Theoretical expression for relationship between height and force

[thakshi]

Homework Statement

I did an experiment investigating what effect height has on force and now i need a theoretical value to compare it to

Homework Equations

f=ma but i don't think acceleration is constant here

The Attempt at a Solution

f=ma
f=0.104x9.8
f=1.01
i got 49 N with the force plate..help

 

[Mandelbroth]

Homework Statement

I did an experiment investigating what effect height has on force and now i need a theoretical value to compare it to

Homework Equations

f=ma but i don't think acceleration is constant here

The Attempt at a Solution

f=ma
f=0.104x9.8
f=1.01
i got 49 N with the force plate..help

We need more information. We're mathematicians and scientists, not magicians.

 

[Simon Bridge]

The height of what?
On what force?

 

[thakshi]

what information do you need?

 

[thakshi]

basically the experiment was dropping a constant mass at different heights and seeing if the force exerted on the Force Plate increases or decreases as height increased. it increased and now i need an accepted value to compare my data to

 

[Simon Bridge]

what information do you need?

The height of what?
On what force?

basically the experiment was dropping a constant mass at different heights and seeing if the force exerted on the Force Plate increases or decreases as height increased. it increased and now i need an accepted value to compare my data to

The force on the plate will vary over the time of the collision - so I'm guessing the plate is recording either the maximum force or the time-average of the force or the specific impulse... which is it?

This is important because there is no such thing as the force of a collision.

The maximum force would be proportional to the change in momentum, which depends on the initial speed and the material properties of the plate and mass. So how does the speed that a mass hits the floor depend on the height it was dropped from?

What you should do it plot your height vs "force" and see what sort of curve it appears to be trying to be.
Compare with the speed vs height graph for the same values... you are comparing the shape not the exact values.

 

[thakshi]

it records max force

 

[thakshi]

well i did not record the time when it fell so i won't be able to do speed

and the graph i plotted-- as height increases, the force increases

i dropped a mass of 104 grams

 

[Simon Bridge]

well i did not record the time when it fell so i won't be able to do speed

You don't need to have timed the fall to get a theoretical value for the final speed - you know the height and you know the acceleration, and you know the kinematic equations.

and the graph i plotted-- as height increases, the force increases

Is this what you would have expected? i.e. do you normally expect objects to hit the ground harder when they fall from higher up?

In what way does it increase?
i.e. is it a line or a curve?
If a curve, what sort of curve?
If a line, what is it's slope?

What would air resistance do to a force vs height graph?

i dropped a mass of 104 grams

How does the mass matter?

 

[thakshi]

is it constant?

is there an equation relating height and force?

 

[Simon Bridge]

is it constant?

Is what constant?

is there an equation relating height and force?

It's not an easy one.
It depends on the material properties of the object falling and the surface it lands on.
Basically it is Newton's second law - but using momentum.

 

[thakshi]

is acceleration constant?

 

[Simon Bridge]

Hah! You tell me ...

What determines acceleration?

(note: I've edited post #9 since you've replied)

 

[thakshi]

oh forget it

 

[Simon Bridge]

Giving up so easily?Perhaps you are expect people to just give you the answers?

This was posted as "homework".
We do not spoon-feed people here - it's your homework, you have to do the work.
What I can do is point you in a useful direction.Perhaps you do not know what determines acceleration?

Hint: Newton's second law. You seem to keep getting stuck on this - perhaps you should revise Newton's Laws?
The one you want is "the force on an object is equal to the rate of change of it's momentum".

Or you could just remember the famous Galileo experiment - dropping balls off the leaning tower of Piza?

 

[thakshi]

well i tried and i don't get it so i came here
thanks though

 

[Simon Bridge]

I'm sorry but I see no evidence of you trying at all.

Either that or you are doing physics that is way above your knowledge level ... you should be able to work out whether the acceleration of a falling object is a constant or not for instance. You should know Newton's Laws already. Stuff like that.

Either way: If you don't answer questions, nobody can help you.

You need to get to grips with the basics before you can expect to be able to tackle experiments like the one you described.
Is this for school or is this something you rigged up for your own interest?
What's the context?

 

[thakshi]

What do you want me to answer?

 

[Simon Bridge]

From post #9:

and the graph i plotted-- as height increases, the force increases

Is this what you would have expected? i.e. do you normally expect objects to hit the ground harder when they fall from higher up?

In what way does it increase?
i.e. is it a line or a curve?
If a curve, what sort of curve?
If a line, what is it's slope?

How does the mass dropped matter?

From post #13:

What determines acceleration?

From post #15:

Perhaps you do not know what determines acceleration?

Hint: Newton's second law. You seem to keep getting stuck on this - perhaps you should revise Newton's Laws?
The one you want is "the force on an object is equal to the rate of change of it's momentum".

Or you could just remember the famous Galileo experiment - dropping balls off the leaning tower of Piza?

From post #17:

Is this for school or is this something you rigged up for your own interest?
What's the context?

The thing is - there is no "standard value" for the max-force of impact for you to compare your results with, but you do need to compare them with something in order to see if the theory is any good. The answers to the above questions will help me figure out something that you can use. You probably won't want to use the kind of thing I'd use...

At the very least you should be able to write a conclusion about the way the force increases with the height dropped from to show the marker that you've understood what you've done.

Note: drop tests using electronic sensors like you did are routine in engineering.
i.e. http://www.sandv.com/downloads/0702metz.pdf

 

[thakshi]

Is this what you would have expected? i.e. do you normally expect objects to hit the ground harder when they fall from higher up?- yes this is what i usually expect

In what way does it increase?- with my errors bars taken into account, the force is directly proportional to the height at which it droppedHow does the mass dropped matter?- as long as its constant, it shouldn't matter?

What determines acceleration?- mass, force? velocity over time?
Perhaps you do not know what determines acceleration?Is this for school or is this something you rigged up for your own interest?- its for a lab due VERY soon, physics class
What's the context?

 

[Simon Bridge]

OK - do you know Newton's second law? F=ma => a=F/m
For your object: what are the forces on it as it falls? Do these forces change with the start-height at all?
If the forces are constant, the acceleration is constant.

It's tough that you are running out of time ... but you appear to have fundamental gaps in your physics.
Best practice is to read through the Lab notes a couple of days before you do the lab to make sure you understand the concepts involved ... just following instructions, as you are finding out, won't help.

Since you have your data you can do some analysis (find the equation of the best-fit line, pref. with errors) ... and you can write down a conclusion right away.

The exact wording depends on what the "aim" says ... but you want to say that the data is consistent with the max impact force being directly proportional to the drop height with relation [the equation of the best fit line].

Notice that you do not want to conclude that the force is directly proportional to the drop-height (it's not actually the case) ... just that your data is consistent with this.

To look extra good you want to comment on this result ... normally this is where you compare with some standard result. Unless you have information I don't, there won't be an easy way to get a standard result ... but you can comment on what the y-intercept of the graph means. Does the best-fit line go through the origin? Is this to be expected? (i.e. what would the theoretical force on the plate be for a zero drop-height?)

That should be good enough for a senior High School physics lab ... and it should scrape by good marks for entry-level college.

Technically the force should depend on the square-root of the height ... but your data does not show that so you'd have to derive it to use it.

 

[thakshi]

yeah i do know the second law
the best fit line does go through the origin cause when the height is zero, there won't be a force.
the comparing with the standard part is the part I am having difficultly with...
"Technically the force should depend on the square-root of the height ... but your data does not show that so you'd have to derive it to use it." how would i derive it

 

[Simon Bridge]

yeah i do know the second law

... well you kept not answering any questions about it so I couldn't assume you did. We get people with all kinds of ability here so if you don't show off what you got we have to assume you ain't got it ;)

the best fit line does go through the origin cause when the height is zero, there won't be a force.

... so you are saying that an object just sitting on the force plate does not exert a force on it?

the comparing with the standard part is the part I am having difficultly with...
"Technically the force should depend on the square-root of the height ... but your data does not show that so you'd have to derive it to use it." how would i derive it

(a) you don't need it - just the observation about the y-intercept should count as a standard result. (Unless the instructions specifically tell you to compare something to a standard result?)
(b) you have to use you knowledge of Newton's Laws ... since you didn't answer the last round of questions I cannot be more specific.

Additional concepts that may help:
- coefficient of restitution
- conservation of energy
- conservation of momentum
- specific impulse

I have the whole derivation sitting here but I cannot give it to you because you'll be tempted to just copy it - if you did that, it will be obvious that you have cheated and you'd be punished. I don't want to do that to you.

 

[thakshi]

thanks so much for your help
its due tomorrow and think i will not include the derivation then
yeah i have to compare it to a standard result!

 

[Simon Bridge]

Since you are told in the instructions that you need to compare it to a standard result - then you will have been provided with an appropriate result to do the comparison with. Most likely a result that can be derived from the data you collected.

Check the "aim" of the experiment and review the resources on hand.
Maybe you have been provided a theory already that you are supposed to use and you missed it? Check.

 

[thakshi]

we were not given anything, it was up to us to do everything

 

[Simon Bridge]

No books lying around the lab?
No lists of properties?

Was there any documentation with the force transducer?
(These things are often calibrated by dropping things on them. The manual may have a chart.)
I suspect you'll kick yourself when you get the marks back and compare with others.

Is this High School or College?
That makes a difference in what you are expected to do yourself.

Have a look at:
http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/BallBounce.pdf
... this is a similar experiment to yours at the professional level.
I'd have thought that varying the height of the drop style experiments would be easier to find :(
But if you look at various ball-drop experiments online, you can get some "typical values" for a lot of your parameters... if you understand the physics that is.

 

[thakshi]

too late for anything. thanks for your help

 

[wreckemtech]

F=ma isn't really the equation you're looking for to determine how much energy the object is developing as it falls. Gravity is accelerating the object over the period of time it falls, to give it kinetic energy.

ke=(mv^2)/2 vf^2-vi^2=2az where vi = initial velocity, vf = final velocity, z=vertical distance.

So you'll need to plug in your acceleration (g constant) and your height and solve for kinetic energy at the instant it hits the ground.

Easy enough, right? Not really. Because just as the object doesn't attain max velocity instantly, it doesn't decelerate instantaneously either when it hits bottom. So you'll need to plug your ke and max force values into something like the impulse equation and calculate how long it takes to decelerate from final velocity back to zero. Which will give you something to plug into the F=ma equation that might agree with what you're seeing on the force meter.

I say 'might' because the force the pad is applying on the object to decelerate it isn't constant. It starts high and trails off as the object loses its momentum over the very short lifespan of the collision.

The equations you need should be in the first 5 chapters of any decent physics text. This is kind of a nuanced problem as most real-world physics tends to be. The other poster giving some good info too. Good luck!

 

[Simon Bridge]

No worries ... now you know the sort of thing you need to be thinking about for next time.

 

[Simon Bridge]

I don't think there's any way you'd be expected to do this - included just to show you what you were asking about. Note: you could have used a more "hand-wavey" argument... but I don't know what level you needed it for.

When the mass hits the force pad, it remains in contact for a short time T, then it lifts off (because it deforms during the impact, and bounces) If the ball and the pad are pretty hard then the force-time graph should be well modeled by an inverted parabola. Something like:

$F(t)=4F_{max}(T-t)t/T$

This is related to the change in momentum of the dropped mass as:
$$\Delta p = \int_0^T F(t)\text{d}t = \frac{2}{3}T^2F_{max}$$
... $T$ depends on the material properties of the dropped mass and the plate - it's probably safe to call T a constant for the setup.

Thus $\Delta p \propto F_{max}$

The momentum of the object just before impact (t=0 in the above equation) is mv, given from the equations of motion for an object falling under gravity ... there's air resistance to consider... but if the distances are less than about three meters for objects of order of centimeters then air resistance can be neglected and if the experiment were conducted within a km of sea level with no big mountains or anything close by then the acceleration due to gravity will be a constant g=9.8m/s/s directly downwards.

By conservation of energy - an object falling distance h from rest, without air resistance, will hit with momentum $p=mv=m\sqrt{2gh}$.

The final momentum we don't know - it depends on the materials involved but it is usually a proportion of the initial momentum ... if the object hits exactly dead-on, then the initial momentum is down and the final momentum is up.

If $p_f=-ep: 0<e<1$ [1]
... then $\Delta p = (e+1)m\sqrt{2gh}$

From there: $$(e+1)m\sqrt{2gh}=\frac{2}{3}T^2 F_{max}\\ \qquad \implies F_{max} = \frac{3(e+1)m\sqrt{2g}}{2T^2}\sqrt{h}\\ \qquad \implies F_{max}\propto \sqrt{h}$$ ... sooooo: if the assumptions (remember what they were) hold, the graph of $F_{max}$ vs $h$ should be a square-root function.

If your graph is indistinguishable from a line, then you may not have had enough data for lower heights.

Approximating the square-root function as a 1st-order Taylor series about h=1m yields a line with slope = $$s=\frac{3(e+1)m\sqrt{2g}}{4T^2}$$
It should be possible to look up "typical" values for T and e that are close to the experimental setup. i.e. I have found some measurements[2] suggesting values: e=0.65/0.77=0.84 and T=0.15ms - for a ball bearing off a pretty hard surface. But I think you have used a softer surface/object because these values get max forces in the order of kilo-Newtons where you had values of tens of Newtons.


----------------------

[1] ... $e$ is called the "coefficient of restitution".
[2] Cross R. (2006) The bounce of a ball - table p224
http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/BallBounce.pdf

 

[wreckemtech]

I don't think there's any way you'd be expected to do this - included just to show you what you were asking about. Note: you could have used a more "hand-wavey" argument... but I don't know what level you needed it for.

When the mass hits the force pad, it remains in contact for a short time T, then it lifts off (because it deforms during the impact, and bounces) If the ball and the pad are pretty hard then the force-time graph should be well modeled by an inverted parabola. Something like:

$F(t)=4F_{max}(T-t)t/T$

This is related to the change in momentum of the dropped mass as:
$$\Delta p = \int_0^T F(t)\text{d}t = \frac{2}{3}T^2F_{max}$$
... $T$ depends on the material properties of the dropped mass and the plate - it's probably safe to call T a constant for the setup.

Thus $\Delta p \propto F_{max}$

The momentum of the object just before impact (t=0 in the above equation) is mv, given from the equations of motion for an object falling under gravity ... there's air resistance to consider... but if the distances are less than about three meters for objects of order of centimeters then air resistance can be neglected and if the experiment were conducted within a km of sea level with no big mountains or anything close by then the acceleration due to gravity will be a constant g=9.8m/s/s directly downwards.

By conservation of energy - an object falling distance h from rest, without air resistance, will hit with momentum $p=mv=m\sqrt{2gh}$.

The final momentum we don't know - it depends on the materials involved but it is usually a proportion of the initial momentum ... if the object hits exactly dead-on, then the initial momentum is down and the final momentum is up.

If $p_f=-ep: 0<e<1$ [1]
... then $\Delta p = (e+1)m\sqrt{2gh}$

From there: $$(e+1)m\sqrt{2gh}=\frac{2}{3}T^2 F_{max}\\ \qquad \implies F_{max} = \frac{3(e+1)m\sqrt{2g}}{2T^2}\sqrt{h}\\ \qquad \implies F_{max}\propto \sqrt{h}$$ ... sooooo: if the assumptions (remember what they were) hold, the graph of $F_{max}$ vs $h$ should be a square-root function.

If your graph is indistinguishable from a line, then you may not have had enough data for lower heights.

Approximating the square-root function as a 1st-order Taylor series about h=1m yields a line with slope = $$s=\frac{3(e+1)m\sqrt{2g}}{4T^2}$$
It should be possible to look up "typical" values for T and e that are close to the experimental setup. i.e. I have found some measurements[2] suggesting values: e=0.65/0.77=0.84 and T=0.15ms - for a ball bearing off a pretty hard surface. But I think you have used a softer surface/object because these values get max forces in the order of kilo-Newtons where you had values of tens of Newtons.


----------------------

[1] ... $e$ is called the "coefficient of restitution".
[2] Cross R. (2006) The bounce of a ball - table p224
http://www.physics.usyd.edu.au/~cross/PUBLICATIONS/BallBounce.pdf

Much better explanation. You're good!

 

[Simon Bridge]

I'm glad you pointed out the conservation of energy bit though - saved me some typing.
The F=ma from earlier was needed for OP to understand whether the acceleration was a constant - OP expressed confusion about that.

Acceleration under gravity is only a approximately a constant - for falling distances that are small compared with the radius of the Earth (as in this case) and where the speeds are too small for air resistance to matter.

The apparently constant slope of the F vs h data suggests that air resistance was not a big factor compared with the experimental uncertainties.

Although, I suspect that if the data were split in half, and a best fit line found for the top (large h) half, and also for the bottom half, the bottom half slope would be bigger than the top half slope ... unless those error-bars are bigger than I'm imagining or there's fewer than, say, ten data points.

There's not really enough information to give any proper help.


Aside: I did try modelling the outcomes of an experiment and found that to get peak forces around 50N required impact times and restitution coefficients more appropriate hard rubber - this sort of experiment is normally done with metal masses and the force-plates are usually high-impact plastic, so I wonder if the force plate was sitting on hard rubber feet, maybe on lino or vinyl flooring.

That link I gave in [2], prev post, - the author's homepage is a lot of fun.

 

[wreckemtech]

True. I didn't really think about it, but g will vary inversely with the distance squared between the center mass of the 2 bodies. But like you say, negligible for this problem. Those were both great links!