How do I find the reduced force on an object due to springs?

[Assasinof6]

Homework Statement

This question is beyond my level, I am doing this for bonus marks. I am unaware of how to do it, I am merely requesting a formula, or a series of steps to help me. I have not yet done calculus.

Imagine that a 1 kg box is falling from a height of 5.2 m, with 4 springs attached to its bottom.
What is the reduced force on the box within the first instant of impact due to the springs?

Please make up various needed variables, such as the spring amplitude, if needed to aid in explanation.

Homework Equations

Unaware of EQn's.

The Attempt at a Solution

Lack thereof explained above.

 

[6Stang7]

I guess my first question is what level of physics are you in? You honestly have no clue of what equations to use?

 

[Assasinof6]

I'm in grade 11... I've tried multiple other forums (such as XKCD's) and sites (such as Answers.Yahoo.com), but either there was insufficient knowledge or bashing. I figure I'd give this forum a try.

As for the actual math, I am aware of its complexity and the fact that calculus may be necessary, but I still want to try.
So, to recap, I am trying to figure out how to go from a box dropping with springs on the bottom (energy equations, in this case mechanical energy eqⁿs), to the diminished force in the first instant of impact (engineering equations, like Hooke's law).

So, please explain the best you can, unless you feel like you're above being taxed by an angsty teen.

 

[6Stang7]

Ah ok. Well, I can tell you right now that you can solve this problem with just algebra if you use pre-determined equations (ie you don't derived equations).

Have you covered Hooke's Law and momentum in your physics class yet (I'm assuming you've done Newton's Laws)?

 

[Assasinof6]

Unfortunately we will not be covering Hooke's law nor other momentum questions.
So (based on your link), are you suggesting that I use the F = dp/dt equation in order to find the reduced force on the box due to the springs?

BTW, is the force itself reduced, or just spread out over a dimension of time?

 

[rock.freak667]

Please make up various needed variables, such as the spring amplitude, if needed to aid in explanation.

Do you have any information about the spring? Such as the spring constant?

 

[Assasinof6]

Because this is an actual physical assignment (an egg drop report), I am given no external data. How would I go about finding the spring constant?

 

[rock.freak667]

Because this is an actual physical assignment (an egg drop report), I am given no external data. How would I go about finding the spring constant?

For a spring, F=kx. So just place a force on the spring (such as a known weight) and measure the deflection 'x'.

 

[Assasinof6]

x is the displacement of the end of the spring from its equilibrium position. So, if I were to press down on the springs with the same force it experiences when hitting the ground, would I measure from where the top of the spring was to where it is now, or from the middle to where it is now? 1

And once I found x, would I divide negative F with x (-F / x) to find the spring constant? 2

FINALLY, once I found the spring constant, what would I do from there?3

Thanks for helping!

 

[6Stang7]

Ok, let's take these equations and see if you can solve it.
F=ma
F=kx
F=m(Δv)/(Δt)

where Δv is the change inl velocity Δt is the change time (thanks for the symbols rock.freak667! :D).

So, from here, what would you do next?

 

[rock.freak667]

x is the displacement of the end of the spring from its equilibrium position. So, if I were to press down on the springs with the same force it experiences when hitting the ground

You don't need to use the same amount of force. Just use standard weights, 1 kg, 2 kg, etc. So just rest the 1 kg weight on one spring.

would I measure from where the top of the spring was to where it is now, or from the middle to where it is now? 1

Where the spring is naturally, that will be your equilibrium position, when the weight is on it, measure from that point to where it is now.

And once I found x, would I divide negative F with x (-F / x) to find the spring constant? 2

FINALLY, once I found the spring constant, what would I do from there?3

From here now, you can use conservation of energy, initially it has gravitational potential energy, which would be converted to elastic potential energy (remember you have 4 springs).

 

[Assasinof6]

Assuming that:
Mass = 1 kg
Acceleration = 9.8 m/s^2
I know that F = 1 N.
Thus, if F = kx, and x is an assumed 0.02m, then:
k = 50.

so... now what? (and what unit is k measured in?)

 

[rock.freak667]

Assuming that:
Mass = 1 kg
Acceleration = 9.8 m/s^2
I know that F = 1 N.
Thus, if F = kx, and x is an assumed 0.02m, then:
k = 50.

so... now what? (and what unit is k measured in?)

Well you need to use the energy equation.

gravitation pe = elastic pe in 4 springs.

 

[Assasinof6]

oh, using the equation (1/2)kx^2, I can figure out the elastic potential per spring, if I discovered x using a 1kg weight on 1 spring. And from elastic potential to...

 

[Assasinof6]

Is the elastic potential equal to the mech energy halfway, or gravitational at the top?

 

[rock.freak667]

oh, using the equation (1/2)kx^2, I can figure out the elastic potential per spring, if I discovered x using a 1kg weight on 1 spring. And from elastic potential to...

The placing of the 1 kg on the spring was to get k.

Now you are dropping the mass, so you need to get new value for 'x' due it being dropped. If you are dropping it from 5.2m, then it has gravitational potential energy. (How do you find gravitation potential energy?)

EDIT:

Is the elastic potential equal to the mech energy halfway, or gravitational at the top?

At the top there is gpe, at the bottom there is only elastic. So equate them.

 

[Assasinof6]

Alrighty, so:

h = 5.2m
m = 1 kg
g = 9.8

E_g = mgh
E_g = 50.96
50.96 = (1/2)kx²
101.92 = 50x²
x = sqrt(5096).
x = 71.39

Now, what do I do? Do I work my way back to force, and find the difference?

 

[PhanthomJay]

What is the reduced[/color] force on the box within the first instant[/color] of impact due to the springs?

At the instant of impact, the spring deflection is 0. So what force does the spring exert on the mass at the instant of impact?

 

[rock.freak667]

Alrighty, so:

h = 5.2m
m = 1 kg
g = 9.8

E_g = mgh
E_g = 50.96
50.96 = (1/2)kx²
101.92 = 50x²
x = sqrt(5096).
x = 71.39

Now, what do I do? Do I work my way back to force, and find the difference?

Right well you have 4 springs, so the bold line should be 4*(1/2)kx².

That will give you the extension (well compression) all the springs undergo. Then if you just use Hooke's law, you will get the force each spring takes. Then multiply that by 4.

Though you are dropping it from 5.2 meters, which is quite high, so I am not sure if your springs will break or deform if the actual spring constant is too small.

 

[Assasinof6]

Right well you have 4 springs, so the bold line should be 4*(1/2)kx².

That will give you the extension (well compression) all the springs undergo. Then if you just use Hooke's law, you will get the force each spring takes. Then multiply that by 4.

Though you are dropping it from 5.2 meters, which is quite high, so I am not sure if your springs will break or deform if the actual spring constant is too small.

I will be using a parachute and crumple zone which will reduce the force dramatically, I just wanted to use a simple example to help introduce myself to this new concept.

So, just to clarify, will the force I find be less than if I just dropped the box?

 

[rock.freak667]

I will be using a parachute and crumple zone which will reduce the force dramatically, I just wanted to use a simple example to help introduce myself to this new concept.

So, just to clarify, will the force I find be less than if I just dropped the box?

The force should be less. But I don't really remember the exact steps to solving such a problem. I am not too sure if my method will work.

 

[Assasinof6]

The force should be less. But I don't really remember the exact steps to solving such a problem. I am not too sure if my method will work.

Well, that's good news!

Not too worry, I will get all the values I can tomorrow, try it out, and then return to this forum to either confirm the results (unlikely), or get more help (very likely).

Thanks to you three guys for helping me!

 

[6Stang7]

So, are you trying to figure out what spring constant you need so that when a box impacts the ground, the force will not exceed a certain value?

 

[Assasinof6]

Correct. I figured out that an egg cannot sustain more than 4.9 Newtons of force.
Unfortunately I will not be able to discover the values today, it will have to wait for tomorrow.

 

[6Stang7]

So ideally, you'd want to know what spring value (k) you need so that, give an egg of a known mass, dropped from a known height, won't exceed your thresh hold of ~5N.

Well, we know that:

F=ma=xk

Since the egg is going to go from some maximum velocity to zero velocity, the change in velocity is equal to the maximum velocity. We also know that all of the kinetic energy of the egg is going to go into potential energy in the spring; that is

(1/2)mV^2=(1/2)kx^2 where x is the amount the spring compresses.

We can re-arrange this so that

x=v*(m/k)^1/2

So we have:

ma=kx=k*v*(m/k)^1/2=F=v*(m*k)^1/2

We can write this in terms of k, getting

k=(1/m)*(F/v)^2

Finally, remember that v=(2gh)^(1/2) where g is gravity and h is how high an object is dropped from your zero potential line (in this case ground)

We now have:

k=(1/m)*(F/[(2gh)^(1/2)])^2

so, give m in kilograms, F in Newtons, g=9.81m/s^2, and h in meters, you know what k has to be in N/m.

 

[Assasinof6]

...Since the egg is going to go from some maximum velocity to zero velocity, the change in velocity is equal to the maximum velocity. We also know that all of the kinetic energy of the egg is going to go into potential energy in the spring; that is

(1/2)mV^2=(1/2)kx^2 where x is the amount the spring compresses...

Alright. Can I safely subtract from the kinetic energy if I take into account that the force is reduced due to crumple zone and parachute, without messing anything up?
For instance if the kinetic force without reducers upon hitting the ground is 6m/s, but with reducers is 5 m/s, therefore (1/2)mV^2, instead of being (1/2)m(6)^2, is (1/2)m(5)^2?


On a partially related note,
You, are incredible. Can't get more succinct than that. Thanks for your help!

 

[6Stang7]

A crumple zone is basically like a spring; it absorbs energy. The difference is that a crumple zone dissipates all that energy into the environment (as heat, sound, etc).

If you're going to have a parachute, then I'd use this equation:

k=(1/m)*(F/v)^2

Where m is the mass of the system (egg, springs, parachute, glue, etc), F is the force that you want the system to "feel", and v is the velocity when the system impacts the ground. The difficult part will be finding out what the velocity of the system is at the point of impact.

 

[Assasinof6]

Not at all. I know:
V₁ = 0m/s
a = 9.8 m/s
\Deltad = 5.1m (reformed, not 5.2)
\Deltat = ?

Thus, using the equation
\Deltad = V₁\Deltat +(1/2)a\Deltat²
I can solve for time:
5.1 = (1/2)(9.8)\Deltat²
\sqrt{}(5.1/4.9) = \Deltat
THerefore \Deltat = approx 1.02 seconds.

That would be the time if air-resistance is negligible.

In real life I would time the actual descent time, and using that figure out the acceleration, and from there v₂.

However, if you don't mind could you please derive the equation you gave me:
k=(1/m)*(F/v)^2
(and give an explanation to go with it) please? Thanks again.

 

[6Stang7]

Not at all. I know:
V₁ = 0m/s
a = 9.8 m/s
\Deltad = 5.1m (reformed, not 5.2)
\Deltat = ?

Thus, using the equation
\Deltad = V₁\Deltat +(1/2)a\Deltat²
I can solve for time:
5.1 = (1/2)(9.8)\Deltat²
\sqrt{}(5.1/4.9) = \Deltat
THerefore \Deltat = approx 1.02 seconds.

That would be the time if air-resistance is negligible.

In real life I would time the actual descent time, and using that figure out the acceleration, and from there v₂.

However, if you don't mind could you please derive the equation you gave me:
k=(1/m)*(F/v)^2
(and give an explanation to go with it) please? Thanks again.

Well, the problem is that your a will be different depending on how high up the object is. Air resistance isn't a constant force; it varies with the speed at which the object is moving until you hit terminal velocity.

As for deriving k=(1/m)*(F/v)^2, I did it in my above post, but in a semi convoluted way. Start trying to derive it yourself and if you get stuck, post where you are getting stuck and what your issues are, and I'll work with you to get through it. I suggest you start by drawing a free body diagram of the system, gathering all your knowns and unknowns, and all relevant equations

 

[Assasinof6]

I finally got the values!
v_i = 0
d = 5.1m
m = 1.00 kg (nice and round!)
t = 0.92

k=(1/m)*(F/v)^2
Assuming the "v" in the formula is the end speed.

a = (vf-vi)/0.92
2a + vi = vf
vf = 0.92(9.8)
vf = 9.016

k = (1/(1.00)) * (F/(9.016)^2

Is F the net force experienced by the box at that moment, or just the gravitational?