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Force of Impact calculations

  1. May 7, 2007 #1
    I am in the middle of physics debate with someone who has some rather, shall we say, unconventional views regarding classic mechanical physics. It started when he posted that he feels 9/11 was a conspiracy because a section of a building falling does not increase the "load" (as he called it) on the lower floors.

    Our rebuttal was well do you think a brick resting on your head exerts the same force as one that falls and hits you in the head and he said YES.

    I replied that yes, of course the force of impact is greater than when it is at rest and gave an example of a 5 kg bowling ball dropping 1 meter and fully stopping within .02m after impact, and I worked out the forces at rest.

    If you put a 5 kilogram bowling ball on a scale, you will get 5 kilos. Its weight (which is a force) 5 * 9.8 = 49 Newtons of force. F = M G

    Now lets take that 5 kg ball and drop it onto the scale from 1m high. Do you really think it will still only show 5kg upon impact? Of course not.

    It will impact the scale at velocity of 4.43m/s2, with a Kinetic energy of 49 Joules. KE = 1/2 MV^2 (4.43^2 *5 /2 ) = 49 joules.

    Here is one of the other concepts you are not understanding, its the Work/Energy principle, that the CHANGE in Kinetic energy is directly related to the amount of work done. You recognized a change in kinetic energy but disregarded it.

    Here is something else you did not take into account. When the object is DECELERATED (acceleration in the opposite direction), the longer the deceleration, the less of an impact force. If you catch something fast, is there a greater impact on your hand if your hand does not move, or it you give with the impact and the hand moves with it?

    So if we take that 49 Joules of kinetic energy, and it is equal to the work being done (work energy principle), so we can substitute KE for Work.

    Force * Distance = Kinetic energy which is 49 joules. We say that 49 Joules of work was expended over a deceleration of .02M upon impact with the scale.

    Force = 49N / .02m.

    Force = 2450 N - average force during impact (when decelerated within .02m)

    Compare that with 49N when at rest on scale.
     
    Last edited: May 7, 2007
  2. jcsd
  3. May 7, 2007 #2

    Hootenanny

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    Welcome to the Forums,

    Forgive me, I don't understand your question...
     
    Last edited: May 7, 2007
  4. May 7, 2007 #3
    His replies regarding it

    (thats also him not me)

    If you have any questions as to the context of his usage,
    http://p214.ezboard.com/fcatfightsfrm18.showMessageRange?topicID=7.topic&start=21&stop=37
     
    Last edited: May 7, 2007
  5. May 7, 2007 #4
    I was in the middle of the posting, I had not finished yet.

    Okay, I assert (well its normal newtonian physics to most everyone including high school students) that an object in motion that strikes another, such as a car hitting an boulder, bowling ball being dropped, has a force of impact (average force exerted during the impact as it comes to a stop), he disagrees with that and says the force is the same as when it is at rest.

    Which goes back to him saying that if 40 odd floors in a skyscaper all fall one floor, there is no additional force or load on the the rest of them since weight is the same. (He does not believe in any additional force as they impact)

    And that he says even though the Kinetic Energies are different, force is the same, which to me is a competely violation of laws of physics and work/energy principle.

    His definition of Force is complete rubbish too. But its impossible to get him to realize it, so I offered to take it to a physics forum. Eventually he will show up for this thread as "Cydoniaquest"
     
    Last edited: May 7, 2007
  6. May 7, 2007 #5

    Hootenanny

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    I'm not commenting on buildings collapsing, but with respect to force of impact I agree with you. Cydoniaquest is mistaken. There are however, some small error(s) in your analysis, but these do not invalidate your point.
     
    Last edited: May 7, 2007
  7. May 7, 2007 #6

    robphy

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    Does that mean that
    this person wouldn't care if a brick
    rested on his head or fell from a second-story window onto his head?

    (At some point, you might want to discuss the idea of "impulse".)
     
    Last edited: May 7, 2007
  8. May 7, 2007 #7
    :rofl: Your friend is not too bright. You should demonstrate this point to him by dropping a brick on his head! Survival of the fittest.
     
  9. May 7, 2007 #8
    I would hope my ideas are valid even if i do word something wrong, and if I was wrong in something, I would want to learn where I was wrong and to better my understanding.

    A couple years ago we had a huge argument trying to fix his understanding of F= MA and well we had no luck as you can see by his force definitions.

    Hopefully we will stop in like he said he would, and maybe, just maybe, he will learn something.
     
  10. May 7, 2007 #9
    I’m not making that claim. You simply did not comprehend my point. I’m saying that the bowling ball in freefall has the same force acting upon it as in rest. (Force of gravity).

    I am also saying (and I realize this may be controversial) that, at the moment of impact (contact)….in that frozen moment in time, there is no force, because force can only truly be measured, BEFORE and AFTER impact, NOT at the moment of impact. In that moment, we have instantaneous velocity and therefore Kinetic energy transferred as per the conservation of momentum principal, but we do not have change in velocity or change in time, because we are looking at a frozen moment in time. Therefore, NO Force is present at the moment of impact.

    So, what I said to you was that the term, Impact Force was a bit of a misnomer….At the moment of impact, we have no actual Force because we have no change of velocity or change of time in that frozen moment.

    Force, then, can only truly be measured as an ACTION, and REACTION, before and after the moment of impact….

    I am claiming that what physics calls an impulse force is actually a force of REACTION….AFTER actual impact takes place.

    I’ll continue with your example of a car hitting a proverbial immovable brick wall;

    Let’s say a car, traveling at constant velocity impacts with the brick wall, and the wall doesn‘t react, while the car absorbs the full energy of the impact…Your question to me is; Find the impact force….. Well, I think we have to define that term a bit. Are we talking about Net force of the action/reaction pair as a whole? OR are we talking about determining reaction force acting upon the mass of the car after impact?

    Let me further clarify my meaning by another example, then I’ll return to the car and the brick wall….

    Let’s say we have a brick at rest on a table. Is there a force acting upon that brick? Well, yes, of course, the force of gravity is acting upon the brick’s mass giving us the value for Weight. BUT….what’s the Net Force of the brick? I maintain that the Net Force is 0, because Net Force is a sum of all force vectors acting in the system. The downward force of gravity acting upon the mass of the book is being directly countered and offset by the opposing upward force of the table….0 Net force then is the resultant…

    So back to our car into the immovable brick wall…I’m saying that, if the collision is perfectly elastic, (in other words, the car sticks to the wall after impact)….we have a 0 net force situation.

    Now, it’s a different story if we are to attempt to determine the individual force component caused by deceleration acting upon the mass of the car as it impacts the wall…That would be like figuring the weight of the brick at rest on the table.


    If THIS is what we mean by Impact force, then what we are really talking about here is finding the individual reaction force vector component, rather than Net resultant force…of the system as a whole.

    To further clarify…let’s slow down the car, in our minds at the moment of impact…the moment the front bumper first makes contact with that wall. Everything that happens AFTER that point is due to a reaction force vector. Meaning…if the car is traveling due East upon impact with the wall, it is suddenly being countered by a reaction force vector in directly the opposite direction, due West…

    As the car begins to crumple after the moment of impact, it is decelerating, which is a negative acceleration (in other words, an acceleration vector in the opposing direction from the car’s original direction of travel, in other words, a REACTION force).

    Now here’s my problem with this scenario…Cause and effect. The reaction force is Caused by the wall AFTER impact of the car. It is acting upon the car causing a deceleration which results in a force imparted to the car…NOT a force imparted to the brick wall.

    After all, the brick wall is not the mass that is decelerating in this scenario. The car is decelerating…Therefore, I don’t think we can say the car is imparting a FORCE to the wall, given that the impact does not result in an acceleration of the wall. F= ma, therefore if the wall does not accelerate after impact, there is no Net Force. There is Kinetic Energy imparted to and absobed by the wall upon the car's impact, but no Net Force, because the wall doesn't move.

    If it did, then we could calculate a Net force imparted by that resulting acceleration. In this scenario, however, the wall doesn’t accelerate after impact.

    In this scenario, the CAR decelerates, (negative acceleration component) and therefore the Force is acting upon the car in a westerly direction, NOT the wall, but in the direction opposite the wall. Again, if the force was acting on the wall, we would have acceleration of the wall in the Easterly direction after impact.

    So maybe it is a question of semantics here...and that's all I am really arguing.
     
  11. May 7, 2007 #10

    russ_watters

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    Sorry, guys, but this is a banned topic. The conspiracy theory mindset is such that it is utterly pointless to try to debate with such people and entertaining such discussions only brings crackpots oud of the woodwork. Needless to say, msu, the crackpot you are discussing this with (whom we now seem to have attracted...) doesn't have a clue about physics or engineering. My advice: don't waste your time.

    Sorry.
     
    Last edited: May 7, 2007
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