How much time does it take to the floor to stop a falling book?

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sorry if the title is not clear, if you have a better idea i'll change it.

say i drop a 10kg book to the floor. when it reach the floor, the floor stops it by "activating" the Normal force on it...
my question is, how much time does it take to the floor to stop the book, and by this time how far does it get into the floor... and what's the magnitude of the normal force regarding those...
obviously that's probably depends on a lot of parameters but just try to give me an approximate answer with average parameters to the floor, the book, and everything else


if don't get it right just tell me..
 

Answers and Replies

  • #2
Drakkith
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Thats going to depend on the makeup of the floor and the book. If you drop it on concrete it will stop faster than it will on carpet.
 
  • #3
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I'm somewhat assuming here... But if the book WOULD have a higher normal force than gravity then wouldn't it bounce for a short time? I'm a level 3 physics student, so I'm just taking a stab in the dark here... And of course the density of the floor would be a factor
 
  • #4
AlephZero
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If you replace the book with a rectangular block (so you dont have to think about what happens to the individual pages if the book lands on its edge) and it landed flat on one face, the deceleration of the block would typically be in the range of 100G up to maybe 10,000G, depending on the materials of the block and the floor.

A typical deceleration used for modelling this type of impact is 3000G.

So if the block velocity was say 10 m/s before impact, the stopping time would be of the order of a few milliseconds, down to microseconds.

The large forces produced by the large decelerations explain why things break when you drop them, of course.
 
  • #5
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So the deceleration force is downward? I can see how it makes sense, objects come to an abrupt stop usually, but it seems crazy that it would decelerate at 3000G, or close to 10 000 m/s2? Sorry my understanding of this is minimal
 
  • #6
Drakkith
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So the deceleration force is downward? I can see how it makes sense, objects come to an abrupt stop usually, but it seems crazy that it would decelerate at 3000G, or close to 10 000 m/s2? Sorry my understanding of this is minimal
It depends on the velocity and the makeup of the floor and the falling object.
 
  • #7
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I think common sense could tell you that, but I'm wondering if there's a specific property or relationship which deals with the complete "stop" time of a falling object. The V2y of the object wouldn't be zero though, so I'm not sure really what could be used
 
  • #8
Drakkith
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I think common sense could tell you that, but I'm wondering if there's a specific property or relationship which deals with the complete "stop" time of a falling object. The V2y of the object wouldn't be zero though, so I'm not sure really what could be used
Hrmm. I'm sure there is, but I don't know the math behind it all.
 
  • #9
AlephZero
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So the deceleration force is downward? I can see how it makes sense, objects come to an abrupt stop usually, but it seems crazy that it would decelerate at 3000G, or close to 10 000 m/s2? Sorry my understanding of this is minimal
Here are a couple of examples that might convince you (randomly picked by Google).

Drop-testing of missiles, as a cheap way to produce the same accelerations as firing the missiles from a gun - about 20,000G. http://www.empf.org/empfasis/2010/September10/tech-tips-910.html

A company that does shock-testing of electronic circuit boards for cellphones, laptops, etc using, with a test machine that can produce accelerations of 5000G. http://www.dfrsolutions.com/drop-testing/

For comparison, assume about the pistons in your car engine are doing simple harmonic motion with a stroke of 0.1m at 6000 RPM (100 rev/sec). The acceleration =
0.05 x (2 pi x 100)^2 = about 20,000 m/s^2 = about 2000G.
 
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