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From: TomGee on 11 Oct 2005 22:54 Randy Poe wrote: > TomGee wrote: > > Randy Poe wrote: > > > TomGee wrote: > > > > Randy Poe wrote: > > > > > TomGee wrote: > > > > > > Randy Poe wrote: > > > > > > > > > > I said this: > > > > > > > I'm saying that work is done by the component of force > > > > > > > in the direction of motion. > > > > > > > > > > > > > > If the force is ALWAYS at right angles to the motion, no > > > > > > > work is done. > > > > > > > > > > And Tom, apparently in reply, said this: > > > > > > > > > > > Ah yes, so then work is being done by the component of force which you > > > > > > claim is a non-existent force? > > > > > > > > > > I can't connect this to what I said since I said nothing about > > > > > (a) work being done (I was explicitly talking about work NOT > > > > > being done) and > > > > > (b) a non-existent force. > > > > > > > > > > > > > > Yes, you did: (a)"I'm saying that work is done by the component of > > > > force in the direction of motion" In English that means _when and if > > > > ever_ work is done, it is then being done by the "force" > > > > > > That's correct. And it's being done by the component of > > > force which is in the direction of motion. > > > > > > > which you mentioned but failed to name. > > > > > > Eh? It's a general statement. When work is being done, it's > > > being done by the component of force in the direction of > > > the motion. What "name" is missing there? > > > > > > > > Oh you're sooo coy! > > No coyness, you're just confused. There are three separate situations > we're talking about, and you keep taking words I say from one > and applying them to another. I'm not being "coy" when I try > to put those words back where they were said. > > The situations are: > (a) No external forces. Velocity and KE are constant. > > Velocity can be constant but aren't you saying that KE does not change? I asked you to support that claim before, to no avail. Reference please, again. > > > There > is no force to do any work, so discussions of "work" > are meaningless. > > I repeat: But you said a force performs work on an object at CV, did you not? You brought up that term, not me. Inertia is resistance to change, and an object in orbit resists falling into the source of the right angle force, therefore the right angle force must be resisted by some other force because it cannot be resisted by a quantity (of momentum). You claim there is no other force involved, so what provides the resistance to the right angle force? > > > (b) External force present, but at right angles to the > motion (e.g. circular motion). Here there is a force > which could do work, but the geometry is such that > no work is done. > > I see you are also confused about gravitation. The force that could do work, what's its name? And how does the geometry prevent it from doing work? > > > Speed and KE are constant. Velocity > is not, in the sense that direction changes. > > (c) External force present, not at right angles to the > motion. Now the force is doing work. > > I see, you mean that any force but a right angle force does work. You're making up your own physics, that's what I see. > > > Speed and KE are > changing. If the force does 1 Joule of work, the KE > changes by 1 Joule unless there are losses (due to > friction). But at any rate the KE can't change unless > the force does work to change it. > > > > Right, no work is being done in that case. > > > > > > > wrt its KE. > > > > > > That's meaningless. What does "work wrt its KE" mean? > > > > > > > > More coyness! > > What's "coy?". "Work wrt KE" is an abuse of the language. > It doesn't mean anything. > > Well, I only repeated what you said. Did you forget those are your words? > > > > You said that the KE of an object does not change under > > the right angle force because under that force no work is being done, > > and we all agree with that. > > Good. This is situation (b) > > > You then supported that statement by > > saying that some other force does work moving a body in its "direction > > of motion". > > Not in situation (b). I said that in contrast to this > situation, there is a *different* situation which I'm now > calling (c). > > Oh I see: You forgot to mark that as "not in situation (b)" when you first said that. > > > > > That is correct of course but you did not think it through well enough > > > > to notice that your unnamed "...component of force in the direction of > > > > motion" is the (b)"non-existent" force which you et al claim is not > > > > needed by sole objects having zero net external forces to move at CV. > > > > > > That's right, I didn't think that there is a component of > > > force in the case of zero force > > > > > Well, duuuh! Yes, you didn't think.... > > And I still don't think that I have to talk about "unnamed > components of force" in situations where the force is zero. > > Well of course you don't, unless you want to. > > > Obviously you think that when there are no forces, there's > a force to talk about anyway. OK, please describe to me > the external nonzero forces in the situation where the > external forces are zero. > > Oh you're whipping the hell out of your pal PD with your funny bizness! > > > > Now that you have thought it over a bit, don't you think that means > > there is a force in the case of nonzero force? > > Yes, but you will notice that when I said "I didn't think that > there is a component of force in the case of zero force" I was > referring to the case of ZERO force. That's different from > NONZERO. See the difference? ZERO... NONZERO. They mean different > things. > > Looks like even PD has deserted you. I asked him to help you out but he ignored my request. Can't blame him; he's got his problems trying to figure out what happened to his previously happy place here. > > > > > When the force is zero, all components are zero. > > > > > What components are those? > > The components in any direction. > > I'm sure you mean something by that, but I can't figure it out. > > > If I have zero force, I > can also say I have zero horizontal force, zero vertical > force, zero upward force, zero northward force, zero > eastward force... > > Do you disagree? > > With what? I've already told you can say anything you wish. > > > > You don't have to keeping that same statement over and over anymore, > > you have said it enough times to where we are convinced you understand > > that zero means zero. > > Good. Yet you claim there is a force present in the case of > no external forces. > > Right. The inherent force claimed by Newton and first defined by Aristotle. > > > > > When no work is being done, as in the case of constant > > > velocity, there are no external forces present. > > > > > Wrooonnnnggggg! There is no work being done on the object by any NET > > external forces. Only in the case of CV can we say there are no net > > external forces acting on the object. > > I just said that. > > You think you just said that, but you left out the necessary word NET again. > > > > > When there are no external forces, there is no work being > > > done, and KE is constant. > > > > > Reference, please. > > It's in the Tipler quote you provided. > > No, it is not. > > > > > > > > What force makes the work in moving the body? > > > > > > > > > > There is no "work" being done in moving a body at constant > > > > > velocity in the absence of any external forces. > > > > > > > > > But you said above that work is done by "the component of force in the > > > > direction of motion". > > > > > > Yes, and in the case of constant velocity, there is no > > > force. > > > > > But that contradicts your statement that work is done by a force in the > > direction of motion because CV has a direction of motion to it! > > Mixing of situations (a) and (c). > > When a force is present, work is done by the force in the > direction of motion. > > Okay. When a force is present, what is the name of the force doing the work in the direction of motion? > > > When a force is not present, there is no force "in the direction > of motion" or in any other direction. > > Well of course, when no force is present. But Newton's inertial inherent force is always present in a body, so there is no time when a force is not present for an object. > > > When I say "work is done by the component of force in the > direction of motion", I am not saying "when there's motion there's > a force in that direction". I'm saying "when there's a force > and it's in the direction of motion, it's doing work." > > No, that's not what you said at first. You're saying that now trying to squirm your way out of your hole but I don't thing readers are so stupid as to let you get away with it.
From: Randy Poe on 12 Oct 2005 23:20 TomGee wrote: > Randy Poe wrote: > > The situations are: > > (a) No external forces. Velocity and KE are constant. > > > > > Velocity can be constant but aren't you saying that KE does not change? If velocity is constant, KE is constant. > I asked you to support that claim before, to no avail. Reference > please, again. What claim exactly? That when velocity is constant, KE is constant? Or that KE and velocity are constant in the absence of external forces? > > There > > is no force to do any work, so discussions of "work" > > are meaningless. > > > I repeat: But you said a force performs work on an object at CV, Situations (b) and (c) muddled. There are situations with force (situation (a)), and situations without force (b and c). When there is no force present, I didn't say anything about what a (non-existant) force does, in terms of work or anything else. I also didn't say a force performs work on an object at constant velocity. If a force is performing work on an object, the velocity is NOT constant. > did you not? You brought up that term, not me. Which term? Force, work, or constant velocity? I brought up force in discussing situations in which force is present. Constant velocity is not such a situation. Most recently, I brought the concept of force and work into the discussion on the question of circular motion, which is emphatically *NOT* either a forceless or a constant-velocity situation. You will recall that Newton said that bodies, in the absence of forces, remain at rest or in straight line motion. Circular motion is *not* rest, and it is *not* straight line motion. > Inertia is resistance to > change, and an object in orbit resists falling into the source of the > right angle force, therefore the right angle force must be resisted by > some other force because it cannot be resisted by a quantity (of > momentum). You claim there is no other force involved, so what > provides the resistance to the right angle force? There is no "resistance to the right angle force". The force is applied and the path changes. > > (b) External force present, but at right angles to the > > motion (e.g. circular motion). Here there is a force > > which could do work, but the geometry is such that > > no work is done. > > > I see you are also confused about gravitation. The force that could do > work, what's its name? If you want to talk about gravity, it would be the gravitational force, in classical physics. But situation (b) refers to ANY circular motion. The force could also be due to a magnetic field, a piece of string, or the walls of a rotating spacecraft. The same principles apply. > And how does the geometry prevent it from doing > work? Because the motion and force are at right angles, so the force has no component in the direction of motion. As it acts it merely deflects the object, neither speeding it up nor slowing it down. > > Speed and KE are constant. Velocity > > is not, in the sense that direction changes. > > > > (c) External force present, not at right angles to the > > motion. Now the force is doing work. > > > I see, you mean that any force but a right angle force does work. Yes. > You're making up your own physics, that's what I see. http://www.physics.ucla.edu/k-6connection/force,wp.htm "Work is done when a force is applied through a distance and the force is in the same direction as the distance moved." http://www.andrew.cmu.edu/course/33-231/F-V.pdf (Equation 25 shows the general definition of work. It shows a vector dot product between force F and displacement dr. That is equal to the component of F along the direction of displacement, and is zero when F and dr are perpendicular). http://www.glenbrook.k12.il.us/gbssci/phys/Class/energy/u5l1a.html (Note the equation W = F*d*cos(theta), and the discussion of what happens when theta is 90 degrees. This by the way is another way of writing the dot product of F and d). http://hyperphysics.phy-astr.gsu.edu/hbase/work.html "Work refers to an activity involving a force and movement in the directon of the force." (Notice that the movement must be in the direction of the force) http://www.mcasco.com/p1wke.html "Work is defined as force multiplied by displacement. Remember in the general case, force and displacement are vectors. Work is a scalar quantity so the multiplication operation we are talking about here is the dot product as discussed back in Vector Arithmetic. If the fellow in the picture at the right was actually holding up his end of that bit of pipe while the machine held up the other, neither of them would be doing any work since the force (upward) is perpendicular to the motion (horizontal)." And I have no doubt that Tipler to will mention the relationship between force, displacement, work and angle, and will also point out somewhere that no work is done when force and displacement are at right angles. Imagine my power! I made this up just a few days ago for our discussion, and already it's found its way into all these websites!!! > > > > That's meaningless. What does "work wrt its KE" mean? > > > > > > > More coyness! > > > > What's "coy?". "Work wrt KE" is an abuse of the language. > > It doesn't mean anything. > > > Well, I only repeated what you said. Did you forget those are your > words? I defy you to find a post where I used the phrase "work with respect to KE". > > Not in situation (b). I said that in contrast to this > > situation, there is a *different* situation which I'm now > > calling (c). > > > Oh I see: You forgot to mark that as "not in situation (b)" when you > first said that. Yes. I contrasted this with circular motion, but somehow you got the impression that I was still talking about circular motion, and that AT THE SAME TIME I was still talking about straight line constant velocity motion. Hope you have it all straightened out now, and realize that when I'm talking about circular motion I'm not talking about straight line motion. > > And I still don't think that I have to talk about "unnamed > > components of force" in situations where the force is zero. > > > Well of course you don't, unless you want to. No, even if I want to, a force which is zero is not going to have a nonzero component in any direction. > > Obviously you think that when there are no forces, there's > > a force to talk about anyway. OK, please describe to me > > the external nonzero forces in the situation where the > > external forces are zero. > > > Oh you're whipping the hell out of your pal PD with your funny bizness! Non-responsive. Please describe to me the external nonzero forces in the situation where the external forces are zero. > > Yes, but you will notice that when I said "I didn't think that > > there is a component of force in the case of zero force" I was > > referring to the case of ZERO force. That's different from > > NONZERO. See the difference? ZERO... NONZERO. They mean different > > things. > > > Looks like even PD has deserted you. I asked him to help you out but > he ignored my request. He's enjoying the show. What exactly do you think I need to be "helped out" with? > > > > When the force is zero, all components are zero. > > > > > > > What components are those? > > > > The components in any direction. > > > I'm sure you mean something by that, but I can't figure it out. Did you ever study vectors? A vector quantity, a thing with a magnitude and an arbitrary direction, can be viewed as being made up (in general) of three components, along the x, y and z axes. These three component vectors, added together, sum up to the original vector. Here's a high-school level lesson on vectors and components which teaches the concept for 2-space (x and y axes only). http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l1d.html I think that's enough new material for now. Enjoy. - Randy
From: TomGee on 13 Oct 2005 21:31 Randy Poe wrote: > TomGee wrote: > > Randy Poe wrote: > > > The situations are: > > > (a) No external forces. Velocity and KE are constant. > > > > > > > > Velocity can be constant but aren't you saying that KE does not change? > > If velocity is constant, KE is constant. > > > I asked you to support that claim before, to no avail. Reference > > please, again. > > What claim exactly? That when velocity is constant, KE > is constant? Or that KE and velocity are constant in the > absence of external forces? > > > > There > > > is no force to do any work, so discussions of "work" > > > are meaningless. > > > > > I repeat: But you said a force performs work on an object at CV, > > Situations (b) and (c) muddled. > > Evasion noted. > > > There are situations with force > (situation (a)), and situations without force (b and c). When > there is no force present, I didn't say anything about what a > (non-existant) force does, in terms of work or anything else. > > Ha ha. > > > I also didn't say a force performs work on an object at constant > velocity. > > Right. You denied that even though Newton said a body has an inherent force that provides its motive power. > > > If a force is performing work on an object, the > velocity is NOT constant. > > Yes, we know that's your position, but you have no way to show that Newton was wrong about that. > > > > did you not? You brought up that term, not me. > > Which term? Force, work, or constant velocity? > > I brought up force in discussing situations in which force is > present. Constant velocity is not such a situation. > > Most recently, I brought the concept of force and work into > the discussion on the question of circular motion, which is > emphatically *NOT* either a forceless or a constant-velocity > situation. > > You will recall that Newton said that bodies, in the absence > of forces, remain at rest or in straight line motion. > > No, that's wrong. He said, "...in the absence of any net external forces". Can you see the difference in the two statements, or do you think they are the same? > > >Circular > motion is *not* rest, and it is *not* straight line motion. > > Oh, did someone say something to the contrary about that? If not, why do you mention something everyone knows and agrees with already? > > > > Inertia is resistance to > > change, and an object in orbit resists falling into the source of the > > right angle force, therefore the right angle force must be resisted by > > some other force because it cannot be resisted by a quantity (of > > momentum). You claim there is no other force involved, so what > > provides the resistance to the right angle force? > > There is no "resistance to the right angle force". The force > is applied and the path changes. > > I see you are still confused about the gravitational force vs a body's inertia. > > > > > (b) External force present, but at right angles to the > > > motion (e.g. circular motion). Here there is a force > > > which could do work, but the geometry is such that > > > no work is done. > > > > > I see you are also confused about gravitation. The force that could do > > work, what's its name? > > If you want to talk about gravity, it would be the gravitational > force, in classical physics. > > And in modern physics it would be what? Curved space? Nah nah, curved space is not a force, remember? You don't know that the gravitational force keeps a body in orbit while the body's inertia tries to keep it going straight? > > > But situation (b) refers to ANY circular motion. The force could > also be due to a magnetic field, a piece of string, or the walls > of a rotating spacecraft. The same principles apply. > > > And how does the geometry prevent it from doing > > work? > > Because the motion and force are at right angles, so the force has > no component in the direction of motion. As it acts it merely > deflects the object, neither speeding it up nor slowing it > down. > > Okay, I see what you meant, but I do not see why you insist that Newton's inherent force in a body is not doing any work since the force is in the direction of motion. > > > > > Speed and KE are constant. Velocity > > > is not, in the sense that direction changes. > > > > > > (c) External force present, not at right angles to the > > > motion. Now the force is doing work. > > > > > I see, you mean that any force but a right angle force does work. > > Yes. > > > You're making up your own physics, that's what I see. > > So, the gravitational force is not doing any work in deflecting the body from a straight path, but Newton's inherent force in the body is being applied in the direction of the body's motion, so isn't it doing work then? > > > http://www.physics.ucla.edu/k-6connection/force,wp.htm > "Work is done when a force is applied through a distance and the force > is in the same direction as the distance moved." > > http://www.andrew.cmu.edu/course/33-231/F-V.pdf > (Equation 25 shows the general definition of work. It shows > a vector dot product between force F and displacement dr. That > is equal to the component of F along the direction of displacement, > and is zero when F and dr are perpendicular). > > http://www.glenbrook.k12.il.us/gbssci/phys/Class/energy/u5l1a.html > (Note the equation W = F*d*cos(theta), and the discussion of what > happens when theta is 90 degrees. This by the way is another way > of writing the dot product of F and d). > > http://hyperphysics.phy-astr.gsu.edu/hbase/work.html > "Work refers to an activity involving a force and movement in the > directon of the force." (Notice that the movement must be in > the direction of the force) > > http://www.mcasco.com/p1wke.html > "Work is defined as force multiplied by displacement. Remember in > the general case, force and displacement are vectors. Work is a > scalar quantity so the multiplication operation we are talking about > here is the dot product as discussed back in Vector Arithmetic. If the > fellow in the picture at the right was actually holding up his end of > that bit of pipe while the machine held up the other, neither of them > would be doing any work since the force (upward) is perpendicular to > the motion (horizontal)." > > And I have no doubt that Tipler to will mention the relationship > between force, displacement, work and angle, and will also point > out somewhere that no work is done when force and displacement > are at right angles. > > Imagine my power! I made this up just a few days ago for our > discussion, and already it's found its way into all these > websites!!! > > Yeah right. > > > > > > > That's meaningless. What does "work wrt its KE" mean? > > > > > > > > > More coyness! > > > > > > What's "coy?". "Work wrt KE" is an abuse of the language. > > > It doesn't mean anything. > > > > > Well, I only repeated what you said. Did you forget those are your > > words? > > I defy you to find a post where I used the phrase "work with > respect to KE". > > > > Not in situation (b). I said that in contrast to this > > > situation, there is a *different* situation which I'm now > > > calling (c). > > > > > Oh I see: You forgot to mark that as "not in situation (b)" when you > > first said that. > > Yes. I contrasted this with circular motion, but somehow you > got the impression that I was still talking about circular motion, > and that AT THE SAME TIME I was still talking about straight > line constant velocity motion. Hope you have it all straightened > out now, and realize that when I'm talking about circular motion > I'm not talking about straight line motion. > > > > And I still don't think that I have to talk about "unnamed > > > components of force" in situations where the force is zero. > > > > > Well of course you don't, unless you want to. > > No, even if I want to, a force which is zero is not going to > have a nonzero component in any direction. > > > > Obviously you think that when there are no forces, there's > > > a force to talk about anyway. OK, please describe to me > > > the external nonzero forces in the situation where the > > > external forces are zero. > > > > > Oh you're whipping the hell out of your pal PD with your funny bizness! > > Non-responsive. > > Please describe to me the external nonzero forces in the situation > where the external forces are zero. > > > > Yes, but you will notice that when I said "I didn't think that > > > there is a component of force in the case of zero force" I was > > > referring to the case of ZERO force. That's different from > > > NONZERO. See the difference? ZERO... NONZERO. They mean different > > > things. > > > > > Looks like even PD has deserted you. I asked him to help you out but > > he ignored my request. > > He's enjoying the show. > > No, I think he's having the same problems as you. > > > What exactly do you think I need to be "helped out" with? > > > > > > When the force is zero, all components are zero. > > > > > > > > > What components are those? > > > > > > The components in any direction. > > > > > I'm sure you mean something by that, but I can't figure it out. > > Did you ever study vectors? A vector quantity, a thing with a > magnitude and an arbitrary direction, can be viewed as being > made up (in general) of three components, along the x, y and > z axes. These three component vectors, added together, sum > up to the original vector. > > Here's a high-school level lesson on vectors and components > which teaches the concept for 2-space (x and y axes only). > http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u3l1d.html > > I think that's enough new material for now. Enjoy. > > I'm sure you think all that has something to do with our topic, but it doesn't.
From: TomGee on 14 Oct 2005 19:09 Randy Poe wrote: > TomGee wrote: > > Randy Poe wrote: > > > The situations are: > > > (a) No external forces. Velocity and KE are constant. > > > > > > > > Velocity can be constant but aren't you saying that KE does not change? > > If velocity is constant, KE is constant. > > > I asked you to support that claim before, to no avail. Reference > > please, again. > > What claim exactly? That when velocity is constant, KE > is constant? Or that KE and velocity are constant in the > absence of external forces? > > > > There > > > is no force to do any work, so discussions of "work" > > > are meaningless. > > > > > I repeat: But you said a force performs work on an object at CV, > > Situations (b) and (c) muddled. There are situations with force > (situation (a)), and situations without force (b and c). When > there is no force present, I didn't say anything about what a > (non-existant) force does, in terms of work or anything else. > > I also didn't say a force performs work on an object at constant > velocity. If a force is performing work on an object, the > velocity is NOT constant. > > Here's what you said, "work is done by the component of force in the direction of motion". Newton's body having no external net forces acting upon it moves in one direction at constant velocity, which supports your and everyone else's assertion that work is done by a "force in the direction of motion". > > > > did you not? You brought up that term, not me. > > Which term? Force, work, or constant velocity? > > I brought up force in discussing situations in which force is > present. Constant velocity is not such a situation. > > Yet in those same situations you insisted there was no right angle force present, so which force were you talking about that was present? > > > Most recently, I brought the concept of force and work into > the discussion on the question of circular motion, which is > emphatically *NOT* either a forceless or a constant-velocity > situation. > > No, not "most recently", but immediately. And above you state that the question of circular motion is "NOT...a forceless situation" (meaning that there is a force present), so if there is no right angle force present, which force is present? I know the answer and you know the answer, and everyone reading this knows the answer - the question is, are you willing to let the correct answer come out of your mouth and get some sympathy from the onlookers for your plight, or refuse to submit to the truth and thus destroy any lingering belief that you may have some common sense after all. > > > You will recall that Newton said that bodies, in the absence > of forces, remain at rest or in straight line motion. Circular > motion is *not* rest, and it is *not* straight line motion. > > No, I recall no such thing. You left out - deliberately I'm sure - the more important words of his statement, which are, "in the absence of any _net external_ forces". No real physicist would wrongly say, as you have, "in the absence of forces...." You want me to be wrong so badly that you have resorted to fudging on quotes, and that's sad. > > > > Inertia is resistance to > > change, and an object in orbit resists falling into the source of the > > right angle force, therefore the right angle force must be resisted by > > some other force because it cannot be resisted by a quantity (of > > momentum). You claim there is no other force involved, so what > > provides the resistance to the right angle force? > > There is no "resistance to the right angle force". The force > is applied and the path changes. > > Oh, of course there is, silly! Inertia is a body's resistance to change when acted upon by a directional force. The inertial force resists the gravitational force. Everyone knows that. > > > > > (b) External force present, but at right angles to the > > > motion (e.g. circular motion). Here there is a force > > > which could do work, but the geometry is such that > > > no work is done. > > > > > I see you are also confused about gravitation. The force that could do > > work, what's its name? > > If you want to talk about gravity, it would be the gravitational > force, in classical physics. > > But situation (b) refers to ANY circular motion. The force could > also be due to a magnetic field, a piece of string, or the walls > of a rotating spacecraft. The same principles apply. > > > And how does the geometry prevent it from doing > > work? > > Because the motion and force are at right angles, so the force has > no component in the direction of motion. As it acts it merely > deflects the object, neither speeding it up nor slowing it > down. > > > > Speed and KE are constant. Velocity > > > is not, in the sense that direction changes. > > > > > > (c) External force present, not at right angles to the > > > motion. Now the force is doing work. > > > > > I see, you mean that any force but a right angle force does work. > > Yes. > > > You're making up your own physics, that's what I see. > > http://www.physics.ucla.edu/k-6connection/force,wp.htm > "Work is done when a force is applied through a distance and the force > is in the same direction as the distance moved." > > http://www.andrew.cmu.edu/course/33-231/F-V.pdf > (Equation 25 shows the general definition of work. It shows > a vector dot product between force F and displacement dr. That > is equal to the component of F along the direction of displacement, > and is zero when F and dr are perpendicular). > > http://www.glenbrook.k12.il.us/gbssci/phys/Class/energy/u5l1a.html > (Note the equation W = F*d*cos(theta), and the discussion of what > happens when theta is 90 degrees. This by the way is another way > of writing the dot product of F and d). > > http://hyperphysics.phy-astr.gsu.edu/hbase/work.html > "Work refers to an activity involving a force and movement in the > directon of the force." (Notice that the movement must be in > the direction of the force) > > http://www.mcasco.com/p1wke.html > "Work is defined as force multiplied by displacement. Remember in > the general case, force and displacement are vectors. Work is a > scalar quantity so the multiplication operation we are talking about > here is the dot product as discussed back in Vector Arithmetic. If the > fellow in the picture at the right was actually holding up his end of > that bit of pipe while the machine held up the other, neither of them > would be doing any work since the force (upward) is perpendicular to > the motion (horizontal)." > > And I have no doubt that Tipler to will mention the relationship > between force, displacement, work and angle, and will also point > out somewhere that no work is done when force and displacement > are at right angles. > > Imagine my power! I made this up just a few days ago for our > discussion, and already it's found its way into all these > websites!!! > > > > > > > That's meaningless. What does "work wrt its KE" mean? > > > > > > > > > More coyness! > > > > > > What's "coy?". "Work wrt KE" is an abuse of the language. > > > It doesn't mean anything. > > > > > Well, I only repeated what you said. Did you forget those are your > > words? > > I defy you to find a post where I used the phrase "work with > respect to KE". > > > > Not in situation (b). I said that in contrast to this > > > situation, there is a *different* situation which I'm now > > > calling (c). > > > > > Oh I see: You forgot to mark that as "not in situation (b)" when you > > first said that. > > Yes. I contrasted this with circular motion, but somehow you > got the impression that I was still talking about circular motion, > > If I got the impression that you were talking about circular motion then you were doing just that, no "somehow" about it. > > > and that AT THE SAME TIME I was still talking about straight > line constant velocity motion. Hope you have it all straightened > out now, and realize that when I'm talking about circular motion > I'm not talking about straight line motion. > > It sounds like you're not sure what it is you're talking about. > > > > > And I still don't think that I have to talk about "unnamed > > > components of force" in situations where the force is zero. > > > > > Well of course you don't, unless you want to. > > No, even if I want to, a force which is zero is not going to > have a nonzero component in any direction. > > > > Obviously you think that when there are no forces, there's > > > a force to talk about anyway. OK, please describe to me > > > the external nonzero forces in the situation where the > > > external forces are zero. > > > > > Oh you're whipping the hell out of your pal PD with your funny bizness! > > Non-responsive. > > Please describe to me the external nonzero forces in the situation > where the external forces are zero. > > > > Yes, but you will notice that when I said "I didn't think that > > > there is a component of force in the case of zero force" I was > > > referring to the case of ZERO force. That's different from > > > NONZERO. See the difference? ZERO... NONZERO. They mean different > > > things. > > > > > Looks like even PD has deserted you. I asked him to help you out but > > he ignored my request. > > He's enjoying the show. > > What exactly do you think I need to be "helped out" with? > > > > > > When the force is zero, all components are zero. > > > > > > > > > What components are those? > > > > > > The components in any direction. > > > > > I'm sure you mean something by that, but I can't figure it out. > > Did you ever study vectors? A vector quantity, a thing with a > magnitude and an arbitrary direction, can be viewed as being > made up (in general) of three components, along the x, y and > z axes. These three component vectors, added together, sum > up to the original vector. > > But up above you said, "And I still don't think that I have to talk about "unnamed components of force" in situations where the force is zero.", yet here you are again doing that very same thing in saying, "When the force is zero, all components are zero."! What made you change your mind?
From: Randy Poe on 17 Oct 2005 14:34
TomGee wrote: > Randy Poe wrote: > > TomGee wrote: > > > Randy Poe wrote: > > > > The situations are: > > > > (a) No external forces. Velocity and KE are constant. > > > > > > > > > > > Velocity can be constant but aren't you saying that KE does not change? > > > > If velocity is constant, KE is constant. > > > > > I asked you to support that claim before, to no avail. Reference > > > please, again. > > > > What claim exactly? That when velocity is constant, KE > > is constant? Or that KE and velocity are constant in the > > absence of external forces? > > > > > > There > > > > is no force to do any work, so discussions of "work" > > > > are meaningless. > > > > > > > I repeat: But you said a force performs work on an object at CV, > > > > Situations (b) and (c) muddled. There are situations with force > > (situation (a)), and situations without force (b and c). When > > there is no force present, I didn't say anything about what a > > (non-existant) force does, in terms of work or anything else. > > > > I also didn't say a force performs work on an object at constant > > velocity. If a force is performing work on an object, the > > velocity is NOT constant. > > > > > Here's what you said, "work is done by the component of force in the > direction of motion". And I said that after introducing circular motion into the conversation. Prior to that point in the conversation, I was talking about straight line motion in the absence of external forces. > Newton's body having no external net forces > acting upon it moves in one direction at constant velocity, which > supports your and everyone else's assertion that work is done by a > "force in the direction of motion". Newton's body moving in the absence of external forces says nothing whatsoever about what happens in the presence of external forces. > > I brought up force in discussing situations in which force is > > present. Constant velocity is not such a situation. > > > Yet in those same situations you insisted there was no right angle > force present, There are two situations I have discussed where force is present. One of them, situation (b), is the case where force is at right angles to the motion. I have never insisted "there was no right angle force present" in circular motion. Your statement above is contrary to fact. > so which force were you talking about that was present? In situations (b) and (c), there is a force present, either at right angles to the motion (b) or not at right angles to the motion (c). > > Most recently, I brought the concept of force and work into > > the discussion on the question of circular motion, which is > > emphatically *NOT* either a forceless or a constant-velocity > > situation. > > > No, not "most recently", but immediately. And above you state that the > question of circular motion is "NOT...a forceless situation" (meaning > that there is a force present), Right. There's a force present in circular motion. > so if there is no right angle force present, There is a "right angle force present". More correctly, there is a force present and it is at right angles to the motion. > which force is present? I know the answer and you know the > answer, and everyone reading this knows the answer - the question is, > are you willing to let the correct answer come out of your mouth What comes out of my mouth and what you SAY comes out of my mouth are two clearly different things. > > There is no "resistance to the right angle force". The force > > is applied and the path changes. > > > Oh, of course there is, silly! Inertia is a body's resistance to > change when acted upon by a directional force. Yes, but "inertia" is not a force. > The inertial force resists the gravitational force. There's not "inertial force". > Everyone knows that. Everyone does not "know that". I hold a rock over a hole in the ground. I let go. Please tell me how the rocks inertial force is resisting the gravitational force. Describe the forces present and how they are acting. > But up above you said, "And I still don't think that I have to talk > about "unnamed components of force" in situations where the force is > zero.", yet here you are again doing that very same thing in saying, > "When the force is zero, all components are zero."! What made you > change your mind? I didn't change my mind. I do not have to consider any components of force, since they are all zero, in situations where the force is zero. You are seeing contradictions where they don't exist. "The northward component of force is zero" does not contradict "there is no northward component of force". "None" and "zero" are synonyms. I did not "change my mind" when I talk about no components in one place and zero components in another. - Randy |