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Old Dec 30th 2013, 02:41 PM   #1
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Red face Force displacement acceleration problem

First time using a forum so apologies if there is any wrong doing.

I can't seem to find a solution to this problem.
If F=MA then why does a fall from 10 meter ladder exert more force on a body than a fall from a 1 meter ladder. Considering that the acceleration on earth is the same for both falls from the ladder and the mass is the same. If F=MA then how is a greater force created, I am assuming it has a lot more to do with energy but can't get my head round it in a mathematical theory way.
Does acceleration increase with distance hmmm??? Please help!

I'm sure there is a fundamental principal that I have failed to consider here and I would be most obliged if someone could point it out to me.

Kind regards.
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Old Dec 30th 2013, 05:29 PM   #2
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In this inelastic collision it is the impulse of the force: mv / delta t .
Also called the change of momentum. And the area under a force / time curve
Impulse of force : integral t1,t2 F dt = delta P = mv2 - mv1

Last edited by morrobay; Dec 30th 2013 at 05:37 PM.
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Old Dec 30th 2013, 05:30 PM   #3
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Originally Posted by Bringitondown View Post
First time using a forum so apologies if there is any wrong doing.
Welcome to the forum. You're doing just fine.

Originally Posted by Bringitondown View Post
I can't seem to find a solution to this problem.
If F=MA then why does a fall from 10 meter ladder exert more force on a body than a fall from a 1 meter ladder.
The force of impact when hitting the ground is related to something called impluse which is defined as the the product of the average force multiplied by the time it is exerted. It's not due to the force of gravity whatsoever. Gravity only plays a role as to who the object obtained the momentum it has when it stikes the ground. The faster you're moving when you hit a solid object the greater your momentum p was just before impact. That means that to stop cold there must be a larger change in momentum to stop the body then when it's moving slower. However that change of momentum depends on the details of impact and are diffcult to calculate.

I don't know what level your education is so you might not be familiar with the definition of force as F = dp/dt = time rate of change of momentum.

Think of hitting a wall moving in on a skate board when you're moving at constant speed not subject to any any force whatsoever (the net force is zero). But you're still subjected to a force when you hit the wall, right? Otherwise you'd be unable to stop.

Originally Posted by Bringitondown View Post
If F=MA then how is a greater force created, I am assuming it has a lot more to do with energy but can't get my head round it in a mathematical theory way.
No. Not energy, momentum.

This is related to impulse. See details at http://en.wikipedia.org/wiki/Impulse_(physics)
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Old Dec 31st 2013, 02:36 AM   #4
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Smile Thank you

Thank you all for your reply's. crystal clear answers thanks, I am now just going to go and dig out my linear impulse and momentum notes now that I know where to look. I only studied this stuff last year. Memory of a goldfish.
Thanks again, great job.
Sean.
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Old Dec 31st 2013, 06:42 AM   #5
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Originally Posted by Bringitondown View Post
Thank you all for your reply's. crystal clear answers thanks, I am now just going to go and dig out my linear impulse and momentum notes now that I know where to look. I only studied this stuff last year. Memory of a goldfish.
Thanks again, great job.
Sean.
You're very welcome. Glad to be of service!
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Old Dec 31st 2013, 05:53 PM   #6
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All you need is algebra here to solve this problem for the impulse of the force
from a 10 meter fall : displacement = y = vot + 1/2gt^2
And v = vo + gt . Falling from rest disregard vo and g = 9.8m/sec^2
So with t = v/g then you can solve for v in displacement equation.
Then plug into impulse of force = mv/delta t
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Old Dec 31st 2013, 07:20 PM   #7
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Originally Posted by morrobay View Post
All you need is algebra here to solve this problem for the impulse of the force
from a 10 meter fall : displacement = y = vot + 1/2gt^2
And v = vo + gt . Falling from rest disregard vo and g = 9.8m/sec^2
So with t = v/g then you can solve for v in displacement equation.
Then plug into impulse of force = mv/delta t
That's incorrect. You have the wrong idea about impulse. The amount of impulse experienced by a body that strikes a surface requires more knowledge than simply the speed of the body when it first comes into contact with that surface. To be able to determine the impulse requires knowledge of the force experienced by the object as a function of time.

The expression you gave for time above tells us nothing. As you've defined it, t is the length of time the body was in free-fall. I.e. according to what you wrote the velocity as a function of time is v = gt so that the time the body took from falling from rest at constant acceleration to obtain the speed v is t = v/g. Your comment
So with t = v/g then you can solve for v in displacement equation.
Then plug into impulse of force = mv/delta t
makes no sense. The velocity does not appear in the displacement equation. Even if it did it wouldn't be of any use.

Consider a steel sphere falling from rest at a distance of 10 m onto a rubber surface. Compare that to the same situation but hitting a steal floor instead. The force experienced by the sphere is much lower in the case of the ball falling on the rubber floor. The numerical value of the impulse is simply the difference in the momentum before and after the collision.

Consider the expression F = dp/dt and solve for p. This gives dp = F dt. Now integrate to obtain

delta p = p_final - p_initial = Integral(t_1, t_2) F dt

To know delta p you have to know F(t) as a function of t. That depends on the material of the object and the material the floor is made of.
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Old Jan 1st 2014, 12:52 AM   #8
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Originally Posted by Pmb View Post
That's incorrect. You have the wrong idea about impulse. The amount of impulse experienced by a body that strikes a surface requires more knowledge than simply the speed of the body when it first comes into contact with that surface. To be able to determine the impulse requires knowledge of the force experienced by the object as a function of time.

The expression you gave for time above tells us nothing. As you've defined it, t is the length of time the body was in free-fall. I.e. according to what you wrote the velocity as a function of time is v = gt so that the time the body took from falling from rest at constant acceleration to obtain the speed v is t = v/g. Your comment

makes no sense. The velocity does not appear in the displacement equation. Even if it did it wouldn't be of any use.

Consider a steel sphere falling from rest at a distance of 10 m onto a rubber surface. Compare that to the same situation but hitting a steal floor instead. The force experienced by the sphere is much lower in the case of the ball falling on the rubber floor. The numerical value of the impulse is simply the difference in the momentum before and after the collision.

Consider the expression F = dp/dt and solve for p. This gives dp = F dt. Now integrate to obtain

delta p = p_final - p_initial = Integral(t_1, t_2) F dt

To know delta p you have to know F(t) as a function of t. That depends on the material of the object and the material the floor is made of.
Maybe a misunderstanding: My delta t is the impact time, ie .1 sec
From a 10 meters free fall 1.42 sec is time to impact.
Im sure I dont have to do the algebra when substituting v/g for t in displacement and solving for v.
v^2 = vo + 2ax
And this is v at impact. 14m/sec. Since object fell from rest the evaluation
of integral = mass * 14m/sec/.1sec
And since there is nothing about materials in OP Im assuming an inelastic collision with hard surface.

Last edited by morrobay; Jan 1st 2014 at 12:56 AM.
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Old Jan 1st 2014, 03:56 AM   #9
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Originally Posted by morrobay View Post
And since there is nothing about materials in OP Im assuming an inelastic collision with hard surface.
That's an invalid assumption and irrelevant to question. The fact is that there is insufficient information given to answer the question and thus any value of force calculated is something you added into the problem (in this case dela t) to find a value for force. In any case your calculation is wrong since you're missing a value of "2" in the expression you gave for force.

I don't see why you're assuming this is an inelastic collision. An elastic collision makes it easier to analyze.. In an elastic collision the kinetic energy is conserved. This has nothing to do with the hardness of the surface but with the nature of the material, i.e. does the material get deformed in some irreversal way? Is part of the energy changed into thermal energy upon impact? etc. In answering questions like this you have to try to put yourself in the place of the person asking the question. In this case the context tells us that Bringitondown was merely confused about the reason for the larger force of impact with speed. That's quite understandable. I'm sure we've all made that assumption at one time or another.

If he understood the problem he was really asking about, i.e. that it all has to do with impluse, then he wouldn't have even asked the question. As stated there's no way to calculate any kind of force. In anycase the expression that you gave force = mv/delta t is wrong.

I'll get back to this with more detail later. In the mean time please think of the collision as if it was a real one, not an artificial one. Model it as if the body interacting with the ground as if it was hitting a spring which becomes fully compressed and then releases, loosing no kinetic energy in the process in thus still being an inelastic collision. The hardness of the collision is then modeled as the stiffness of the spring.

I'll get back to this later when I have more time to give a more in depth answer.

Last edited by Pmb; Jan 1st 2014 at 07:52 PM.
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Old Jan 1st 2014, 11:54 AM   #10
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[I decided this would be much easier to explain with diagrams. See attached diagram. I gave two examples of the potential energy U(x) of two different surfaces.

All real surfaces have an effective potential associated with them that describes the surface. A soft surface has a potential energy function like the one on the left hand side of the diagram. Since force F is related to U by F = -dU/dx you can see that a particle hitting the soft surface will experience less of a force. If the particle has the same speed in each case then the depth the particle hits the surface depends on the kinetic energy since the particle won't stop until U = K. As you see this means that the particle penetrates deeper into the soft surface than the hard surface. However since the force is given in terms of potential energy the energy of the particle is conserved in each case making the collision with the surface elastic in each case. However please notice that this has nothing to do with the maximum force that the particle hitting the surface will experience.
Attached Thumbnails
Force displacement acceleration problem-surface.jpg  

Last edited by Pmb; Jan 1st 2014 at 07:12 PM.
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