of force", as force is already the abruptness of application of IMPULSE. $\endgroup$
    – pete
    Jun 18, 2017 at 21:15
  • $\begingroup$ @pete Does my answer help? $\endgroup$
    – Mark H
    Jun 22, 2017 at 0:58

2 Answers 2

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The key here is they are talking about an impact where you are talking about essentially static forces.

Impacts can cause far greater damage than a slowly applied force. This is because the material takes some time to react to the force. If you do it slowly, things have opportunities to easily dissipate some energy as sound/heat etc. When you do it very fast, this energy has less time to dissipate and may instead cause more of the energy to go into deforming the material.

If it's something brittle like bones; that extra deformation could cause a fracture or some other form of material failure.

Your measure of 8 pounds is "static' because the body has plenty of time to react as it is slowly lowered.

Finding the exact effects relative to speed would actually require some very in-depth analysis.

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  • $\begingroup$ Will those nuances truly make a big difference for something hard like a bone? I am trying to dispel a misconception about what "force" means. The person I'm arguing with thinks that an "abrupt application" of 8 pounds of force makes an "immense" difference. I'm saying that dropping an 8-pound weight on something actually confers a lot more than 8 pounds of force but he isn't having any of it. He seems to be mistaking force for impulse. $\endgroup$
    – pete
    Jun 5, 2017 at 19:05
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    $\begingroup$ Yes, your friend would interested in the calculation of impulse forces (change in momentum divided by time in which momentum transfer occurs) which will be different than the weight of the object. This is the difference in the Falcon 9 simply landing on the barge vs. the equally entertaining Falcon Punching of the barge resulting in explosions, damage, and ultimately millions of dollars. The legs of the Falcon 9 must be designed to handle the impulse force during landing and not just the weight alone. $\endgroup$
    – Rob S.
    Jun 5, 2017 at 20:08
  • $\begingroup$ @pete Along with what Rob said, you can also look at the potential/kinetic energy, the one that drops obviously has more energy than one just resting on your stomach. You can also just consider the one with a velocity has to decelerate to stop. This deceleration will have to come from your bones, and after (and even while) it decelerates it will still apply it's weight. My answer is really only relevant to the question in the title, as this isn't about 8 pounds of force. It's about the impact force of a dropped 8 pound mass. $\endgroup$
    – JMac
    Jun 5, 2017 at 20:15
  • $\begingroup$ Correct, I'm trying to illustrate the fact that an 8-pound mass can confer any amount of force, and dropping one confers much more than 8 pounds. But more importantly that it didn't make sense for him to say things like "force distributed over a large period of time", because force is an instantaneous measure; he was thinking about "impulse distributed over a large period of time". If you could comment on the Quora thread that would be helpful too. $\endgroup$
    – pete
    Jun 5, 2017 at 21:48
  • $\begingroup$ In fact he cited as reasoning: “In mechanics, an impact is a high force or shock applied over a short time period when two or more bodies collide. Such a force or acceleration usually has a greater effect than a lower force applied over a proportionally longer period.” Even within this quote the second scenario has "lower force" by his own admission, so... again, it's a confusion caused by the word "pounds" which can be used as a weight OR a force. $\endgroup$
    – pete
    Jun 5, 2017 at 21:53
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A part of the difficulty in the discussions of this question is the misuse of vocabulary. The reference.com article states that "it takes approximately 7 pounds of pressure to break a human clavicle." Seven pounds of pressure does not make any sense, since pounds is a measure of force, not pressure. It's like saying a car has 18 feet of speed. It would make more sense to say that it takes seven pounds per square inch of pressure to break a human clavicle--though this seems very low to me. A human can exert about 200 pounds of force with their jaw when biting with the molars. Our clavicle may be a weaker bone than the jaw, but not 15 times weaker.

The way to think about stiff objects under load, at least as an approximation, is as a spring. If you apply a force to an object, that object deforms in response: a spring compresses, a bone bends, a table sags. More force means more deformation. The relationship between force and deformation is approximated by Hooke's Law: $F = kx$, where $F$ is the force applied, $x$ is the distance of the sag, compression, or bend, and $k$ is a measure of the stiffness of the material. Granite will have a much higher value of $k$ than rubber. Another thing to note is that, by Newton's Third Law, the material under load exerts an equal magnitude force against the load.

Now, in a real material, there is a maximum amount of deformation before something in the internal structure breaks and the deformation becomes permanent or the material breaks into pieces. The existence of a maximum deformation implies that there is a maximum amount of force that the object can take. If you put too much weight on a table, it breaks.

Here's a video of somebody ax-kicking a desk with hilarious consequences. The kick happens at 1:08. But notice that at the very beginning of the video and at 0:36, somebody stands on the desk with no harm done (I like how somebody tells the guy to put both feet on the desk, as if that would put more weight on it). This is about 100-200 pounds of force, so how can a single foot traveling at speed actually break the desk?

Because the foot has mass, it takes a force to stop it. Because the desk cannot create an infinite force, the foot will keep traveling into the desk after the initial impact. Because the foot and the desk cannot occupy the same space, the desk deforms to make way for the foot. In order for the desk to survive the kick, it has to stop the foot before it reaches the breaking point described two paragraphs ago. The same goes for clavicles.

Let's consider the moment of impact, when the heel first hits the desk. At this point, the desk has not deformed at all, so it exerts no force on the foot. The foot keeps moving at the same velocity. An instant later, the desk has started to bend, and so it puts a force on the foot, slowing it down. But, the foot is still moving down. As the desk bends more and more as the foot continues to move down, the force the desk exerts on the foot grows larger (Hooke's Law and Newton's Third Law), so the foot slows down faster and faster. This is a race between:

  1. the force increasing enough to stop the foot and
  2. the foot traveling far enough to break the desk.

If the force does not increase fast enough, either from the foot being too massive or the initial velocity too high, then the foot will still be moving when it has traveled though the desk's maximum deformation, causing it to break.

Why does standing on a desk not break it? In this case, the desk only has to stop the load from accelerating in the first place. If the weight doesn't cause breaking deformation, then it can resist it. Stopping a moving object in a short distance can require an arbitrarily large amount of force, independent of the weight of the moving object. This is why dropping something on your foot hurts more than placing it on your foot. A greater force is required to stop the object than to prevent it from moving, and a greater force causes greater compression of your foot.

See the Technical Section below for the math.

Clarification

I assumed that "abrupt application of force" means an impact, which implies a collision of two objects at speed. If you meant simply changing a force very quickly with no motion, then the answer is no, it will do no more damage than a static load.

To see this, imagine a bowling ball hanging from the ceiling by a rope. You place your hand on the bottom side of the bowling ball so that it is touching, but with no upward force. If the rope is suddenly cut, you can tense your muscles and stop the bowling ball from starting to fall without moving your hand. Your hand is fine despite the sudden application of force to it. If you tried to do the same thing (stop a falling bowling ball with your hand held still) but with the bowling ball starting at a height above your hand, the consequences are obvious.

For a practical application, imagine firing a shotgun in two stances. In the first (and wrong) stance, you hold the butt of the gun a small distance away from your shoulder; in the second (correct) stance, you press the butt of the gun firmly against your shoulder. The first stance will be subject to all of the analysis above because the gun is impacting your shoulder with an initial velocity, resulting in injury to your shoulder depending on the speed of the gun recoil. With the second stance, the force on your shoulder is limited by the force of the gunpowder on the bullets. Depending on the size of the force, it may still leave a bruise, since the $k$ of flesh is less than that for bone, but there is an upper limit to the force, unlike the impact of the gun in the first stance.

Technical Section

Since the kick needs to be stopped in a certain distance, the correct measure of damage potential is kinetic energy, not momentum. The foot has an initial kinetic energy at impact of $$K=\frac{1}{2}mv^2$$ where $K$ is the kinetic energy, $m$ is the mass of the foot, and $v$ is its velocity. This is equal to the amount of work the desk has to do to stop the foot, which for a spring is $$W = \frac{1}{2}kx^2$$ where $W$ is the work (same units as energy) and $k$ and $x$ are the same quantities from Hooke's Law above. Since there is a maximum amount of deformation ($x_{max}$) before breaking, we have the following equation to describe the condition for breaking the desk: $$\frac{1}{2}mv^2 > \frac{1}{2}kx_{max}^2$$ Solving for $v$: $$v > x_{max}\sqrt{\frac{k}{m}}$$ From this we can see that there is a velocity that can break the desk, regardless of the mass of the foot. If this inequality is true, then the desk cannot do enough work to stop the foot before breaking. To put this in terms of forces, let's substitute with Hooke's Law into the original equation: $$\frac{1}{2}mv^2 = \frac{1}{2}\frac{F_{max}^2}{k}$$ where $F_{max}$ the force exerted by the table at maximum deformation. I've switched to equality since I want to know what happens when the desk survives, that is, $W=K$. Solving for $F_{max}$ $$F_{max} = v\sqrt{km}$$ From this we can conclude that the equivalent static load on a desk from an impact can be arbitrarily high based on the velocity of the projectile.

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  • $\begingroup$ This is precisely what I was trying to demontrate on the Quora thread. It doesn't matter if 8 pounds of force is caused by an 8-pound weight sitting still or a 1-pound weight crashing slowly into it or a ping pong ball shooting with tremendous velocity; it will still read a maximum of 8 pounds of force. And 8 pounds dropping will be way more than 8 pounds of force. So if he ridicules me for my "Resting 8-lbs weight" critique of the 8-pounds claim because "a rapid force has more effect than a gradual force", then he doesn't really understand what force is at all. $\endgroup$
    – pete
    Jun 22, 2017 at 6:22

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