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Springs - How important are they, really?

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  • #31
    Compress a spring with 10 pounds. Measure amount it has compressed. Now it does not mater what you use, one scale and a weight, your hand, or another scale to compress spring to the second scale, 10 pounds is 10 pounds and spring will compress same amount. In the case with two scales both scales will show ten pounds not 5 or 20.
    Not true, MRO. You will agree that it takes a certain amount of energy to compress a spring. Where is this energy(force) coming from? If I take a steel rod and push it against the wall with a force of 10 pounds, I will be pushing against the wall with ten pounds of force. We all know this.
    But now, if I try to push a spring against the wall, the spring will absorb some of this force by compressing, no? So... if I want to push against the wall with ten pounds of pressure, I may have to use 15 pounds of force, as it may take five pounds of pressure to first compress the spring. Energy must come from somewhere, and go somewhere. The spring absorbs and stores energy.
    "Damn it Jim, I'm a doctor, not a mechanic!' ('Bones' McCoy)

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    • #32
      I give up!!!!

      When you get it all worked out...............



      mro

      Comment


      • #33
        In reply to Pro's reply to my reply to Pro's...

        Pro- I still hold my contention. Let me see if I can illustrate my point without use of illustration (being semi-computer illiterate).


        So... if you'd indulge me once again, good sir,(... please answer, and explain(in terms I can understand,...) my following questions:
        "By an under-extension of the distal digits, proximal to the opposable phalangal counter-structure, can the quasi post-modern designed ceramic NaCl containment vessel be propelled in an ever increasing arc, it's radius not to excede "R+2", and it's velocity at a stable, yet ever ampifying rate of uniform acceleration, be brought to an alternate state of rest, the coordinates of X/Y..."
        "The Hookean rheological element (spring)..." ...Umm, I don't think I'm smart enough to be able to explain it in those terms... (I looked that one up in a book)

        Okay, here I go again...

        Firstly, I believe that when one is standing on a bathroom scale, it (the scale) is registering pounds of mass (lbm) rather than pounds of force (lbf). Newton's 2nd Law is F=ma [Force=(mass)x(acceleration)] The force on the scale is a result of the mass times the acceleration due to gravity. Since gravity is constant, these two different variables are often interchanged.

        The spring/Pro/scale situation, Pro would weigh the same amount (exert the same downward force) in this gravitational field because his mass does not change, thus he'll exert the same amount of force whether he's standing on the scale directly or standing on a spring on the scale. When the spring is between Pro and the scale, it is exerting a downward force due to it's mass as well. It also now has a Potential Energy due to being slightly compressed. Pro also has a new additional Potential Energy because of his proximity to the scale. Gravitational Potential Energy is proportional to (mass)x(height). You're higher now. I think that these two PE's will counteract.

        But anyhow, on to the question at hand...

        Assumption #1: a scale (such as a bathroom scale) is essentially an extension of the physical structure upon which its bottom side is touching.

        Assumption #2: The scale is only a measurement of exerted force and will not add any additional interaction past what is initially exerting the force.

        Assumption #3: The spring is not all the way compressed at any point in the experiment.

        Here is my scenario: Take two (bathroom type) scales and a spring. Hold the scales on the palms of your hands. Now make a scale spring sandwich and hold it horozontally. Picture a martini shaker. (from left to right, we have palm of hand, scale, spring, the other scale, and then the other hand)

        Now squeeze everything together with enough force to register 10 lbs on one scale and hold it there. The other scale will also register 10 lbs. Since the scales are only extentions of the hands (that happen to register force), each hand is exerting ten pounds. That's a total of twenty on the poor little spring.

        Therefore, each end of the spring is pushing back with 10 pounds. Also, if something is pushing something else with any amount of force, that something else has to push back with the same amount of force or else it will move. (one of Newton's laws -not sure which). Since neither end of the spring is moving, each
        has to be counteracting each hand with 10 pounds of its own force.

        So Pro (and anyone else who had the patience to read the entire thing), do you buy that explaination?
        -Do what makes you happy.

        '79 Honda CB 750 K (2)
        '78 XS 11 E - "Rhona"
        ...and a 2nd E, for the goodies on it.

        Comment


        • #34
          ...been doin' more thinkin'...

          I did consult a Materials book last night to see if I could find any supporting data to my claim, and ran across the statement that the force applied to a spring is instantaneously transferred to its other end. That got me thinking to the why. (I don't like to just say "But it said so in a book")

          Before, we were saying that it'd take time to transfer the force from one end of it to the other. Here's the explaination a co-worker gave me when I asked him. Since the other end is fixed and doesn't move (well, I supposed if I were fixed, I wouldn't feel like moving around too much either. ...but I digress...), any movement (i.e. compression) at all in the spring would be a result of the initial force (the cam on the bucket) transferring its energy to the spring. The spring won't compress unless the other end (the engine's head) pushes back.

          So my answer to the original question remains that it wouldn't matter if the tighter coils of a progressive spring are at the top or at the bottom; the spring just compresses. And that happens at the place in the spring where the coils are loosest.

          HOWEVER, if the spring is tapered (I'm not sure if progressive springs are tapered or not), then the larger diameter coils should be down. The larger diameter makes for a more stable base.

          Okay, I'll shut up now.
          -Do what makes you happy.

          '79 Honda CB 750 K (2)
          '78 XS 11 E - "Rhona"
          ...and a 2nd E, for the goodies on it.

          Comment


          • #35
            ...been doin' more thinkin'...

            Some 600 scientists and engineers from 75 institutions in Canada, China, France, Germany, Italy, the Netherlands, Norway, Russia, Spain the United Kingdom, and the United States are working on "spring theory " . While it has been determined that it is NOT instantaneous, they have found it is easier to achieve the desired results if a co-ed is drunk!


            instantaneously???
            do you really want a response???



            mro
            Last edited by mro; 09-22-2006, 09:07 AM.

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            • #36
              "Spring-break theory?"

              they have found it is easier to achieve the desired results if a co-ed is drunk!
              Finally, something we can all agree upon. (But I think more experimentation is in order!)
              Pro also has a new additional Potential Energy because of his proximity to the scale. Gravitational Potential Energy is proportional to (mass)x(height). You're higher now.
              Am trying to figure out what it is you're saying here. Actually, if I follow you correctly, this is incorrect. Gravatational potential energy would be (mass)x(distance) from gravatational source. The further from the source, the less you weigh. Hence weightlessness once a certain distance from earth, no? So actually, standing on the spring, I would be further from the earth, and weigh less, (but this is insignificant for our application.)
              (Actually, had a medical technician tell me that the scales in my upstairs bathroom and downstairs bathroom register different, too)
              Damn... where are those "Moon Shoes" I used to have as a kid when I really need them? You know the ones... had those big springs on the bottom. And MRO... what of your "Pogo Stick"? Shouldn't you be up in the attic searching fer it?
              Now squeeze everything together with enough force to register 10 lbs on one scale and hold it there. The other scale will also register 10 lbs. Since the scales are only extentions of the hands (that happen to register force), each hand is exerting ten pounds. That's a total of twenty on the poor little spring.
              I disagree.
              Oh, I agree with your scenario, but your numbers are off.
              It will be more than twenty on the little spring. You will agree that to compress a spring requires force.(At least, you should)
              And this force must come from somewhere and be accounted for.
              Now, in your above sandwich scenario, lets put a block of steel between the two scales and press in from the sides. If each scale reads 10 pounds, then yes, indeed, there are twenty pounds pressing in on the steel block. (Ten from each direction).
              But, we know that a spring requires force to compress it. And this particular spring, if compressed with ten pounds of force, will collapse one inch. I now try to compress the spring with ten pounds of force with one hand... and watch as it flies across the room. (So now I'll use two hands)
              I want to compress this spring one inch, so I must apply ten pounds of force. Five from my right hand, and five from my left hand, and the spring compresses one inch(It is also resisting my force, five to the right, and five to the left) No bathroom scale was involved in this experiment. I can get a spring to compress one inch by applying ten pounds of force. And the spring will not have absorbed ten pounds of force till it has compressed one inch. The spring is also trying to push my hands apart with ten pounds of force. Once the spring hits the one inch compression mark, these two forces balance. My hands no longer are moving together, and the spring is no longer pushing my hands apart. ten in, and ten out. Nothing moves. Conservation of energy.
              Now... I belive that I have demonstrated that it takes ten pounds of force to collapse this spring, the spring absorbing ten pounds of force with this compression, and then everything equals itself. The spring, at its ten pound compression mark has collapsed one inch. It will collapse no further, and for all purposes, is as an uncompressable steel block.

              Now... to correct a statement that was made in an earlier post,(I dare not search for the author as my computer is starting to freeze up) that if you place a spring on a scale and apply ten pounds of force, the scale will read ten pounds.
              This cannot be so. If I put ten pounds of force to a steel block on a scale, it will register ten pounds, as the force transfers through the solid block directly. But with the spring, it must first collapse(which requires ten pounds of force) before it 'turns to an unyeilding block' and can transfer force to the scale. So... to press on a spring in order for the scale to register ten pounds, I must first use ten pounds to collapse the spring(zero pounds register on the scale) and then an additional ten pounds of force to get the scale to read ten.
              Of course, the reading on the scale will be going up all the time as force is applied, but in increments of 1/2 force applied overall. (for every one pound of force applied in this experiment, 1/2 is taken up by the spring, and the other 1/2 by the scale) Twenty pounds of force is needed when pressing on a scale with a spring for the scale to read ten.

              And, as my teachers always thought of me as a backward child... since A+B=C, it is also true that C- A=B.
              A scale was mounted to my bedroom wall and I have my hand pressed against it. My parents have stuck a fish hook through my hand and this is attached to the opposite wall by a spring. My punishment for failing 6th grade science class is to get this scale to register ten pounds. My spindly arm is capable of exerting ten pounds of force on the scale, but I can't... as I don't have enough strength to supply the ten pounds neccesary to overcome the restraining spring.
              (They left me there like that for three days. Oh, I can talk about it now... I'm better, really... but it explains a lot of things)
              "Damn it Jim, I'm a doctor, not a mechanic!' ('Bones' McCoy)

              Comment


              • #37
                originally posted by mad scientist from somewhere in the North West
                Damn... where are those "Moon Shoes" I used to have as a kid when I really need them? You know the ones... had those big springs on the bottom. And MRO... what of your "Pogo Stick"? Shouldn't you be up in the attic searching fer it?
                Had to sell pogo stick at an early age, was found that when jumped on XSively at the top of the arc one would become light headed..............................

                And, as my teachers always thought of me as a backward child... since A+B=C, it is also true that C- A=B.
                A scale was mounted to my bedroom wall and I have my hand pressed against it. My parents have stuck a fish hook through my hand and this is attached to the opposite wall by a spring. My punishment for failing 6th grade science class is to get this scale to register ten pounds. My spindly arm is capable of exerting ten pounds of force on the scale, but I can't... as I don't have enough strength to supply the ten pounds neccesary to overcome the restraining spring.
                (They left me there like that for three days. Oh, I can talk about it now... I'm better, really... but it explains a lot of things)
                Since the ten pounds force needed to pull spring (expanding, oposite of compressing) would cancel out the equal force needed to compress the scale 10 pounds,

                net sum(-10 pounds and + ten pounds) of force would be Zero.



                mro

                Comment


                • #38
                  Hmm....

                  Man, now the gears in my brain are really smoking. Let's see if I can understand what yuo're saying and weigh that with/against what I'm saying. Here goes...
                  What I said:
                  Pro also has a new additional Potential Energy because of his proximity to the scale. Gravitational Potential Energy is proportional to (mass)x(height). You're higher now.
                  What Pro Replied:
                  Actually, if I follow you correctly, this is incorrect. Gravatational potential energy would be (mass)x(distance) from gravatational source. The further from the source, the less you weigh.
                  The key word from what I said was proportional. Potential Energy (PE) =mgh where m=mass, g=gravitational constant, h=height above the earth. I took out the constant, thereby rendering the equation a proportionality rather than an equality. But you're right, that's (mostly) besides the point.

                  I think I may know what we're disagreeing on. I believe that we are looking at it from different viewpoints. I from a Force POV, and you from a Potential Energy POV.

                  What I'm saying is that there is a transferance of force through the spring. What you're saying (I know you'll correct me if I'm mistaken ) is that there is energy absorbed by the spring, and that's why the scale would register less force.

                  You asked for no techno-babble, but I gotta bust out some equations here.

                  I think the crux of it is that we're at the produce stand and I'm looking at the apples and you're looking at the oranges and we're trying to mash them into an orple (or an appange- whichever you prefer).

                  Force=mass x acelleration; F=ma
                  Force needed to compress a spring = (k)(x) where k= the spring constant and x= displacement
                  These two equations are analogous.
                  Potential Energy of a spring = U = (.5)(k)(x squared)

                  Units: F= lb(f), m=lb(m), a=ft/sec/sec, k=lbs/ft, x=ft, U=(ft)(lb)

                  Ignoring the bathroom scales and all other such measuring devices, let's assume that it is what it is, with no verification necessary, as they introduce additional variables and I think they confuse the situation.

                  When you compress a spring with your hand on a table, you apply a force [lb(f)], say 10 lbs. To quantify the example, let's say that we choose a spring that has a spring constant of k=100 lb/ft for ease of math, because we want to compress the spring 0.1 ft. There's the displacement, x.

                  The PE of the compressed (or stretched for that matter) spring is U. U=(.5)(100)(0.1)(0.1) = 5 ft/lb.

                  I think that this is the 5 that you're coming up with. (again, if I'm wrong, set me straight) Force is an ingredient of Potential Energy. [U=(.5)(F)(x)] By applying a force to the spring, the spring will absorb some Potential Energy, but the force applied, no matter how small, will be transferred through any solid body.

                  So the table will be pushing back with its own force, which is the force applied to the other end which was transferred through. Force is just force. If you're holding it slightly compressed in a static situation, the forces will cancel out to make the sum of all forces zero.

                  But there is still potential energy stored up in the spring. As soon as you lessen the force with which you are pushing, the Potential Energy in the spring will be exerting more force on your hand and displace your hand upward until it expends all of its Potential Energy and the spring is fully extended. If you decreased your downward force, then the spring will push until equilibrium is reached again. The displacement of the hand end of the spring is less (not as squished as before).

                  Also, as far as your parents teaching you a lesson with the fishhook, I absolutely agree that you'd have to exert 20 lbs of force in order to make the scale read 10. (assuming, of course, that the spring constant of the spring attached to the hook in your hand is the same as the k of the scale's spring. But I think that's what you were implying anyhow)

                  So what do you think of my new explaination? Is this where we've been disagreeing? Library closes soon, so I won't be able to read any reply till Mon.
                  -Do what makes you happy.

                  '79 Honda CB 750 K (2)
                  '78 XS 11 E - "Rhona"
                  ...and a 2nd E, for the goodies on it.

                  Comment


                  • #39
                    Yo Prometheus

                    Compressive/tensile forces you’ve applied to this thread, especially to the tibia/fibula of readers ((BS * W) over - C squared) XSive pronation leads to hypermobility and instability, resulting in metatarsus primus varus and associated unilateral juvenile hallux valgus (bunion) deformity( LLD). That causes problems with ambulation and may result in “walking in circles” .



                    mro
                    btw
                    I prop my eyes open now and then to help make sense of the incomprehensibilities that surround me and to fool people into believing I'm actually awake and conscience.

                    Comment


                    • #40
                      "Mental exercise has reached logical conclusion"

                      Progressively tighter circles, MRO.

                      Thank you, Erik the Red, (and all other's concerned... even those's who didn't give a "Fiddler's F@#k)
                      Last night in the wee hours, after much coffee and many trips to the bathroom to go wee, I finally got it.
                      Not that I've been blind to all that had been said, I just couldn't grasp many of the principles involved in the manner that they were stated. (Equations... Bah!)
                      But, I receive solice in the knowing that the "Puzzle of the Spring" has long plagued mankind.
                      Serious students of the arts will remember Michelangelo's early work "The Creation of Prometheus", in which the spring is clearly noticeable.
                      IMG][/IMG] (he later removed the spring, added the religeous stuff, and changed the name to Adam) (For our female readers... Umm... I must add that, an unknown painting restorer, when cleaning this great early work of 'Prometheus', mistakenly "reduced" a certain part on the original, possably to save on pigments or something)
                      I rethought the meaning of work... ERGs, BTU's, Horse Power(Which started with "Acreage", the measured area that one horse can plow in one day), and then it hit me. Distance, the missing element was distance.
                      The solid block transfers the force directly... but the spring gives up it's resistance, by yielding the distance across which that force must be applied. In this instance, Force required is related to Distance moved. Leverage, and such. (The extra force required to compress the spring is offset by the spring yeilding ground, which absorbs and balances this force) No extra force is needed, nor consumed... Mass, as it relates to time/space.
                      I thank you all for indulging me in my whims. I can now agree on how a spring behaves, though with my limited powers of deduction and equation skills, would be hard pressed to prove it.

                      (Now... do "Gremlin Bells"really work? Do I find no Gremlins when looking in my carbs 'cause they quickly scurry to the cylinders to hide? )
                      "Damn it Jim, I'm a doctor, not a mechanic!' ('Bones' McCoy)

                      Comment


                      • #41
                        I gotta say, I enjoyed the entire discussion. A couple times you had me second guessing myself, and I was FORCED to return to my old school notes for assistance. I think that was the most thought I'd put into any single thing in a long time. (the gerbil up there running the wheel nearly had a heart attack trying to turn the gears)

                        I was going to attempt filling this with puns and such, but I'll have to leave that to the pun gurus, as I'm finding I'm not all that creative today. Well, I'd say we thouroughly discussed that one into the ground.
                        -Do what makes you happy.

                        '79 Honda CB 750 K (2)
                        '78 XS 11 E - "Rhona"
                        ...and a 2nd E, for the goodies on it.

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