inkliing

More than one person has objected to "unwanted heat loss", which is apparently the wrong thing to say in this context. I'm referring to the idea that heat transfer through a finite temperature difference can contribute to the irreversibility of processes (Moran, John (2008). "Fundamentals of Engineering Thermodynamics", p. 220. John Wiley & Sons). I should have been more clear in the OP. I'm talking about a real process (not idealized or quasistatic, etc.) of heat conduction across some finite temperature difference (not infinitesimal), and therefore not reversible. What I mean by unwanted heat losses is heat loss that causes an irreversible increases in entropy, either in the system or the surroundings, and that these heat losses that cause irreversible entropy increases imply that heat, even if it's being treated as some sort of idealized fluid, is not a conserved quantity. A long-needed review of thermodynamics started me thinking about this. Of the many ways that work and heat are different, one that struck me was that work in a conservative field could be path independent, whereas it seems that heat is never path independent in a real process. What seemed to me to be the only process which might come close to path independent heat was Fourier's conduction law: a vector quantity = -gradient (a scalar), which is similar in form to the definition of a conservative field: field vector=-gradient(potential). So I wondered if the vector quantity in Fourier's law, conductive heat flux, can be considered to be a conservative field with potential= the scalar quantity in Fourier's law =temperature. It seems to me that all heat transport texts, when they discuss conductive heat transport, treat heat in this fashion - as an idealized, conserved fluid that flows from higher temperature surfaces to lower temp. surfaces. But if real heat transport across a finite temp. diff. is irrev. then heat must be lost to irrev. increases in entropy (syst. or surr.), implying heat is never a conserved quantity, even when we try to pretend it's an idealized fluid. I realize that there are irrev. processes, which therefore cause irrev. increases in entropy, but involve no heat in/out of the system- e.g., free expansion vs a vacuum. But free expansion isn't a heat transfer process (conduction, convection, radiation). It seems to me that any real heat transfer process across a finite temp. diff. must be irrev., and therefore incur irrev. entropy losses (in the syst. or surr.), and that these irrev. entropy differences must be cause by heat in or out, which I was calling unwanted heat losses. I'm not stating these ideas as fact. In fact, I do not completely understand them, and I'm trying to understand them. That is why I posted.

For Fourier's law [latex]\vec{J}=-k\nabla{T},[/latex] where [latex]\vec{J}=[/latex] conductive heat flux in [latex]\frac{W}{m^2}[/latex] k=thermal conductivity in W/m-K T=temperature, [latex]\frac{1}{k}\vec{J}[/latex] certainly appears to be a conservative field with temperature as potential and for any path from position a to position b, [latex]\int\limits_{path}\vec{J}\cdot d\vec{r}=-k(T_b-T_a)[/latex], i.e, conductive heat flows from an equipotential surface at [latex]T_a[/latex] to an equipotential surface at [lower] [latex]T_b[/latex] This is analogous to: [latex]\vec{F}_{grav.}=-\nabla{\mbox{[potential energy]}}[/latex] where [latex]\int\limits_{path}\vec{F}_{grav}\cdot d\vec{r}=-(P.E._b-P.E._a)[/latex] and mass flows (falls) from a to b. Note: The path from position a to position b is a path in three dimensional space with coords x,y,z, not a path in a thermodyn. phase space, such as p-V diagram, in which heat is always path dependent. But this path-independent heat seems to only be useful when heat is considered to be an idealized conserved quantity, i.e., 1. heat is a perfect fluid - no meaningful internal or thermal energies, no chemical interaction, nuclear interations, partical collisions, particle velocity distributions, etc. 2. heat is conserved - no unwanted heat losses. Which is a clearly limited model. On the other hand, isn't it true that, whenever we do any relatively-straightforward engineering analysis involving conductive heat flow, such as the very-common one dimensional application of Fourier's law, that we are treating heat as a ideal conserved quantity? But on the other hand, heat transfer requires [latex]\Delta T\Rightarrow[/latex] heat transfer never occurs in thermodyn. equilibrium [latex]\Rightarrow[/latex] heat transfer is always irreversible [latex]\Rightarrow[/latex] heat transfer always involves unwanted heat losses [latex]\Rightarrow[/latex] heat transfer can never involve a perfectly idealized conserved quantity. So I think the conclusion is: We do pretend that heat is an idealized conserved quantity when we do relatively-straightforward conductive heat transport analysis, even though the 2nd law [latex]\Rightarrow[/latex] there'r always unwanted heat losses. Does this seem reasonable, or am I missing something important?

PeterDonis at https://www.physicsforums.com/threads/inertial-reference-frames.797217/ gave an excellent explanation, definitely worth a read. I'm no longer confused about these ideas.

I'm reviewing physics after ~30yrs of neglect, starting with Halliday & Resnick (and the internet). Here's what I understand to be standard Newtonian/classical inertial frames: 1. There exists a set of reference frames, called inertial frames, in which mass, time, force, acceleration, etc. are (Galilean) invariant but position, velocity, translational and angular momentum, work, kinetic energy, etc. aren't. 2. Measurements in one inertial frame can be converted into measurements in another frame by a Galilean transformation of coordinates. 3. All inertial frames are in relative rectilinear motion. 4. Any two frames are inertial if and only if they measure the same accelerations for all particles. 5. A frame is inertial if and only if a perfect accelerometer at rest measures no acceleration. Also: 6. A frame is inertial if and only if physical laws are observed to be in their simplest forms. When I got to the Equivalence principle: 1. Inertial mass = gravitational mass. 2. No experiment can determine whether you're in a free falling elevator or an elevator floating out in space. It is understood that the free falling elevator is just above the Earth's atmosphere and that the elevator in space is far from any gravitational source. Also, the free falling elevator is small enough, and falls for a short enough time period, that tidal deviations are below the level of experimental detection. I had the following problem: I realized that #5 above was not true in the classical sense. Given two observers, each with a damped-spring accelerometer, one observer is in free fall above Earth's atmosphere and the other is floating out in space, then each observer will see that both accelerometers measure zero acceleration, even though they are accelerating with respect to each other. Each observer knows that his zero acceleration measurement proves that he is in an inertial frame, but also can see that the other observer's accelerometer aslo measures zero acceleration, that that therefore the other observer must also be in an inertial frame. And yet each observer clearly measures, in thier own inertial frame, that the other observer is accelerating. I had assumed I would be able to do classical physics without having to worry about relativity, and that I could do relativity when I was ready. I guess was wrong. I tried several things to get around the problem: 1. #5 above isn't really Newtonian, it must actually be an idea from relativity. This doesn't work since it can clearly be seen that #3-#5 above are all more or less equivalent. 2. I reasoned that the speeds and gravitaional fields involved were not relativistic, that somehow then the problem should go away and that the frames involved should be able to be described as noninertial in the classical sense. But this leads nowhere, if my accelerometer reads zero. I'm in an inertial frame. 3. I tried to convince myself that the 2 observers can't read each other's accelerometer, i.e., a signal sent between them would somehow be messed up by some "relativistic effect," but I knew this was rediculous since the speeds and field strengths involved weren't relativistic. I gave up and opened my old general relativity book (Gravitation by MTW). The 1st chapter describes inertial motion: Following a geodesic (free fall, orbit, etc.) = natural (weightless) motion of a particle = local inertial (i.e., Lorentz) frame = particles move at constant speed in straight lines. Physics (physical laws) is simple when viewed locally. So I tried to revise my definition of an inertial frame: 1. For any particle moving in Nature, there exists a set of local reference frames, called inertial frames, in which mass, time, force, acceleration, etc. are (Lorentzian) invariant but position, velocity, translational and angular momentum, work, kinetic energy, etc. aren't. 2. Measurements in one local inertial frame can be converted into measurements in another local frame by a Lorentzian transformation of coordinates (Galilean at low speed). 3. All local inertial frames are in relative rectilinear motion. 4. Any two local frames are inertial if and only if they measure the same accelerations for all particles. 5. A local frame is inertial if and only if a perfect accelerometer at rest measures no acceleration. Also: 6. A local frame is inertial if and only if physical laws are observed to be in their simplest forms. This seemed simple enough and I hoped to continue with classical physics, but I was bothered that "local" is opposed to "global" and that global should refer to the shape of the whole universe, not merely the difference between free falling near the Earth and floating relatively nearby in space, which seems "local" with respect to the entire universe. Then I realized I still had the above problem. If one obserever is free falling above Earth's atmoshpere and the other observer is in low Earth orbit, and if the free falling and orbiting observers pass through the same event, i.e., the same point in space at the same time, then, at that event, i.e., that point on the spacetime manifold, both observers should have the SAME set of local inertial frames, since they are passing through the same point in space at the same time and both of their accelerometers read zero. But they can't be in the same inertial frame since each observer measures the other as accelerating. So it seems I can't avoid certain general-relativistic ideas, even when trying to avoid them and just stick to classical physics. Any help with the resolution of this paradox will be greatly appreciated. Thanks in advance. P.S. I would like to avoid the mathematical formalism of special or general relativity, if possible, in the resolution of this problem. Intuitively, I strongly suspect it isn't needed, as these ideas are fundamental to an understanding of inertial frames, and as such should be relatively simple to explain.

elfmotat, Thanks for the suggestion. I did try essentially what you suggest, realized how tedious it would be, and gave up. I only attempted the original problem because I thought it would be straightforward, lol. I will back-burner this one till a get back to higher-level dynamics. Thanks.

This isn't homework, I'm reviewing physics after many years of neglect. I thought this particular problem would be simple - it isn't. 2 particles of mass m<<M fall from rest along the same straight line directly down toward gravitational source M. The particles are initially a distance [latex]2R_i[/latex] apart and their center of mass is initially a distance [latex]r_i>>R_i[/latex] from M. Therefore, just to be clear, the upper particle, particle 2, falls from an initial height of [latex]r_i+R_i[/latex], and the lower particle, particle 1, falls from an initial height of [latex]r_i-R_i[/latex], both falling straight down along the exact same straight line toward M. How much time is needed for their separation to double, to [latex]4R_i[/latex]? Let the unprimed frame be the inertial frame in which M is stationary at the origin. Let the primed frame be the non-enertial frame in which the C.M. of the 2 particles is stationary at the origin. Therefore the primed frame falls with the particles. Let us call the primed frame 'the local frame' and the C.M. of the 2 particles the 'local C.M..' Therefore the tidal force on particle 1 [latex]=-\frac{2GMmR}{r^3}[/latex] implies that the eqn. of motion of particle 1 in the local frame is [latex]\ddot{R}-\frac{2GMR}{r^3}=0[/latex](eqn.1) and the eqn. of motion of the local C.M. in the unprimed frame is [latex]\ddot{r}+\frac{GM}{r^2}=0[/latex](eqn.2). The soln. to eqn.2 is well known, [latex]t=\frac{r_i^\frac{3}{2}}{\sqrt{2GM}}[\sqrt{\frac{r}{r_i}}\sqrt{1-\frac{r}{r_i}}+cos^{-1}\sqrt{\frac{r}{r_i}}][/latex]. Keep in mind [latex]\frac{\ddot{R}}{R}=\frac{2GM}{r^3}[/latex] and [latex]\dot{r}=-\sqrt{2GM}\sqrt{\frac{1}{r}-\frac{1}{r_i}}[/latex]. Good luck.

This isn't homework. I'm reviewing physics after many years of neglect. Since a simple harmonic oscillator is a conservative system with no energy losses, then a driven undamped harmonic oscillator, once the transient solution has died out, can't be receiving any energy from the driving mechanism, else the system's M.E. would increase without bound. But when a simple harmonic oscillator is driven at the system's natural frequency, the amplitude of the position, and therefore the system's M.E. increases without bound, which can't happen unless the system is receiving energy from the driving mechanism. Specifically: Let [latex]\omega_{\circ} = \sqrt{k/m}[/latex], driving function = [latex]f(t) = \frac{F_{\circ}}{m}\cos\omega t[/latex], therefore [latex]a + \omega_{\circ}^2x = \frac{F_{\circ}}{m}\cos\omega t[/latex], therefore, ignoring the transient solution, [latex]x = \frac{F_{\circ}}{m(\omega_{\circ}^2-\omega^2)}\cos\omega t[/latex], a steady state solution of clearly constant amplitude. Therefore the system's mechanical energy, [latex]E = U + K = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 =[/latex] a constant. This is true for all input frequencies [latex]\neq\omega_{\circ}[/latex]. Therefore, at all input frequencies [latex]\neq\omega_{\circ}[/latex], the driving function can't be inputing energy into the system, since the system's M.E. = a constant. That is, since there are no energy losses, if the driving mechanism were inputing a steady stream of energy, the system's M.E. would go to infinity as time goes to infinity. Now let [latex]\omega = \omega_{\circ}[/latex]. Therefore [latex]a + \omega_{\circ}^2x = \frac{F_{\circ}}{m}\cos\omega_{\circ} t[/latex], therefore, ignoring the transient solution, [latex]x = \frac{F_{\circ}}{2m\omega_{\circ}}t\cos\omega_{\circ} t[/latex], a steady state solution with an amplitude which clearly [latex]\rightarrow\infty[/latex] as [latex]t\rightarrow\infty[/latex]. Therefore the system's mechanical energy, [latex]E = U + K = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 \rightarrow\infty[/latex] as [latex]t\rightarrow\infty[/latex]. Therefore, the driving function must be inputing energy into the system when [latex]\omega = \omega_{\circ}[/latex]. So which is it? Either the driving mechanism is supplying the system with energy or not. If energy is being input, then why doesn't the system energy tend to infinity at all driving frequencies? If energy is not being input, then why does the system energy tend to infinity when driven at the natural frequency? Both must be explained.

I've talked to a few people and made progress. Ignore Ex. 1. The rocket eqn. (1) is derived for systems in which the exhaust and the rocket (the total closed system, if you will) are affected by exactly the same external forces. But in Ex. 1, whether one considers the pile to be the rocket and the rising chain to be the exhaust or vice versa, there exists a force, namely the driving force that is raising the chain, which is exerted only on the rising chain, and not on the pile. Therefore eqns. (1) and (2) don't apply. Therefore the issues I was having with frame dependence were meaningless. For Ex.2, there exists a force, that of the luggage handler or the wall (if thinking of the problem in terms of sand) which is exerted only on the "rocket,' i.e., only on the mass on the belt, not on the 'exhausts' (the input and output streams). So, again, eqns. (1) & (2) don't apply directly but must be re-derived from conservation of momentum. So that clears up the confusion I was having with frame dependence vs. frame invariance, which was the hard part. The only problem I'm still having is showing that the force needed to drive the conveyor belt at constant speed in Ex. 2 is exactly [latex]2\dot{m}v[/latex] (btw, someone asked about the difference between M and m. M is the mass of the system [whether open or closed] at time t, [latex]\dot{m}[/latex] is the mass flowrate into and out of the system, which are equal when the system is at steady state.). Everyone I've talked to seems to agree that the total driving force on the belt is [latex]2\dot{m}v[/latex]. No one seems to be able to show it exactly, tho'. Here's what I have: Let the system be just the unput flow (falling vertically down, therefore no horiz. mom.) + the mass on the belt at steady state. Therefore no stopping force or output flow. It is straightforward (eqns. (1) & (2) apply) to see that the additional force needed to drive the belt with with mass flowing onto it (as compared to the force needed to drive an empty belt) = [latex]\dot{m}v[/latex]. Call this driving force [latex]F_{in}[/latex]. Therefore [latex]F_{in}=\dot{m}v[/latex]. Now let the system be just the mass on the belt + the output stream at steady state. Remember that the mass, whether luggage or sand, is stopped, while still on the belt, by an external force, before it leaves with no horiz. momentum. Following a fixed number of particles as they move along the belt, i.e. a closed system, at time t we have m+[latex]\Delta[/latex]m moving on the belt at speed v. At time t+[latex]\Delta[/latex]t we have m still moving on the belt at speed v but [latex]\Delta[/latex]m has been stopped by the external force, call it [latex]F_{wall}[/latex], and has left the belt with no horiz. momentum. Call the force, here, driving the belt, [latex]F_{out}[/latex]. Therefore [latex]\sum F_{ext}=F_{out}-F_{wall}=\dot{p}=\frac{d}{dt}[mv-(m+\Delta m)v]=-\dot{m}v<0[/latex]. Therefore [latex]F_{wall}=F_{out}+\dot{m}v \ and \ F_{wall}>F_{out}[/latex]. Now let the system be the whole system with input and output streams at steady state. At time t mass m+[latex]\Delta m_{out}[/latex] is on the belt moving at speed v, [latex]\Delta m_{in}[/latex] is entering vertically with no horiz. mom., and at time t+[latex]\Delta[/latex]t, m+[latex]\Delta m_{in}[/latex] is moving on the belt at speed v and [latex]\Delta m_{out}[/latex] is falling with no horiz. momentum. Steady state [latex]\Rightarrow\Delta m_{in}=\Delta m_{out}=\Delta m[/latex]. Therefore [latex]\sum F_{ext}=F_{in}+F_{out}-F_{wall}=\dot{m}v+F_{out}-(F_{out}+\dot{m}v)=0=\dot{p}=\frac{d}{dt}[(m+\Delta m)v-(m+\Delta m)v]=0[/latex]. But we can't solve for [latex]F_{out}[/latex] since it cancels. Now let the system be the whole system but with no conveyor belt and with everything floating out in space in an inertial frame far from any external forces (except for [latex]F_{wall}[/latex], clearly). It should be clear that the only driving force necessary here is that needed to accelerate the incoming particles from 0 to v, which then move at constant v to the wall, which stops them. Therefore clearly [latex]F_{\text{wall in space}}=\dot{m}v<F_{wall}[/latex] since [latex]F_{wall}[/latex] must also overcome the frictional force between the belt and the mass. The best hand-waving argument I can come up with is that [latex]F_{\text{total driving force}}=F_{in}+F_{out}=\dot{m}v+F_{out}=2\dot{m}v[/latex] since that would mean [latex]F_{wall}=2\dot{m}v[/latex] which exceeds [latex]F_{\text{wall in space}} \ by \ \dot{m}v[/latex] which seems to make sense since the difference between [latex]F_{wall} \ and \ F_{\text{wall in space}}[/latex] is that [latex]F_{wall}[/latex] must overcome the frictional force between the belt and the mass. So all I need is a direct mathematical argument showing that [latex]F_{out}=\dot{m}v[/latex].

Sorry, but this will be a long post. This isn't homework, I'm reviewing physics after many years of neglect. Halliday & Resnick, 4th Ed., section on variable mass and rockets, refers the interested reader to an article, "Force, Momentum Change, and Motion," Martin S. Tiersten, Am. J. Phys., Jan. 1969. I'm trying to apply Tiersten's methodology (which seems to me to be more straightforward than other methods) to 2 problems. The usual representations of Newton's 2nd law for open mass systems: [latex]M\frac{d\vec{v}}{dt} = \sum\vec{F}_{ext.} + (\vec{u}-\vec{v})\frac{dM}{dt} = \sum\vec{F}_{ext.} + \vec{v}_{rel.}\dot{M}, (1)[/latex] [latex]\frac{d}{dt}(M\vec{v}) = \sum\vec{F}_{ext.} + \vec{u}\frac{dM}{dt} = \sum\vec{F}_{ext.} + \vec{\phi}_{\text{net mom.in}}, (2)[/latex] where M = system mass, [latex]\vec{v}[/latex] = system velocity, [latex]\vec{u}[/latex] = velocity of incoming/outgoing mass at time t. Please note: 1. Tiersten helpfully points out: (let the sums be taken over all of the particles in the system at time t) [latex]\vec{v} = \frac{\vec{p}}{M} = \frac{\sum m_i\vec{v}_i}{M}\mbox{. M not constant implies, in general, }\vec{v}\neq\vec{v}_{cm} = \frac{d}{dt}\vec{x}_{cm} = \frac{d}{dt}(\frac{\sum m_i\vec{x}_i}{M})[/latex] [latex]\vec{a} = \dot{\vec{v}} = \frac{d}{dt}(\frac{\vec{p}}{M})\mbox{. In general, }\vec{a}\neq\frac{\sum m_i\vec{a}_i}{M} = \frac{\sum\vec{F}_{ext.}}{M}\mbox{ and }\vec{a}\neq\vec{a}_{cm} = \frac{d^2}{dt^2}\vec{x}_{cm}[/latex] 2. For rockets, [latex]\vec{v}_{rel.}\dot{M}[/latex] is usually called momentum thrust. (1) is Galilean invariant simply because [latex]\vec{u}-\vec{v}[/latex] is a relative velocity and therefore clearly invariant in any inertial frame (M, [latex]\dot{\vec{v}}, \sum\vec{F}_{ext.}[/latex] are clearly invar.). 3. (2) is frame dependent (clearly [latex]\dot{\vec{p}},\phi[/latex] are frame dependent), though [latex]\dot{\vec{p}} - \vec{\phi}[/latex] isn't. The examples given in the article are explained well, but there are 2 from other sources which are confusing me. Ex. 1. A force lifts a length L mass m uniform chain at constant speed, v, vertically from a pile. Let N = normal force, F = lifting force, x = length of chain already lifted at time t. 1. Let S = the system = the entire chain = closed. [latex]\sum F_{ext} = F+N-Mg = \dot{p} - \phi = \frac{d}{dt}((m/L)xv) - 0 = (m/L)v^2\Rightarrow F+N = (m/L)v^2+Mg = constant.[/latex] It seems clear that (a) system closed [latex]\Rightarrow\sum F_{ext} = \dot{p}[/latex] in all frames and therefore [latex]\dot{p}[/latex] is frame invariant and (b) system closed [latex]\Rightarrow\phi = 0[/latex] in all frames. 2. Now let S = just the pile = open system. [latex]N-(m/L)(L-x)g = \dot{p} - \phi = 0 - 0[/latex] gives, when combined with the above result, the expected [latex]F = (M/L)(v^2+xg)[/latex]. Here's where the confusion begins. [latex]\dot{p} = 0[/latex] because we are observing the system from the pile's restframe and therefore [latex]v_{pile} = 0 \ and \ \dot{v}_{pile} = 0. \ \phi = 0[/latex] because I intuitively know that the rising chain applies no thrust to the pile, i.e., [latex]\phi \ (= u\dot{M})[/latex], when combined with [latex]-v\dot{M}[/latex] (= 0), should give zero thrust. But this seems mathematically insatisfactory because mass leaves the pile at speed u = v, suggesting that [latex]\phi = u\dot{M}\neq0[/latex]. I suppose we could reason this away by observing that the mass doesn't actually leave the pile at any speed, rather it simply disappears, and therefore the speed of the disappearing mass will always be zero relative to the pile. But let us now keep the same open system = just the pile, but observe it from another inertial frame, say the restframe of the rising chain. Now [latex]\dot{p}[/latex] seems to be [latex]\neq0[/latex] since [latex]v_{pile}\neq0[/latex], i.e., [latex]\dot{p} = -v\dot{M}[/latex]. And if we remain consistent, insisting that the speed of the disappearing mass is zero relative to the pile in all inertial frames, then [latex]\phi[/latex] still = 0, giving N-(m/L)(L-x)g = [latex]-v\dot{M}[/latex], which is incorrect. We can't have it both ways. Either the speed of the disappearing mass relative to the pile is zero in all frames ([latex]\Rightarrow\phi = 0[/latex] in all frames [latex]\Rightarrow\sum F_{ext} = N-(m/L)(L-x)g = \dot{p}\neq0[/latex] in all frames other than the pile's restframe) or the speed of the disappearing mass relative to the pile is not zero in all frames ([latex]\Rightarrow u_{\text{disappearing mass}}\neq0[/latex] in the pile's restframe [latex]\Rightarrow\phi\neq0[/latex] in the pile's restframe [latex]\Rightarrow\sum F_{ext} = N-(m/L)(L-x)g\neq0[/latex] in the pile's restframe). Either way implies [latex]\sum F_{ext}\neq0[/latex] in some inertial frame. Ex. 2. Halliday & Resnick, 4th ed., ch.9, problem 55. "To keep a conveyor belt moving when it transports luggage requires a greater driving force than for an empty belt. What additional driving force is needed if the belt moves at a constant speed of 1.5 m/s and the rate at which luggage is placed on one end of the belt and removed at the other end is 20 kg/s? Assume that the luggage is dropped vertically onto the belt; persons removing luggage grab hold of it and bring it to rest relative to themselves before lifting it off the belt." The intuitive and correct answer is [latex]2v\dot{M}[/latex] since if no mass ever left the belt, the driving force would = [latex]v\dot{M}[/latex]. Halliday & Resnick evidently agree, since the answer provided in the back of the book is 60N. But I can't see the exact steps in deriving this solution. And forcing this example to conform to the constraints of [latex]\sum F_{ext} = \dot{p} - \phi[/latex] is giving me the same trouble as Ex. 1. I prefer to think of this problem as sand instead of luggage, since finer particles make it easier to define the system. Therefore assume sand falls vertically onto the belt, moves a certain distance at constant speed, then hits a vertical wall, which stops the sand while it's still on the conveyor belt. But the sand doesn't accumulate in a pile, rather it falls vertically through a hole. 1. Let S = a closed system = a fixed # of sand grains for a time interval [latex]\Delta t[/latex] during the belt's steady state operation. Assume the system boundaries extend a negligible hight above and below the sand on the belt. Therefore the force exerted by the belt-sand on the still-falling input sand, which accelerates the falling sand from 0 to v, and the reaction force of the falling sand on the belt-sand comprise an internal action/reaction force pair. Similarly for the force of friction between belt and belt-sand. Therefore the external forces on the system are the driving force which moves the belt (call it [latex]F_d[/latex]) and the stopping force of the vertical wall ([latex]F_s[/latex]). Assume the fixed # of sand grains comprise a small amount of still-falling sand, call this [latex]\Delta[/latex]m', + the sand on the belt, m, + a small amount of falling output sand, [latex]\Delta[/latex]m". At time t, [latex]\Delta[/latex]m' is still falling and therefore has no horizontal momentum, m+[latex]\Delta[/latex]m" is on the belt moving with speed v. At time t+[latex]\Delta[/latex]t, [latex]\Delta[/latex]m' has already fallen onto the belt, so that m+[latex]\Delta[/latex]m' is moving on the belt at speed v, and [latex]\Delta[/latex]m" has now fallen through the hole at the end and is falling vertically with no horizontal momentum. Steady state [latex]\Rightarrow\Delta[/latex]m'=[latex]\Delta[/latex]m"[latex]\Rightarrow p(t)=p(t+\Delta t)\Rightarrow\dot{p}=0[/latex]. System is closed [latex]\Rightarrow\sum F_{ext} = F_d - F_s = \dot{P} = 0.[/latex] 2. Now let S= sand on belt + negligible height above and below = open system with sand flowing in and out. [latex]\dot{M}=0 \ and \ \dot{v}=0\Rightarrow\dot{p}=0[/latex] in all frames. But [latex]\phi = \dot{m}'_{in}v_{x,in}-\dot{m}'_{out}v_{x,out}\neq0[/latex] in any frame other than the ground frame. So I need some helpful suggestions on how to apply [latex]\sum F_{ext}=\dot{P}-\phi[/latex] consistently in every inertial frame for these 2 examples, and a well as something more than hand-waving to justify getting from [latex]\sum F_{ext}=\dot{P}-\phi \ to \ F_d=2v\dot{M}[/latex] in ex.2. This is nonrelativistic, non-Lagrangian, simple Newtonian physics. Please do not overcomplicate anything.

Thanks! A and B's rest frames are not inertial.

This isn't homework. I'm reviewing physics after many years of neglect. As with most of my posts, I made this problem up. Let object A have mass [latex]m_A[/latex] and object B have mass [latex]m_B[/latex]. One of A's surfaces is flat, as is one of B's. These flat surfaces are in contact and slide relative to each other in a straight line for a distance, x, experiencing kinetic friction. The kinetic friction, of magnitude f, is constant and is the only force acting on A or B in the direction of motion, and the only other forces acting on A and B are the normal forces pressing the surfaces together. The normal forces are assumed to be equal and opposite, so that the only acceleration of the objects is in the direction of friction. No net force acts on the center of mass. During a finite time interval, the speed of each surface relative to the other decreases from [latex]v_i[/latex] to [latex]v_f[/latex] due to friction. The nonconservative work done by each surface on the other is [latex]W_{noncons} = -fx[/latex] since each surface experiences the same displacement and the magnitude of the frictional force exerted by each surface on the other is the same since the two frictional forces comprise a force/reaction force pair. Therefore the change in kinetic energy of A, [latex]\Delta K_A[/latex] as seen from B's restframe is equal to the change in kinetic energy of B, [latex]\Delta K_B[/latex], as seen from A's restframe, since we are assuming that no other forces accelerate the objects. Therefore [latex]\frac{1}{2}m_A v_{f}^{2} - \frac{1}{2}m_A v_{i}^{2} = \Delta K_A = \Delta K_B = \frac{1}{2}m_B v_{f}^{2} - \frac{1}{2}m_B v_{i}^{2} \Rightarrow m_A = m_B[/latex], an absurd result. Something's wrong and it's driving me crazy! Please help me find the error in this reasoning, thanks.

Thx for the helpful responses. It now seems clear to me that, since rigid rods do not exist, the rod's strain determines its tension or compression, leaving 2 equations (sum of forces on the masses) and the 2 unknown static frictions. In other words, if the angle in the ramp is small enough such that both masses, when standing alone with no rod connecting them, experience static frictional forces strictly less than their respective maximums, then, in general, at the same angle, when the masses are then connected by a flexible rod, the rod will experience a nonzero tension or compression resulting in a nonzero tensile or compressive strain of the rod. If this strain is known then the system is determined.

This isn't homework...I'm reviewing physics after many years of neglect. Given 2 masses, [latex]m_1, m_2[/latex], connected by a rigid, massless rod, stationary with respect to a ramp which makes an angle of [latex]\theta[/latex] with the horizontal, with coefficients of static friction between the masses and the ramp = [latex]\mu_{s1}, \mu_{s2}[/latex] respectively, what is the magnitude of the tension or compression in the rod, and what are the magnitudes of the static friction, [latex]f_{s1}, f_{s2}[/latex], acting on each mass? This assumes [latex]\theta[/latex] is small enough that the masses do not lose traction, i.e., [latex]\theta \leq \arctan \frac{\mu_{s1} m_1 + \mu_{s2} m_2}{m_1 + m_2}[/latex] Note that if the masses are assumed to already be moving, then the problem is straightforward: [latex]T = (\mu_{k1} - \mu_{k2})\frac{m_1 m_2}{m_1 + m_2}g\cos\theta[/latex] [latex]f_{k1} = \mu_{k1} m_1 g\cos\theta[/latex] [latex]f_{k2} = \mu_{k2} m_2 g\cos\theta[/latex] Where T is the tension in the rod (T<0 implies compression). Such problems are common in basic physics, e.g., Halliday, Resnick, & Krane, 4th Ed., chap.6, problem 29. So I assumed it would be straightforward to do the same problem, but with the masses stationary. But I get 2 equations (sum of the forces for each mass) and 3 unknowns (T, [latex]f_{s1}, f_{s2}[/latex]). I find it surprising that such a simple situation is undetermined, and assume I've missed something simple. Note: I've looked through several physics texts and can find only problems in which the masses are already moving. Also, it's driving me crazy!

Thank you for the reply. I think you're right. Obviously, different paths will give different amounts of work, and so the force is nonconservative. I just want to be able to see clearly 2 things: 1) why does the 1-1 nature of x(t) in the overdamped case seem to imply path independence? 2) why, exactly, is it not always possible to find a function, call it [latex]\phi(x)\mbox{, such that }\phi_{x}(x(t)) = (\frac{dx}{dt})^2[/latex] ?

This isn't homework. I'm reviewing calculus and basic physics after many years of neglect. I want to show that a damped harmonic oscillator in one dimension is nonconservative. Given F = -kx - [latex]\small\mu[/latex]v, if F were conservative then there would exist P(x) such that [latex]\small -\frac{dP}{dx} = F[/latex]. I want to show that no such function, P(x), exists. The easy way would be to find a closed curve around which the integral of Fdx would be zero, but since Fdx is a 1-dimensional 1-form, this doesn't seem to be a meaningful way to do it. So I think brute force has to prevail. It should be true that: [latex]\small W=\int_{x_1}^{x_2}Fdx = \int_{x_1}^{x_2}(-kx-\mu v)dx = \frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{x_1}^{x_2}\frac{dx}{dt}dx = \frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{t_1}^{t_2}\left(\frac{dx}{dt}\right)^2 dt[/latex] So let [latex]\small\omega_{\circ}=\sqrt{k/m}\mbox{ , }\zeta=\frac{\mu}{2\sqrt{mk}}\mbox{ , }\omega_1=\left\{\begin{matrix}\omega_{\circ}\sqrt{\zeta^2-1},&\zeta>1\\\omega_{\circ}\sqrt{1-\zeta^2},&\zeta<1\end{matrix}\right.[/latex] For underdamped [latex]\small\zeta<1\Rightarrow x=e^{-\zeta\omega_{\circ}t}(C_1 cos\omega_1 t + C_2 sin\omega_1 t)[/latex] [latex]\small\Rightarrow W=\frac{1}{2}kx_1^2-\frac{1}{2}kx_2^2-\mu\int_{t_1}^{t_2}e^{-2\zeta\omega_{\circ}t}[(-\zeta\omega_{\circ}C_1+\omega_1 C_2) cos\omega_1 t + (-\omega_1 C_1-\zeta\omega_{\circ} C_2) sin\omega_1 t]^2 dt[/latex] Therefore x(t) is not 1-1 [latex]\small\Rightarrow \int_{x_1}^{x_2}vdx[/latex] is multivalued implies W is not a function implies p(x) doesn't exist (since W=-[latex]\small\Delta[/latex]P) implies F is not conservative. Similarly for [latex]\small\zeta=1[/latex]. But in the overdamped case, [latex]\small\zeta[/latex]>1, x(t) is a non-oscillating decaying exponential which never crosses equilibrium, implying x(t) is 1-1, implying W is a function, implying F is conservative. But how can this be? How can a frictional damping force, which dissipates energy as heat, ever be conservative?