Physics Help Forum Falling object force for starting lever

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 Feb 18th 2018, 02:23 PM #11 Junior Member   Join Date: Feb 2017 Posts: 18 After a little thought, my last question seems silly. In the first image, the box is irrelevant because we get some value of L2 and then adjust the spring rate to match accordingly. Whatever mgh/d comes out to, we adjust the rate to move the spring the distance we want to absorb the energy over on the spring side. But in the second model where the forces are not linear, the reason they aren't linear is because of forces interacting inside the linkage components. We'd have to adjust the linkage components and the spring rate to get the right spring travel for any given mgh/(d1+d2+d3). So, the best way I see to figure out how much mgh will result in c in the second model is to work backwards, which I mentioned earlier. Since the forces in the box are fixed, for a given case, I can only choose a spring rate to handle the forces on the right. So, I choose a spring rate. Then I can say if the spring is compressed 100%, the lever will be fully down and I know d1+d2+d3. If I can know L4 at this point I will be able to make sure I have a spring rate sufficient to return the load. I don't see how L4 could equal mgh/(d1+d2+d3) since moving d2 to d3 could take 3 times the work to move d1 to d2. But how do I find L4 when the loads needed to move the lever differ throughout the distance from L1 to L4?
Feb 18th 2018, 03:20 PM   #12
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 Originally Posted by builder89 After a little thought, my last question seems silly. In the first image, the box is irrelevant because we get some value of L2 and then adjust the spring rate to match accordingly.
If this thread is not getting enough responses its because most of what you write doesn't make sense. For example; there is no such thing in physics regarding linear or nonlinear forces. The term is used by some people but there
s no definition to be found in physics texts. It may mean something to those who use it but in general its never defined in physics textbooks nor taught in a physic classroom. Also the term "spring rate" is not a term used by most, if not all, physicists. I.e. if you looked in the index of any college level physics text you won't find that term which means the text makes no mention of it. In fact I've been a physicist for over 35 years now and today is the first time I ever heard of it. I had to Google it to find out what it means.

Feb 18th 2018, 06:09 PM   #13
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 Originally Posted by Pmb most of what you write doesn't make sense
HAHA!!! Well, I appreciate the honest opinion. Thanks anyway for reading Pmb. I'll understand if you don't get it. I just asked Siri on my phone and she couldn't give me an answer either. I'm pretty sure she has a lot of indexes memorized though.

Maybe we could both learn from each other in this problem? :-)

Obviously I am not a physicist but I am solving a real world problem that sits in front of me, not in a text book.

If there are any terms you need me to explain, please just ask. "spring rate" is generally the "script k" in the spring equations maybe you are familiar with. I thought I explained "non-linear" pretty well. The additional force required to move the lever down another 10cm differs depending on where you are in the travel distance of the lever. Unlike a "linear" spring (not all springs are linear believe it or not), where each additional 10cm moved requires the same additional force applied.

Seriously... I am really appreciative of the time you guys are giving to help people like me understand.

Feb 18th 2018, 07:52 PM   #14
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 Originally Posted by builder89 HAHA!!!
What's so funny?

 Originally Posted by builder89 Well, I appreciate the honest opinion.
Glad to hear it. I always try to be honest. I've run into way too many dishonest trollers on the internet who either prefer not to be corrected or they can't fathom that they've make a mistake. So instead they attack the people who pointed it out to them. They're very horrible people who are doing a great amount of damage .
 Originally Posted by builder89 Thanks anyway for reading Pmb. I'll understand if you don't get it.
I can understand perfectly when a problem is properly described. Otherwise I choose to let others dig out meaning.

 Originally Posted by builder89 If there are any terms you need me to explain, please just ask. "spring rate" is generally the "script k" in the spring equations maybe you are familiar with.
This is a good example of where a term has a different meaning which depends on who is using it. I looked up the term using Google before I ever mentioned it, as I always do, and I found this

https://www.qa1.net/technical-suppor...ring-rate-tech
 What is spring rate? Spring rate refers to the amount of weight that is needed to compress a spring one inch.
which is not the same thing you used it to mean.

 Originally Posted by builder89 I thought I explained "non-linear" pretty well.
In fact you never explained it at all. All you did is use it and never actually defined it. Do you realize how many physicists such as myself who try to help people understand physics run into people who define their own terms and think that their "definition" counts as a sound one? I lost count. In this case you never provided a definition at all. Where did you get the notion that was what linear meant?

A definition can almost always be placed in the form

"term" - X is said to be "term" if "description"

For example; An linear operator is an operator L such that when operating on a vector V = a M + b N the following equality holds true for all vectors M and N and scalars a and b

L (T)( = a L (M( + bL (N)

The term "vector" as user here has a very general meaning, much greater than you're probably familiar with. For example; in quantum mechanics a quantum state can be represented by a vector. However vectors in this case have the following notation |a> which is called a "ket".

 Originally Posted by builder89 The additional force required to move the lever down another 10cm differs depending on where you are in the travel distance of the lever. Unlike a "linear" spring (not all springs are linear believe it or not), where each additional 10cm moved requires the same additional force applied.
Then you used the term "linear" incorrectly.

 Originally Posted by builder89 Seriously... I am really appreciative of the time you guys are giving to help people like me understand.
I understand. Unfortunately for me I have been unable to concentrate as much as I'd like to for several years now. You see, I am disabled from chronoc pain. If you're patient then I can help someday in the future. Right now I can't concentrate enough to help you.

Last edited by Pmb; Feb 19th 2018 at 05:08 AM.

 Tags falling, force, lever, object, starting

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