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Bear on a bench

  1. Feb 19, 2014 #1
    I'm sitting on a bench in a park. I exert a force of H newtons and the bench exerts a reactive force of H newtons back and we're in equilibrium. Next to me a 10ft tall short-faced bear sits on an identical bench with B newtons. After about 20 seconds the bench gives in and collapses.

    I postulate that the bench could not exert enough reactive force to match the weight of the of the bear and so consequently collapsed. To show this, I climb a tree knowing that I will impact the bench with the same force as when I was sitting on it, thus the bench should not collapse. The bench collapses.

    Flustered and confused by this, I investigate further. I find another non-broken bench, and climb up an even higher tree where I know I will reach terminal velocity before reaching the bench. Upon contacting the bench, there is no net force acting down on the bench yet it still breaks.

    Even more confused and in alot of pain, I then think it may be the kinetic energy or momentum that caused the benches to collapse. However, the mass of the bear had no velocity, so how can it have had kinetic energy or momentum? And indeed, how could there have even been a force that the bear exerted on the bench when the mass had no velocity and therefore no acceleration?

    Seeing that the park officers are distracted by the 10ft tall bear, I take the opportunity to sneak away avoiding vandalism charges but still left very confused and injured...
     
  2. jcsd
  3. Feb 19, 2014 #2

    collinsmark

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    What a bizarre homework problem. :uhh:
     
  4. Feb 19, 2014 #3

    collinsmark

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    Seriously though...

    You have mass, and during the short time when you are touching the bench before it collapes you decelerate. That means you have acceleration in the "up" direction.

    What is Newton's second law of motion?

    Same as above. When you initially touch the bench there is acceleration involved. That's true even if you were previously falling with a constant velocity. Once you hit the bench your velocity is no longer constant.

    That's an excellent way to approach the problem! :smile: Do a little research on a physics term called "Impulse." Impulse is force times time. It's also the change in momentum.

    You are in contact with the bench for short interval of time. Although this interval of time is short, it is not zero. During this interval your momentum changes. With that you can calculate the average force due to change in momentum.

    Add that to your weight and you have the final answer you were looking for.

    :smile:
     
    Last edited: Feb 19, 2014
  5. Feb 20, 2014 #4
    Sorry I'm being thick and I can't understand your explanation :uhh:
     
  6. Feb 20, 2014 #5

    collinsmark

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    I'll try to be more specific.

    Why would you think that the bench would exert the same force when falling on it? It wouldn't. It's force would not only counter your weight, but in addition it would also cause you to accelerate (albeit for a short interval of time). Thus it would exert more force than it would if you were simply sitting on it.

    No net force? Of course there is a net force. The force from the bench is causing you to accelerate (albeit for a short interval of time). Same as above.

    When the bear starts to fall through the bench, it accelerates. The net force is not quite zero. The magnitude of the normal force from the bench is slightly less than the bear's weight.

    Newton's second law of motion.

    The normal force from the bench is not quite that of the bear's weight. The bear's momentum changes (from 0 to something not quite 0). Net force times time equals the change in momentum.

    Even before the bear fell through the bench, and was just sitting there before the bench broke, the normal force from the bench was equal and opposite the bear's weight (until the bench breaks).

    Newton's third law of motion.
     
  7. Feb 20, 2014 #6
    The tension in whatever the bench was made of when the bear sat on it, was too big for the material to handle it.
    If you stand on something fragile, such as glass, it might not break if your Weight isn't big enough. If you were falling on the same glass, it now might break, because your weight is bigger than before (weight is not the same as mass). Take a hammer and set it on the glass, nothing happens. Smash the glass with the hammer, the glass breaks.
    Different material, but similar idea - the material breaks due to overload.
     
  8. Feb 20, 2014 #7
    I had always believed that whatever height you fell from, you would always hit the ground with same H newtons because force is mass x acceleration therefore (if there is no air resistance) your acceleration g is constant, your mass is constant so your force would be the same from whatever height. If the force that is your weight (in a physics context) is just your mass x gravitational acceleration my weight would be the same regardless of the height I fell?

    So for the bear's bench, we are saying that for the first 20 seconds the bench's reactive force is the same as the weight of the bear. But over time is something happening to the molecular bonds or fibres of the material that is weaking the reactive force? Is there some sort of energy exchange happening here?

    When I am impacting the bench at terminal velocity, is it correct to say that there is in fact a downwards force acting on the bench not a zero net force like when I was in the air because simply, there is no air anymore, once I contact the bench, to give air resistance?
     
  9. Feb 20, 2014 #8

    haruspex

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    The question of "how much force is there on impact" is a regular on this Forum. The short answer is, you can't know. Going from falling to stopped in a very short time implies a very large upward acceleration, much more than g. Compare the time spent falling with the time spent stopping!
    An impact is a sudden change of momentum. The relationship between force and momentum is that momentum change is the integral of force over the time for which the force is applied: I = ∫F.dt. The height you fall from determines your impact speed. Since you must come to rest, that multiplied by your mass gives the change in momentum. To determine a force we need to understand how the force will vary over time.
    In reality, all impacts involve an initial elastic phase, perhaps followed by a plastic deformation when the elastic limit (a maximum force) is reached, then falling to zero when rupture occurs. That is, the force increases linearly from 0 up to a maximum, then may stay constant for a while, before becoming zero. Or it may survive the plastic phase and enjoy a return to elastic as the force declines. The momentum change is the area under this graph.
    For a wooden bench, it will be reasonably elastic up to the point of failure. Any plastic phase would be very short. So the graph looks like a triangle, the right hand side being vertical.
    If we knew the modulus of the bench and impact momentum we could compute the slope and the maximum force. Rest assured, it will be a lot more than mg.
     
  10. Feb 20, 2014 #9

    collinsmark

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    Well, the acceleration due to gravity, g, can be considered constant. But that's not the only force involved!

    The force of gravity acting on your body is the same at any height (within reason -- we're assuming that you always remain relativily near the surface of the Earth). And the force of gravity is the only force involved when you are in perfect freefall. But once you hit the ground there are other forces involved on your body.

    The force of gravity is constant, yes. But that doesn't mean the other forces are constant.

    Let's examine Newton's second law. The sum of all forces equal to mass times acceleration.
    [tex] ma = \sum_n F_n [/tex]
    Now, let's assume that you fall out of a tree that is 5 meters high. We can calculate the velocity of that which you will hit the ground, using [itex] v_f^2 = v_i^2 + 2as [/itex]. If we roughly assume g = 10 m/s2, we calculate that landing velocity to be [itex] v_f = \sqrt{0 + 2(10 \ \mathrm{m/s^2})(5 \ \mathrm{m})} = 10 \ \mathrm{m/s} [/itex]

    Now let's assume that from the time your toe first touches the ground until you finally stop moving (let's assume you don't bounce) happens in 0.2 seconds. The average acceleration for the impact with the ground can be calculated using
    [tex] a_{ave} = \frac{\Delta v}{\Delta t} [/tex]
    or in this case, [itex] a_{ave} = \frac{(10 \ \mathrm{m/s} \ - \ 0)}{0.2 \ \mathrm{s}} = 50 \ \mathrm{m/s^2} [/itex]

    Now, back to Newton's second law of motion. There are two forces involved. There is the normal force from the ground and the force due to gravity.
    [tex] ma_{ave} = F_n - F_g [/tex]
    [tex] = F_n - mg [/tex]
    [tex] m(50 \ \mathrm{m/s^2}) = F_n - m(10 \ \mathrm{m/s^2}) [/tex]
    [tex] F_n = m(60 \ \mathrm{m/s^2}) [/tex]

    So in this example, the average, impact force on the ground is 6 times the force that it would be with you just standing there. Similarly, due to Newton's third law of motion, the same applies to you from the ground.

    Yes, very good. :approve:

    Yes.

    It's not so much that there is no air anymore. The extra force comes from the fact that when you hit the bench, it causes you to accelerate "up" during the impact. And that extra force is also felt on the bench itself due to Newton's third law of motion.
     
  11. Feb 20, 2014 #10
    Ah thank you!! So deceleration in one direction and acceleration in the opposite direction are exactly the same thing! And this reactive force is decreasing as the object comes to rest. So this reactive force isn't just reacting to mg but also the speed so that it can 'know' when to decrease to mg newtons?
     
  12. Feb 20, 2014 #11

    collinsmark

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    Yes. they are the same thing. :approve:

    Following the same reasoning, you might have noticed that I put a minus sign on my mg, for the gravitation force, when I gave the falling to the ground example. The convention I was using is that "up" is positive. Since the force of gravity points down, I put a negative sign on the term to indicate direction.

    Suffice it to say that it changes as the object changes acceleration.

    As haruspex points out in a previous post above, the actual acceleration and time of impact are not so clearly defined in most practical situations.

    I made many assumptions in my example, such as the impact taking exactly 0.2 seconds, the fact that no bouncing happened, etc.

    But whatever the case, yes, the force changes as the acceleration changes.

    In the case of static equilibrium (such as when you or the bear are just sitting on the sturdy bench), the sum of all forces equal zero. In other words,
    [tex] 0 = \sum_i F_i [/tex]
    [tex] 0 = F_n - mg [/tex]
    But in the case where things are not in equilibrium, the sum of all forces is not zero. it's ma.
    [tex] ma = \sum_i F_i [/tex]
    [tex] ma = F_n - mg [/tex]
     
    Last edited: Feb 20, 2014
  13. Feb 21, 2014 #12
    That's very interesting that we don't know exactly how the object decelerates over time upon impact because not knowing that means we can't accurately determine if the bench will break or not. More specifically, I'm thinking in terms of energy where the bench has x amount of time to give a strong enough average reactive force that the object will come to rest quickly enough before its energy (1/2mv^2) translates to enough downward distance (Work = force x distance)through the bench (like Hooke's law) and causes a big enough crack to break the bench.

    Am I thinking right? If so, I'm now thinking how it would look on a time-distance graph. I'm falling at g and we get the curve flicking up. Then we have the reactive force of the bench which, for simplicity's sake let's can't exceed mg. If the forces are equal, we get a straight line with the gradient I impacted the bench with. However the distance is still progressing so the bench breaks! Of course, if the bench didn't break the reactive force is greater but decreasing so we get a new curve towards a horizontal line and it is the distance travelled in the time of that curve that is critical for saying how much of a dent or crack we make in something.

    Thanks again for everyone's posts!
     
  14. Feb 21, 2014 #13

    haruspex

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    Yes, energy can be a useful way to approach it. The energy an object can absorb in the elastic phase is called its resilience; the energy that it can absorb before failure is its toughness.
    But you have to be careful with impacts at high speed. If the impact does not give time for the whole object to deform, you might punch a hole through it with less energy than that.
    Yes.
     
  15. Feb 23, 2014 #14
    So because we can't tell what the reactive force is at a time during the impact, the manufacturer of the bench would find it difficult to calculate how high I could let N newtons land on the bench before it breaks (without experimental testing).
     
  16. Feb 23, 2014 #15

    haruspex

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    It could be calculated from known properties of the wood and dimensions of the bench.
     
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