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Calculation of force from deceleration?

  1. Sep 30, 2011 #1
    Hello people

    In this thread I am interested in how to calculate Newtons or force from a falling object being stopped by someone or something such as a machine.

    In my example below is a scenario-

    A 6000 ton ball of iron moves from directly above this person who is roughply 2 meters tall and quite broad. Its moving at 11 meters/second and travels this 11 meters before hitting this person, who decelerates it to barely moving, its on their back and balanced with them straining, sweat running down their face etc.

    What is the calculation to find out how much force in newtons was required to stop the ball moving?

    I assume its similiar to Force=massxacceleration? but instead of accleration is it deceleration? Is it more complicated that this?

    For example, does the mans own mass, weight etc play a role in absorbing some of the iron balls own force, therefore reducing the force he requires to stop it with his own strength? I am curious on how to calculate this, please help.

    Thank you.
  2. jcsd
  3. Sep 30, 2011 #2
    There is no difference.
  4. Sep 30, 2011 #3


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    To a physicist -- and in the "mass x acceleration" formula you wrote -- acceleration refers to any rate of change in velocity. Deceleration is just a special case of acceleration, where the speed of the object is decreasing.
  5. Sep 30, 2011 #4

    Ken G

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    It is important to keep the bodies separate. If you want to know the force on the falling mass, just consider the falling mass. Write F=ma just for that mass-- it doesn't matter where that force comes from. Note the problem you are describing requires more information-- F here is the force from the man upward, minus the force of gravity downward, but the difference only tells you how rapid the speed comes to rest-- so you also need to know how long that takes to solve for the force needed.
  6. Sep 30, 2011 #5
    You need to know how long it took your oddly-phrased broad, straining dude with sweat running down his face to stop the ridiculously heavy mass that would in reality squash him to reddish-pink pulp. F = dp/dt.
  7. Oct 1, 2011 #6
    Ah I see thanks guys, ok the time it takes for him to stop this moving mass is about 0.5 seconds.

    So what is the calculation, F=massxacceleration? How do I get acceleration from the deceleration, I am probably just confusing myself but heres the information you guys may need written in summary;

    6k ton ball of Iron moving at 11 meters a second for 1 second dropping down on;
    2 meter tall man who is now straining/stooped low so maybe he went down a meter or so? and stopped its movement in 0.5 seconds, so it fell on him and it stopped almost straight away.

    So I want to know the force the man required to stop it so hopefully cleveerr people than I can at least give me a calculation, preferably not in abbreviation but in full e.g. Force=MassxAccleration rather than F=mx etc etc. Then maybe I can find the Force he applied to catch the moving object.

    Also do I add gravity at some point? or is gravity taken into account with the Iron balls downward movement speed?

    Thank you in advance. I am sorry if the example is strange but what I am trying to caculate in effect is the force in newtons the man is excerting to stop it, and I am questioning how force is calculated differently if at all, if deceleration is present rather than base acceleration.
  8. Oct 1, 2011 #7
    v=v0+at says (v-v0)/t = a, and F=ma says F = m(v-v0)/t. Which is precisely F = dp/dt. 6000 kg mass, 5 seconds, and speed difference is 11 m/s minus 0 m/s = 11 m/s, so that's 330,000 newtons. What that number means to you, I don't know, but that's how you calculate a force given the information provided.
  9. Oct 1, 2011 #8
    I see, thank you. I dont understand all the abreviations, but I am probably just slow. If you dont mind can you write it out in full?

    So, F= Force? V=Velocity etc? and the mass was 6000 tons, so 5 443 108.44 kilograms.

    sorry to bother you further but again, I am not too savvy but I sort of understand what your doing if that makes sense. Thank you for your work sir.
  10. Oct 1, 2011 #9


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    F = force
    m = mass. Can also be an abbreviation for meters.
    a = acceleration
    v = (final) velocity
    v0 = initial (starting) velocity
    t = time
    s = abbreviation for seconds
    m/s = abbrev. for meters per second
    m/s2 = abbrev for meters per second-squared

    Here is an example problem; you may of course substitute your own numbers instead:

    If the velocity were initially 51 m/s, and the object comes to rest then it's final velocity is 0 m/s. That's a change in velocity of 0-51 = -51 m/s.

    If that change occurs over a time of 3 s, then the acceleration is

    a = (v - v0) / t = (-51 m/s) / (3 s) = -17 m/s2

    So the acceleration is -17 m/s2. Or you could say the object decelerates at a rate of 17 m/s2.

    I can think of two ways to calculate the force:

    1. Note that 17 m/s2 represents 17/9.8 = 1.7 times the acceleration due to gravity, or 1.7 g's. Therefore, the force on a 2.0 ton mass would be 2.0·1.7=3.4 tons of force. You may convert that force to Newtons.

    Or 2. Convert the object's mass to kg, and use F=m·a to calculate force in Newtons.

    I suspect it's best to leave out calculus-level explanations here.
  11. Oct 1, 2011 #10
    Thank you Redbelly, I gave it a go below using my numbers, do you think I did this correctly=

    So, 11m/s , final velocity is 0m/s. 0-11

    a = (v - v0) / t = (-11 m/s) / (1 s) = 11 m/s^2

    Acceleration -11m/s^2 or deceleration 11 m/s^2

    Now a try at your two methods of finding Newtons;

    1. 11/9.8= 1.12 g's so;

    6000 ton mass multiplied by 1.12= 6720 tons of force. Not entirely sure how to turn that much tons of force to newtons? Using a calculator online which sounded like it can do the work gave me 65900688 Newtons, is this right?

    2. objects mass in KG is 5 443 108.44 multiplied by 1.12= 6096281.4528 newtons?

    I feel I may have done some things wrong but I get a better idea of what I am doing.
  12. Oct 2, 2011 #11


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    That's fine if it taks 1 s to decelerate the object. But you had said 0.5 s in an earlier post...

    Yes, if "ton" means a metric ton or 1000 kg.
    Yes, if "ton" means an imperial ton (2000 lbs or 907 kg).

    You're pretty much on the right track. Double check how much time you want to object to be stopped in (0.5 s or 1 s or other?). Also whether you want to go with metric or imperial tons, which changes the answer by about 10% -- not a big difference.
  13. Oct 2, 2011 #12
    dp/dt means change in momentum (momentum = mass*velocity) divided by change in time, which is how long it takes to stop. Mass isn't changing, so dp = m*dv. I'm too lazy to bother using the greek delta symbol instead of "d", but d generally means the limit of an infinitesimal change (calculus) and delta means the change you're talking about, i.e. velocity goes from initial to final in a specified period of time. It's just another way of writing the same formula.
  14. Oct 2, 2011 #13
    Jeff would you also agree with what I stated? Above? I think 60 million newtons sounds good and close enough to what I was thinking before I did it. Maybe ill change the numbers around a bit to make them as accurate as I want but if everyone agrees this is the right calcluation for this method then I thank you all.
    Last edited: Oct 2, 2011
  15. Oct 2, 2011 #14
    I misread your numbers earlier, but:
    6000 tons = 12 million pounds = 5.45 million kilograms = m
    11 meters/second = delta-v
    0.5 seconds = delta-t
    Therefore force = 120 million newtons.

    I have no idea why this number is important, but that's what I get using your values above.
  16. Oct 2, 2011 #15
    I see, well that makes sense. Using 0.5, my calculation was 1 second instead so using 0.5, it makes sense on the 120 million newtons.

    Thanks everyone, now I have some understanding on the matter I can calculate it myself. If I get into any problems I may return.

    Edit- Looking back I got 6 million newtons didnt I? Does it happening in 0.5 seconds deceleration from 1 second really change it by over 100 million newtons or did I get the calculation wrong then in the second calc?
    Last edited: Oct 2, 2011
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