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Hooke's law (my brain just reached it's elastic limit) Answered

I was very daring and entered the intro to engineering class as my elective in my first year of highschool. It turns out that I am in over my head. I am always up for a challenge, so I am sticking with the course I am one of the 3 freshmen in the class. On the first day of class our teacher was introducing the class and brought up that we would have to use some trigonometry and other math that the freshmen have no learned yet but he is willing to help us out with that... Anyhow today at the end of class we were given a hand out about "hooke's law" because we had just done a lab where we attached weights to a spring. The paper is pretty straight forward about hooke's theory and the formula F= K (delta)L (Force = K times the change in length) it defines k as the "proportionality constant" which I do not understand. What is a proportionality constant? wikipedia says its a force or spring constant which I still do not understand. Wiki also gives the formula F= -K x (with arrows over the k and x). help please!


It is the measure of how easy it is to deform the material.

The large k is, the more force it takes to extend a material by a given amount.

It only holds for the first part of the graph you produced, when the line was straight. It is the gradient of the straight line.

(Sometimes the k is a lambda)

If K is low, then K is not a very strong (spring, in this case). If that's what you meant....

well somewhat, but in the equation Force = K times the change in length so K must be a number, how do I get that number

. That number will either be a given (eg, you know the specs of the spring and want to compute how far it will stretch under a given load) or to be solved for (eg, you know the mass and amount of stretch).

Oh. That's a good question, I'll get back to you on that (should be a few minutes, if I can figure it out.)


Reply 10 years ago

From what I've found, the "number" for K is N/m, or Newtons/meter.

Does that help?

it does, so basically I need would need to find out how many newtons it takes to pull down the spring say 1 centimeter. And I can get my newtons using mass in kg x 9.8 because 1kg = 9.8 newtons

From what I know, yup. Hope all works out for ya.


10 years ago

If Y is proportional to X, that means that you can make a nice equation for the value of Y that involves X multiplied by some constant number (without more complicated factors like powers, logarithms, etc.) The "proportionality constant" is just a fancy name for that constant. So in your classic algebraic equation for a line
  • Y = m X + b
you could call m the "proportionality constant" instead of the slope.

Relating such constants to intuitive real-world characteristics can be challenging. As Kiteman says, for Hooke's law it's related to the stiffness or strength of the spring...

(catching up with comments from while I was typing) To figure out "k" from data gathered in the lab, plot your data, draw a "best fit" straight line through the points, and measure the slope of the line.

so is k actually m in a linear graph using the x/y values in my lab? because we did graph it out x was the weight in newtons and y was the length of the spring

i the formula f=k delta L... If I took out my results from the experiment where my results are Y1=9.8 Y2=30 and my X1=0.49N X2= 1.47N. I would do 30 - 9.8 = 20.2 and plug that in as my delta L. which x value would i plug in for my F.
Or could I just do 30-9.8/1.47-0.49 to get K

um. It looks like you have your X and Y backwards? X1 and X2 (in newtons?) are your f, and Y1 and Y2 are distances ? So you're trying to find k in X = k Y rather than Y = k X ? Do you have more than two points?
Do it backwards would just mean you'd get 1/k instead of k...

x1 and x2 are in newtons, and the y's are distances I have 5 points for each. There are two lines one for each spring we tested, ignore the lower line with the slope of 10 . Here are the coordinates (0.49 , 9.8) (1.47 , 30) (2.45 , 49.8) (3.43 , 68.7) (4.41 , 86.6) What would I make my formula I am somewhat lost now

Picture 52.png

If your data were "perfect", any two points would give you the same slope (try it: your data looks pretty close to perfect.) However, it's more common to plot the points on your graph, draw the best line you can, and then measure the slope from arbitrary points on the line (just by looking at the graph.)

There are nice numerical methods for figuring the equation of the EXACTLY best fit line, but they're not high-school level. If you know any statistics, you can think about calculating the error for each point from the line you've picked, and tweaking the line so that some combination of those errors (probably the standard deviation) is minimized. But you're surely not expected to do this till college (and probably in a numeric methods class rather than a physical sciences lab, or you'll have software that does it for you.)

The way you have your graph, you are plotting L = (1/k) * F rather than F = k * L - I guess you attached known weights and measured length, rather than picking know lengths and measuring force. That should work out fine as long as you keep it in mind.

ok, but say I would be put in a situation where I would need to use f=k*delta L (whatever that may be) how would I handle it because he gave us this hang out today and told us we have a quiz on it tomorrow. would the case be X1 is my force delta L is y2-y1 and then i would plug that in and get k that way?

also thank you so much for you help, really you are doing me a huge favor I'm sorry if I'm being difficult

Or could I just do 30-9.8/1.47-0.49 to get K

I believe that is correct, as from what I understand westfw to mean, is that m (slope) is y2 - y1/x2-x1.

I'll talk to my teacher tomorrow at lunch or show up early in the morning for some help just to be safe. He's somewhat sympathetic for the freshmen because we were not told that we would be using advanced formulas and trig... If I pass this course I will be quite proud of myself lol

Think how nice it will be to be in math class going "I need to learn this so I can do my engineering homework" instead of "why are we learning this? When am I ever going to need to use this?" :-)

Looks like you have this under control, but I will point out that there should not be an arrow over K; it should be over F. F and x are vectors, and have both a magnitude and a direction; k is a scalar and has only a magnitude.