hooke's law

[dhart]

I am doing a science fair project similar to one on your website - "deep knee bends". I have researched Hooke's Law, but I am confused about finding the spring constant for my "quadricep" spring. From what I have read, I understand that I first must find the spring constant with a simpler experiment, hanging weights from the spring, etc. When I do this, I find the Force by multiplying the mass of the weight by acceleration. My question is: do I use gravity (9.8 m/s) for my acceleration? Will the "k" or constant found then work with the set-up for my actual experiment (the knee model)? Please help! Thank you!
-Cassidy

[davidkallman]

Hi dhart,

Some of the resources to calculate the string constant are:

http://www.pa.msu.edu/courses/2002sprin ... pring1.pdf
http://www.rowan.edu/colleges/lasold/Ph ... kesLaw.pdf
http://www.glenbrook.k12.il.us/gbssci/P ... ectoc.html

The first reference gives a direct method to calculate the string constant.

I don't know if the above will be useful to you.

Other resource are available by providing "how do you calculate the string constant in hooke's law" to answers.com

[davidkallman]

Hi Cassidy,

Rereading your post, I'm not sure if I answered your question.

I have two related questions:
1. Do any of the references I listed help?
2. Do you have additional questions?

Thanks.

[Craig_Bridge]

I understand that I first must find the spring constant with a simpler experiment, hanging weights from the spring, etc. When I do this, I find the Force by multiplying the mass of the weight by acceleration. My question is: do I use gravity (9.8 m/s) for my acceleration? Will the "k" or constant found then work with the set-up for my actual experiment (the knee model)?

When you model things, you must first understand that you are coming up with a simplification or aproximation of the real thing.

Assuming that you have chosen a particular mechanical steel wire spring to model the "quad" and that you are attempting to experimentally determine the steel spring constant for your spring, then you need to understand that steel wire springs are NOT linear (do not conform to the Hooke mathmatical model) in two regions. Under extremely light forces less than what engineers typically call the "preload" force, they do not behave linearly (e.g. the value for the spring "constant" varies with force). Once the force exceeds this "preload" force, Hooke's Law applies and you get a repeatable constant called the "spring constant" up until the point where you exceed the elastic modulus of the spring and start permanently altering its shape. Once you have exceeded the elastic modulus and changed the shape and size of the spring, you have a "new" spring. Its preload force, spring constant, and new elastic modulus break down point typically have all changed from that of the original spring.

Now to Mass and Weight. These two terms are very easy to confuse. You should look them up in a dictionary periodically to review their precise meanings. It is very hard to measure mass directly. On the earth, scales typically measure or compare weights. Weight is the gravitational force exerted between an object and the earth. The balance also indirectly measures or compares mass. For example, a one pound object has a mass such that the gravitational force between the earth and that object is one pound of force. If you are comparing objects on a platform balance and the known objects have known masses, then you will indirectly know the mass of your unknown by adding up the mass of all of the known objects required to balance out the gravitational forces. In the English system of units, the weight of an object and its mass are numerically identical because the gravitational constant is precisely unity (1 or one) which horribly blurs the distinction. Further confusion occurs because one pound of gravitational force is associated with one pound of mass on earth. To convert the mass to weight (e.g. gravitational force), you need to multiply by the gravitational constant (aka gravitational acceleration). In scientific units of Killograms, Newtons, meters, and seconds, the gravitational constant is NOT unity but 9.8 as you indicated. If I've confused you, sorry, English units are confusing and are tied to the earth.

The spring constant you derive by determining the change in spring length with respect to a change in force (as supplied by different weights) will be valid as long as you remain in the linear region of the spring (e.g. greater than the preload force and less than the maximum safe force before exceeding the elastic modulus of the spring).

I've tried to go into a very long explaination of something that is much simpler if you stay within the bounds of where the model applies so that hopefully you will understand that the simplifications have their bounds and to look out for the boundary conditions beyond which they do not apply.