traveler Posted October 8, 2008 Author Posted October 8, 2008 Ed, We are talking about a two body experiment, correct? I am talking about how gravity works, not what happens if another separate third object that has a different velocity, different acceleration rate, and different mass, with different inertia interacts on the second (totally different in every way) object . I understand the universe is a system, and that there are not just two massive bodies in the universe that interact with each other. NET FORCE causes acceleration. When you talk about tying objects together, you are only saying "increase the mass of the small object," as there is only 2 objects, not more than two interacting on each other. The universe as I see is nothing more than a mass, distance, and time soup of infinite dimension.
mooeypoo Posted October 8, 2008 Posted October 8, 2008 Apparently this is a 'pick and choose' type of debate, where you just nitpick which posts to answer and which not to. Twice already Bignose pointed out you are ignoring his responses. His very *relevant* responses. posts #99 and #87 are awaiting your response, traveler. debater or troll, traveler? Pick one and stick to it.
traveler Posted October 8, 2008 Author Posted October 8, 2008 Then please accept the help that is being offered in this thread. People are trying to help you. They are trying to tell you that your equation doesn't even have the correct units. They are trying to help you understand that having the correct units is essential in any kind of calculation. You know this. When you cash your paycheck, you expect to be paid in units of money -- not bananas or peacocks or telephone poles. When the cop pulls you over for speeding he doesn't tell you you were going 42 pounds per pizza in a 35 vortexes per milliliter zone. Units matter. That is the help that you appear to be asking for in the quote above, but you are refusing. You are asking for help, and yet you are refusing it so far. Accept the help. Unless this is just an empty request, in which case you are just a troll. Sorry to be so blunt, but that's the long and the short of it. There is lots of help in this thread, if you'd just accept it. I specifically asked DH to help me rewrite the formula to correctly reflect my idea, and he has yet to respond. Does that mean a formula can not be made for my idea if it were reality? So then what if it were reality? I've shown at least reasonable doubt, as the acceleration can not be a constant, as when the distance decreases the acceleration increases. I've also shown how I derived the correct units, I just didn't make it part of the formula, for the above reasons.
mooeypoo Posted October 8, 2008 Posted October 8, 2008 I've also shown how I derived the correct units, I just didn't make it part of the formula, for the above reasons. But then you were shown how that's NOT the correct units. Notice: It was *shown* not to be the correct units. It was 'suggested' that it 'might' be not 'quite right'. It was shown. Proven. Demonstrated, even, if you look carefully. So you did get an answer, you just didn't like it.
Bignose Posted October 8, 2008 Posted October 8, 2008 Correct units are completely, 100% necessary to match your words. If your equation doesn't have the correct units for an acceleration -- length per time squared -- then it isn't an acceleration. Period. Call it something else. Anything else. Call it "addeleration" or "superflying" or "travelling" whatever. But, it is NOT an acceleration. The term accleration has a very specific meaning, and inherent with the meaning is the correct units -- length per time squared. This is the most basic of concepts. I explained this to you using many examples, some farcical, some not. You know this, unless you were literally born just yesterday. When someone asks you how far it is to the post office, you don't tell them "750 lumens" do you? Or "1.6 Hertz"? Or "52 Ohms"? No, you answer in a unit of length "3 blocks" or "2.2 miles" or "3.1 km". It is the same thing. Your formula returns equally nonsensical answers. You call is an acceleration, but when you query your formula by putting in actual values, it returns something that is NOT a length per time squared. It doesn't matter if the acceleration is meters per second squared, kilometers per year squared, miles per nanosecond squared, leagues per fortnight squared, or parsecs per Gaussian year squared. It has, has, HAS, to be a length per time squared. Your equation returns a mass per length. Something like a kilogram per meter, or a pound mass per yard, or a dram per nautical mile or a batman per light-year. (How cool is that name of a unit of mass? http://en.wikipedia.org/wiki/Batman_(mass) ) You equation might as well return bunches of bananas per hogshead or deciliters per millenia, because the units that get returned are just as meaningless. Unless it has the units of length per time squared, it is NOT an acceleration. Just to head it off at the pass, it is also important to note that just because some combination of variables end up with the units of length per time squared, does not necessarily mean that that combination represents an acceleration. A good example is that units of torque are Newton*meters and a Joule is a Newton*meter. Torque and energy are two very different things, they just happen to have to same units. But, it is the very first step any good physicist/scientist does to an equation -- make sure it is dimensionally sound. So, the first step is to check to see if the equation yields the necessary dimension, And then the second step, if and only if the new equation passes the first step, is to investigate how accurate the equations is. Your equation fails this first step. Period. It needs to be remedied, because until it is remedied, the rest of the conversation is just a waste of time. You can argue about how awesome you think your equation is until you are blue in the face, but it is NOT an acceleration, and you cannot call it as such.
traveler Posted October 8, 2008 Author Posted October 8, 2008 (edited) Yup, I guess it's done, but that has nothing to do with the reality of it all that the acceleration due to gravity is proportional to the net force and inversely proportional to the distance apart. Additionally, as the distance decreases the acceleration increases! BTW, you can argue until you are blue in the face, but when I'm standing on a scale my acceleration is not 9.8m/s^2, it's 0 m/s^2. My velocity is zero m/s towards the center of the Earth, and it will remain 0 m/s, which means my acceleration is also 0 m/s^2. So standing on a scale has zero acceleration, so don't call it acceleration, because it's not! You say the velocity is increasing at the rate of 9.8 m/s^2??? Prove it! Isn't it funny how a formula that has no units () can produce the exact accelerations of two different objects at any specific distance away, and the difference in times of those two different mass objects to the precision of 10^-30 seconds? Amazing. Edited October 8, 2008 by traveler
mooeypoo Posted October 8, 2008 Posted October 8, 2008 Yup, I guess it's done, but that has nothing to do with the reality of it all that the acceleration due to gravity is proportional to the net force and inversely proportional to the distance apart. Additionally, as the distance decreases the acceleration increases! (I bolded the sentence, for a reason, it's not in the original) No. Reality of it all: F=mg *ON EARTH* [math]F=G\frac{m1m2}{r^2} (N)[/math] (notice, the G is bigger here, different constant altogether!) Read!: http://en.wikipedia.org/wiki/Universal_gravitation So your last sentence (The Bold) is not supposed to have the word "aditionally". It's supposed to have the word "THEREFORE" The force of attraction between any 2 massive bodies is directly proportional to the product of their masses and inversely proportional[/b'] to the square of the distance apart of their centre of mass. source: http://www.studentxpress.ie/educ/physics/physics4/physics4.html (Bold not in source, again) The term "inversly proportional" means that the relation is OPPOSITE: that when one gets larger, theother gets smaller, and vice versa. The relation between the distance and the masses is INVERSELY proportional. THEREFORE, when the distance decreases, the acceleration (part of force!) increases. The sentence above is just a phrasing of the equation for Newton's Universal Law of Gravitation. BTW, you can argue until you are blue in the face, but when I'm standing on a scale my acceleration is not 9.8m/s^2, it's 0 m/s^2. My velocity is zero m/s towards the center of the Earth, and it will remain 0 m/s, which means my acceleration is also 0 m/s^2. Fine, I'll be blue then. When you stand on the scale, it shows you the force that is applied on the scale. If there is force, there is acceleration. F=ma if a=0, then F=0. The force that is applied on the scales is your mass times the acceleration, which is your mass * the constant g = ~9.8 m/s^2 Notice, now, why units are important: Force = Body (kg) * 9.8 m/s^2 = 9.8*Body ( Kg * m/s^2) The unit that represents (Kg * m/s^2) is called Newton, which is a unit of force. The only reason you do not "feel it" is because the surface - the ground - is exerting THE SAME FORCE UPWARDS, and therefore your body is in equilibrium. Take away the surface, and you will start moving with g acceleration. Think of a trap door. You're standing on a scale that is on top of a trap door and look at the value on the scales. Then, you open the trap door. By your notion, you have no acceleration until the doors open. In that case, you have to answer the question of what STARTED this acceleration? what is the source that "poofed" into existence the second the trapdoors opened, and wasn't there before? But what does happen is that you will fall towards the ground at F=mg, and v=v0 + gt the reason you will start falling is because the quilibrium is "broken", by the fact that there is no longer a stable surface that exerts the same amount of force UPWARDS, and therefore your body starts moving in direction to the force, which is towards the center of the earth. Until your body meets the ground. If the ground's firm, it will exert N (normal force) that is equal to F=mg . If the ground's something like quicksand, it will exert N that is probably LESS than F=mg, and you will, again, keep moving downwards (just at a lesser pace). Do you see the difference? Gravity is Acceleration Gravity is Acceleration, indeed Yes, yes indeed, Gravity is acceleration. You do have gravity when you stand on the scales, you just evolved not to "notice" it as much as you notice a plane take off. It's biology. I suggest you review your basic physics. ~moo
traveler Posted October 8, 2008 Author Posted October 8, 2008 Mooeypoo, You talked alot about acceleration, but you never mentioned velocity. How could that be? How can you talk about the acceleration of an object without talking about the velocity of the object??? ACCELERATION IS THE RATE OF CHANGE OF VELOCITY! Zero velocity+ zero change in velocity=zero acceleration.
mooeypoo Posted October 8, 2008 Posted October 8, 2008 (edited) Do you agree that the general form F = ma is specialized on Earth-Body relationships as F = mg ? Because if you do, then you will see that the acceleration DOES EXIST, it is just "cancelled out" in the equation of rigid body in equilibrium. The fact the body is in equilibrium does NOT mean that the g stopped. g still exists. You are *ALWAYS* pulled towards the earth at a force that equals approximately 9.8 m/s^2 times your mass. Have you read my explanation about the trap door? How do ou explain it? When the doors suddenly open - what is the source of the sudden movement -- sudden acceleration? Do you have a little rocket engine on your head pushing you downwards? I would guess not. Do you have the tentacles of the Giant Spagetti Monster pulling you downwards? I would assume not. What, then, is it, that starts your movement downwards at mass*9.8 kg*m/s^2 ? Answer that, and perhaps it will clarify a few very important basic concepts in physics. The fact there is no movement (Velocity, and I *did* relate to that, I just called it movement instead of relating to it as velocity, thinking it might be easier not to think of it as CONSTANT, which is ISN'T), plainly means that your body is at equilibrium. Equilibrium is a physical concept that does NOT mean lack of forces, it just means that the sum of the net force is zero. You could have a dozen different forces acting on you from all different directions, pulling, pushing, lifting, dragging -- as long as the NET VALUE (the accumulative value, the total force) of *all* forces in all space is zero (hence, they cancel each other out), then you have an equilibrium state. The fact acceleration is cancelled by counter-acceleration does NOT mean acceleration is nonexistent. In any case, you still have a few other questions to answer, including some very important unit errors that render your equations unscientific and unphysical, which makes your theory is bunk before it even started. Fix those, and then we can continue examining how the math supports your theory properly. == EDIT == btw, here's a good reference too: http://en.wikipedia.org/wiki/Apparent_weight -- apparent weight explains about that "N" I was talking about (the normal force, the 'counter' force) - exerted by the earth ON YOU. Canceling out the *effect* of gravity, but NOT canceling out gravity itself. ~moo Edited October 8, 2008 by mooeypoo
big314mp Posted October 8, 2008 Posted October 8, 2008 Now I'm confused. I was under the impression that acceleration was rate of change of velocity. Therefore, with no change in velocity, there is zero acceleration. F=ma. If a is zero, this implies that F is zero. Since F is the net force on the object, then this implies that there is zero net force on the object. Zero net force can be achieved by zero total forces acting on the object, or all forces acting on the object cancel out. A person standing on a scale falls into the latter category. The force applied by the trap door and scale is the same as the force applied by gravity, so they cancel out. Should the trap door open, this counter force is gone. Then there is a net force on the object and the object experiences acceleration. The point of all of that was that it is logical to consider forces to cancel with each other. I'm lost when you speak of canceling accelerations of objects, as this seems to imply an object trying to accelerate in two different directions. Basically, if we made gravity go away, would the counter acceleration launch the person into space? Obviously not. We may be saying the exact same thing, in which case this is all moot.
mooeypoo Posted October 8, 2008 Posted October 8, 2008 Now I'm confused. I was under the impression that acceleration was rate of change of velocity. Therefore, with no change in velocity, there is zero acceleration. F=ma. If a is zero, this implies that F is zero. Since F is the net force on the object, then this implies that there is zero net force on the object. Zero net force can be achieved by zero total forces acting on the object, or all forces acting on the object cancel out. A person standing on a scale falls into the latter category. The force applied by the trap door and scale is the same as the force applied by gravity, so they cancel out. Should the trap door open, this counter force is gone. Then there is a net force on the object and the object experiences acceleration. The point of all of that was that it is logical to consider forces to cancel with each other. I'm lost when you speak of canceling accelerations of objects, as this seems to imply an object trying to accelerate in two different directions. Basically, if we made gravity go away, would the counter acceleration launch the person into space? Obviously not. We may be saying the exact same thing, in which case this is all moot. Okay, listen, I supplied links for a REASON. There is a lot in those links. A lot about why the Earth's gravity (g) which is NOT "stopping" out of the whim of the object that it's affecting is much like acceleration. Links that also explain the concept of equilibrium. Mainly, I believe we are talking about two different Fs. When you say there's no acceleration, you speak of the TOTAL F -- the *sum* of the forces applied, and in that aspect you are correct. But that doesn't mean that the other forces 'disappeared', they still exist, individually, they just cancel each other out in the BIGGER picture. So *g* still exists in the difference forces that apply, and therefore acceleration DOES exist in the different forms that apply. Maybe this will be easier to see. We have a body in equilibrium, so we write down all the forces that apply on that body: F1 = mg DOWNWARDS and F2 = -mg UPWARDS (Normal Force, exerted by a rigid surface) Because the system is in equilibrium, it means that the TOTAL force must be zero. So the total force: F(total) = F1 + F2 = 0 And so: F(total) = mg + (-mg) = mg - mg = 0 So the TOTAL foce is, indeed, 0, but that individual foces EXIST. Please, the references I supplied explain this much better, with diagrams and more resources. I didn't just throw them in for the kick of it, they are actually helpful. The point of this entire thing is to make sure that we understand that gravity (g) *never stops* --- when you stand on a scale, you are affected by gravity. The number you see on the scale is the Normal Force that is exerted BY the scales (rigid, hard, stable, nonmoving, surface) *upwards*, and since you are not falling down or moving up, both forces are equal. But they both still exist. When you now calculate OTHER forces - like movement forward, backwards or any other mass that might be affecting or interjecting the system, you *must* be aware of *all* the forces on the bodies and the system, regardless of what the total force is. At a certain point it's 0 (equilibrium, nonmovement), and at another it can be a certain value. You see my point? ~moo
big314mp Posted October 8, 2008 Posted October 8, 2008 I did in fact read your link, and my post was basically my best understanding of your link + what I have learned in physics. I think I misunderstood your point on this: "If there is force, there is acceleration." I interpreted force as total force, and acceleration as change in velocity. Thank you for clarifying.
D H Posted October 8, 2008 Posted October 8, 2008 Now I'm confused. I was under the impression that acceleration was rate of change of velocity. Therefore, with no change in velocity, there is zero acceleration. Later in your post you have no problem with viewing multiple forces acting on a body that together yield a net force of zero. Force and acceleration are both vectors, related by [math]\mathbf F = m\mathbf a[/math]. What then is the problem with viewing multiple accelerations acting on a body that together yield a net acceleration of zero? [math]\aligned \mathbf F_{\text{net}} &= \sum_i \mathbf F_i \\ \mathbf a_{\text{net}} &= \frac{F_{\text{net}}} m = \sum_i \frac{\mathbf F_i} m \endaligned [/math] There is no reason mathematically that one must compute the net force first and only then compute the acceleration. There is no difference between computing the net force and then the acceleration versus first dividing each force by the mass to yield the acceleration due to that force and summing these acceleration terms to yield the net acceleration. The two approaches are mathematically identical, so both views are equally valid. In the case of somebody standing on a scale, the gravitational force results in a downward acceleration, the normal force results in an upward acceleration. The two cancel, yielding a net acceleration of zero. Basically, if we made gravity go away, would the counter acceleration launch the person into space? Obviously not. Obviously not, because the counter force (and hence the acceleration) would go away. The normal force is a constraint force. If gravity were not pulling the person down, there would be no normal force. 1
big314mp Posted October 8, 2008 Posted October 8, 2008 I understand the mathematics behind it (somewhat better, now that you have put it in sigma notation), but the notion of a body experiencing acceleration in two different directions seemed nonsensical to me, so I assumed it was just an artifact of the equation, if that makes any sense. I see that they are equivalent, but I think I'll stick with thinking in terms of forces. Thanks for the lesson though!
mooeypoo Posted October 8, 2008 Posted October 8, 2008 I did in fact read your link, and my post was basically my best understanding of your link + what I have learned in physics. I think I misunderstood your point on this: "If there is force, there is acceleration." I interpreted force as total force, and acceleration as change in velocity. Thank you for clarifying. BTW, re-reading my remark, I noticed I might've sounded a bit rude. I didn't mean to, I just felt like my explanations weren't as good as the links I provided, and seeing as traveler has a tendency to occasionally ignore things, I got a bit frustrated. In any case, yes, I meant to say that gravity doesn't vanish, even when the TOTAL force is zero, that was mainly my point. ~moo I understand the mathematics behind it (somewhat better, now that you have put it in sigma notation), but the notion of a body experiencing acceleration in two different directions seemed nonsensical to me, so I assumed it was just an artifact of the equation, if that makes any sense. I see that they are equivalent, but I think I'll stick with thinking in terms of forces. Thanks for the lesson though! I think I have an idea of how to create an example that you may visualise - it might help you understand how it can happen. Instead of a body resting on the floor, think of yourself pushing a cart full of stones on top of an ice rink (I don't want friction, that's why ). You're using force forward to push, and the cart is moving forward at F1=ma .. we won't relate to whatever friction is this time, so theoretically, the cart will accelerate at F=ma. Now.. you keep exerting the SAME amount of force F1, only this time I come to the other side of the cart and start pushing it too, against your movement. The *total* force on the cart would be F(total) = F1 + F2 Let's, for the sake of sanity and reference, pick your forward (the initial movement) as the positive movement. That would mean that my force is negative (it's on the opposite side). So, if my push - my Force F2 - is less than yours, the cart will still move on the positive "sign" (hence, away from you - your 'forward') but with lesser *total* force than before. If my F2 is higher than yours, then the total force would have a negative sign, and you'd be pushed back.. And if both our forces are equal, the cart will STAY in its place. But you're still pushing. And so am I, with all my might. Regardless of the movement or lack thereof of the cart, the *forces* still exist. Did that help?
traveler Posted October 8, 2008 Author Posted October 8, 2008 (edited) I was under the impression that acceleration was rate of change of velocity. Therefore, with no change in velocity, there is zero acceleration. F=ma. If a is zero, this implies that F is zero. Since F is the net force on the object, then this implies that there is zero net force on the object. Zero net force can be achieved by zero total forces acting on the object, or all forces acting on the object cancel out. A person standing on a scale falls into the latter category. The force applied by the trap door and scale is the same as the force applied by gravity, so they cancel out. Should the trap door open, this counter force is gone. Then there is a net force on the object and the object experiences acceleration. The point of all of that was that it is logical to consider forces to cancel with each other. I'm lost when you speak of canceling accelerations of objects, as this seems to imply an object trying to accelerate in two different directions. Basically, if we made gravity go away, would the counter acceleration launch the person into space? Obviously not. We may be saying the exact same thing, in which case this is all moot. Great post. That is exactly my thoughts also. There is a slight difference, though. That infinitesimally small difference is also a mistake of infinite magnitude. Edited October 8, 2008 by traveler
DrP Posted October 8, 2008 Posted October 8, 2008 Great post. That is exactly my thoughts also. . But after that post - the others explained it to him - now he gets it and he thanked them for clarifying no? Anyway, Newton I think: " For every action there is an equal and opposite action" Therfore F=mg down is met with Force UP which is EQUAL and OPPOSITE - thus the NET forces and accelarations are zero. They are still there in both directions (Which I'm sure has been said already - sorry to jump mid conversation)
traveler Posted October 8, 2008 Author Posted October 8, 2008 (edited) But after that post - the others explained it to him - now he gets it and he thanked them for clarifying no? Anyway, Newton I think: " For every action there is an equal and opposite action" Therfore F=mg down is met with Force UP which is EQUAL and OPPOSITE - thus the NET forces and accelarations are zero. They are still there in both directions (Which I'm sure has been said already - sorry to jump mid conversation) What are you measuring when you measure the velocity of the object, and which object's velocity are you measuring? Like I said, the velocity is zero, acceleration is zero, and is NOT 9.8 m/s^2. Edited October 8, 2008 by traveler
mooeypoo Posted October 8, 2008 Posted October 8, 2008 What are you measuring when you measure the velocity of the object, and which object's velocity are you measuring? Like I said, the velocity is zero, acceleration is zero, and is NOT 9.8 m/s^2. Didn't I just explain this, with resource and references, and a nice visual example, along with D H ? So what we have here is that you say one thing while reality says another. And you're not supporting what you are saying with (shriek) evidence (shriek) or (shriek) proper math. I explained the abvove, you just didn't like the conclusion. Your starting to live up to your Troll.... and that's not a good thing.
traveler Posted October 8, 2008 Author Posted October 8, 2008 (edited) Didn't I just explain this, with resource and references, and a nice visual example, along with D H ? So what we have here is that you say one thing while reality says another. And you're not supporting what you are saying with (shriek) evidence (shriek) or (shriek) proper math. I explained the abvove, you just didn't like the conclusion. Your starting to live up to your Troll.... and that's not a good thing. DH has not answered my questions yet. He never responded to me. We are measuring the distance and time of an object. An object that is not getting closer to the center of the Earth (that is what you are measuring, correct?) has a distance of zero, regardless of the elapsed time measured. The object's velocity is ZERO m/s. If I wait 10 minutes and recheck the distance, and the distance is unchanged, the acceleration is ZERO m/s^2. My acceleration due to gravity while I'm standing on the Earth is not, in any way, shape, or form 9.8 m/s^2 like you say it is. Edited October 8, 2008 by traveler
D H Posted October 8, 2008 Posted October 8, 2008 Anyway, Newton I think: " For every action there is an equal and opposite action" Therfore F=mg down is met with Force UP which is EQUAL and OPPOSITE - thus the NET forces and accelarations are zero. They are still there in both directions (Which I'm sure has been said already - sorry to jump mid conversation) You have a very common misperception of Newton's third law here. Newton's third law talks about forces exerted on a pair of bodies. If body A exerts some force on body B, Newton's third law says that body B exerts a force on body A that is equal but opposite to the force A exerts on B. For a person standing on the Earth, the Earth as a whole is pulling on the person gravitationally and the little patch of Earth under the person's feet is pushing the person up. This is not a third law interaction. Suppose you skydive from a stationary helicopter. There is no wind resistance when you first jump out. The only force acting on you is gravity. So what is the third law interaction here? Simple: The Earth is falling toward you gravitationally! Let's say for the sake of argument that your mass is 75 kilograms. The Earth is exerting a force of 735.75 newtons on you. You are exerting a force of 735.75 newtons on the Earth. While the forces are equal but opposite, the resultant accelerations are anything but. You fall toward the Earth at 9.81 m/s^2, and the Earth falls toward you 1.23×10-22 m/s^2. While the Earth's acceleration toward you is immeasurably small, the effects are observable for a much larger mass -- the Moon, for instance. The exact same thing is happening when stand on the Earth. The Earth is still exerting a gravitational force of 735.75 newtons on you, and you exert a gravitational force of 735.75 newtons on the Earth. Yet another third law interaction occurs when you are stand on the Earth. It is called the normal force. You aren't really touching the surface of the Earth. Your feet are hovering a tiny, tiny distance above the surface. The electrons in your feet repel the electrons in the little patch of Earth just below your feet. Pick up a heavy book and that hover distance will shrink a tiny bit. The slightly closer distance means the Earth is exerting a slightly larger normal force on you. By Newton's third law, you are exerting a counter force on the Earth. If you instead stand on a scale, things get even more complex. The third law interactions now are You and the Earth attract each other gravitationally You push down on the top of the scale and the top of the scale pushes up on you (normal force) The top of the scale pushes down on the top of the spring and the top of the spring pushes the top of the scale up (rigid forces) The top of the spring moves downward, compressing the spring (elastic force). The spring elastic force makes the top of the spring push upward and the bottom of the spring push downward. The bottom of the spring pushes down on the bottom of the scale and the bottom of the scale pushes up on the bottom of the spring (rigid forces) The bottom of the scale pushes down on the surface of the Earth, which pushes up on the bottom of the scale (normal force). Whew! If you poke a bit deeper, those things I called normal force, rigid force, and elastic force all arise from the electromagnetic force.
traveler Posted October 8, 2008 Author Posted October 8, 2008 Fine, I'll be blue then. When you stand on the scale, it shows you the force that is applied on the scale. If there is force, there is acceleration. You're turning purple. The bold is where you make your mistake. That is an incorrect statement.
mooeypoo Posted October 8, 2008 Posted October 8, 2008 (edited) DH has not answered my questions yet. He never responded to me. Well, I did... is there a difference between his physics and my physics? We are measuring the distance and time of an object. An object that is not getting closer to the center of the Earth (that is what you are measuring, correct?) Doesn't matter. Force is applied. Multiple forces are applied. Only when you calculate the total force it's gone. If you've bothered to read my visual example, then think of two situations with the cart of stones on an ice rink: 1. The full cart is standing, alone, in the middle of the rink, without moving. 2. The full cart is standing in the middle of the rink while I push on one side and big314mp is pushing from the other side at the same force. If you don't see the physical, practical and realistic difference between two cases (which it seems your statements show you don't), then you have some basic physics studying to do, my friend. There is a huge difference between the statement "there is no force" and the statement "the forces cancel each other out". has a distance of zero, regardless of the elapsed time measured. Actually, that too depends what you are looking at. Physics is about perspective. If you look at the object while you're next to it that might be true (that there is no movement, not that there is no force), but if you look at it from a space shuttle, that is absolutely NOT true; the object moves along with the Earth. The moon's a great example. If you look from the surface of the Earth during the night, then the moon never rotates, it just orbits the Earth, but does not rotate - you can never see its "dark side" (meh) - it's far side is never ever facing the Earth. Practically speaking, it's not rotating. But that's not so. The moon *does* rotate, it just completes a rotation cycle the same time it completes an orbit cycle, which causes it to 'always face' the Earth. That doesn't mean it's not rotating.... The object's velocity is ZERO m/s. TROLL. I explained this, D H explained this, I even gave out links (3 in the beginning, one more later) that EXPLAIN why gravity is acceleration, and why even when there's no speed there is still force, and why that observation is very important. You ignored it, you keep asking a question that was already answered. That's called TROLLING. Go back up and read it again, I have already answered, and I've used physics to answer you. If I wait 10 minutes and recheck the distance, and the distance is unchanged, the acceleration is ZERO m/s^2. My acceleration due to gravity while I'm standing on the Earth is not, in any way, shape, or form 9.8 m/s^2 like you say it is. Well, technically, you've travelled with the ball, so the distance RELATIVE TO YOU, hadn't changed, but that's physics 201, and you seem to have troubles with the basics, so let's ignore that for a moment. Your definitions are simply, quite bluntly, wrong. There's nothing more to it than that. Acceleration is part of force and it is existing whenever there is a force. If you choose to view the system as *net force only* (that is, to only state that the F=0, while ignoring the forces that make up F) then you are right, but you are also not doing very good physics, since you're ignoring the *real* situation. So while you can think of things in terms of Fnet ONLY, you can't just jump back and forth from "what is it really there" assertion to "but I ignore what's really there and treat the total as a whole" assertion. If you go into details, you go into details, you can't just ignore whatever you feel like. Well, you can, but that's trolling. I'm putting a lot of time and thought into my answers to you, trying to explain what physics really is, as opposed to what you seem to think it is. Ignoring whatever you want is not making your assertions any tru'er, and to be frank, it's quite rude. ~moo You're turning purple. The bold is where you make your mistake. That is an incorrect statement. Troll. Edited October 8, 2008 by mooeypoo multiple post merged
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