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Impact Force vs. Weight

  1. Nov 19, 2009 #1
    I am trying to calculate the difference in force between a body standing still on the earth's surface and that same body falling from a distance of 2.7 meters (approx 9 feet).

    Specifically, I am trying to determine what percentage increase in force is attained by this fall.

    Thank you.
  2. jcsd
  3. Nov 19, 2009 #2
    Do you want to find like the precentage difference on the amount of force at 2.7 meters compared to the ground?
    well you need two things for this problem, the radius of the earth and the mass of the earth. G, the acceleration due to earth is approx. 9.8m/s2 at close to the surface of the earth. So what you do is you find out the acceleration at your height times your mass and that will be your force for the height. Then you have both forces so you should be able to do the precentages. do you know the formula for gravitational attraction?
    Last edited: Nov 19, 2009
  4. Nov 19, 2009 #3


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    welcome to pf.
    this is trickier than you think.

    When you are standing on the earth there is a force (your weight) pressing down on the ground through your feet. This is easy to work out, it only depends on your mass, the mass of the earth and its size. Or in simple terms, your mass in kg * 9.8 gives the force in Newtons.

    When you impact something the force depends on how quickly you stop, which depends on the surface - which is why landing on concrete is more force than landing on rubber. The rubber bends and slows you down over a longer period of time so less de-aceleration and so less force.
    it's very difficult to calculate what this slowing down rate would be for different surfaces - you normally have to just measure the force by dropping something
  5. Nov 19, 2009 #4
    Ok. So, for example, if the mass is 60kg this would give me a force of 588 N (60kg X 9.8 m/s/s). Is this correct?

    If so, assuming that there is zero deceleration (hypothetically at least), what would be the impact force, in newtons, of that same 60kg object after falling from 9 feet or 2.7432 meters?

    Thank you.
  6. Nov 19, 2009 #5


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    Yes, that's the weight - in physics mass and weight aren't the same thing.

    Two parts to this, first of all how fast is something dropped from 9ft going when it hits the ground
    The equation for this is: final speed ^2 = initial speed ^2 + 2 g distance
    So in this case 7.3 m/s - Notice how this doesn't use the 60kg - all objects fall at the same speed, assuming no air resistance, if the object you are dropping is a feather it's going to be going slower than an a cannonball.

    Now the force when it stops, force = mass * acceleration, and acelaration is speed/time
    (in physics acceleration means slowing down as well as speeding up)

    So if you think it takes 0.1 second to stop (imagine it had fallen into soft foam) then the force is
    F = 60kg * 7.3/0.1 = 4380N

    But imagine it fell onto concrete and took 1/1000 of a second to stop (because concrete is much harder)
    then the force would be 100x greater.
    if you assume it took no time at all to top then the force would be infinite!
    Last edited: Nov 19, 2009
  7. Nov 19, 2009 #6
    I dont know what force percentage there's on the fall, Im looking up for that one, but I can tell you that there's a great actual difference like explained before.
  8. Nov 19, 2009 #7
    the impact force, meaning the force that the object will be pushing on the ground, will be esentially the same. 588 N. Im not sure what you are asking but when you talk about collisions the words Momentum and Impulse are concerned.

    Do you want to know the Impulse of the collison or the Momentum of the object the instant it hits the ground?
  9. Nov 19, 2009 #8
    Yes, that's the weight - in physics mass and weight aren't the same thing.

    Wow! Ok, so the deceleration makes quite a bit of difference. But, now that I understand, it makes complete sense per the equation.

    Let me offer a few more details to my hypothetical to get to the answer I truly seek

    Suppose that the 60kg object is a person. The person is on a pogo-stick, which works on a spring mechanism to keep the bottom of the stick protruding out of the bottom. When additional force is applied, the spring compresses. 14462d1229061454-show-me-picture-pogo-stick.jpg .
    Additionally, assume that the length of travel of the stick and spring in approximately 17 inches (or approx .4318 m), before the stick would "bottom out."

    How much force would the spring need to be "pre-charged" with to prevent the stick from bottoming out? Or, said another way, how much force must the spring have to recoil just before bottoming out.

  10. Nov 19, 2009 #9


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    Easiest way is conservation of energy.
    When you are up in the air, potential energy = mass * g * height change

    Moving this becomes kinetic energy (don't worry about this)

    Because when you are stopped at the bottom all of this KE has become spring energy = 1/2 k X^2
    Where k is the spring stiffness and X is the compression of the spring.
    So simply:
    m g h = 1/2 k x^2

    if you don't have 'k' for the spring (it's not going to be written on it) you can use that the spring is Force = k x
    So put a known force (ie. the 60kg person) on the pogo and measure the change in length (= x)
    Last edited: Nov 19, 2009
  11. Nov 19, 2009 #10
    Ok so here we can use Conservation of Energy. Lets say it is at 2.7m the starting height, as stated before. So you say PE(potential energy)=SE(stored-spring energy) The equation for PE is Mgh, where M is the mass, g is the accel. due to gravity, and h is the height. Equation for SE is 1/2kx2 where k is the spring constant and x is the distance the spring is compressed/streched. So lets plug in all the values. Mgh=1/2kx2, (60kg)(9.8m/s2)(2.7m)=1/2k(.4138m2)
    now when you plug all that in and do some simple algebra you get a value for k, which is around 18,600 N/m. Which by the way is really strong ha ha.
    With the spring force at that, the pogo stick will compress the full 17 inches and come to a stop before stretching once more.
    That is the answer for which you desire. But I hope you understand it. Springs dont really have a "pre-charged" force assosiated with it. if you think about it, the force that it pushes back or if you stretch it pulls back with, depends on how much you compress/stretch it.

    But this is falling away from your original statement. If you still have questions about it let me know
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