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Physics Question - ma = mg
bostonman:
Okay.... is it safe to just say Mass remains the same for purposes of discussion to answer the question in hand?
IanB:
--- Quote from: bostonman on May 14, 2021, 03:47:02 am ---Okay.... is it safe to just say Mass remains the same for purposes of discussion to answer the question in hand?
--- End quote ---
Let's say you confirm experimentally that F = ma by applying a known force to an object of known mass and measuring the acceleration. And for all combinations of forces and masses this holds true.
Now let's say you drop a known mass from a height and measure its acceleration in free fall. And let's suppose that you measure that acceleration in repeated experiments and it is always 9.81 m/s2. You may suppose from the previous formula that the force acting on it due to gravity should be F = ma, with known mass and measured acceleration.
Furthermore, suppose you place the same mass on a spring scale, and confirm that the weight of your given mass on the scale is exactly the same as the force of gravity you calculated from the free fall experiment.
By these various experiments you now have all the measurements matching up and agreeing with each other. As a result, you may conclude that F = ma = mg until an experiment shows otherwise.
thermistor-guy:
--- Quote from: bostonman on May 14, 2021, 03:13:53 am ---I'm somewhat glad my question wasn't so easily answered because I thought it was too simple. As someone pointed out above, some of this has deviated and dove deep into physics.
Although the basic plug and chug formulas make things easy, sometimes they don't make sense after trying to analyze them.
...
Also, if I place a 100kg block on a table, is the block a Weight with the table acting as a Force in the upward direction, or is the block also a Force? During this discussion, I began realizing, I'm confused as to when to use a Weight and when to use a Force (unless we are talking about say a car driving into an object, then that's a Force).
...
--- End quote ---
Here is how I think of it.
You, a 100 kg mass (rest mass), stand still on your bathroom scales. There is net zero acceleration.
The mass of the Earth bends the space near it, and accelerates objects towards it (its centre of gravity) at, say, 9.8 m/s/s.
Your scales push you back. You can feel the force, that push-back, under your feet.
That force produces an acceleration that nets out (cancels) the acceleration due to the Earth's bending of space.
The force required to do that is 980 N (100 kg * 9.8 m/s/s). In everyday language, you say "I weigh 100 kg today".
You are in a NASA training aircraft, for astronauts, to experience "weightlessness". The plane goes into a parabolic trajectory.
Everything inside the plane seems to float, including you. You are accelerating towards the Earth, at 9.8 m/s/s because of the bend in space, but you feel no force.
In everyday language you say "I feel weightless." There is no force to stop you from accelerating towards to Earth.
You're at home, outside, and you stand on a strong table. You and the table tend to accelerate towards the Earth, because the Earth bends space towards it.
But the net acceleration of both you and the table is zero. The table pushes against your feet with enough force, 980 N, so you have net zero acceleration.
The Earth pushes back against the table, a little bit harder, so that the table plus your 100 kg mass have net zero acceleration.
The way I think of it, is that you can have acceleration due to a large mass bending space, and you can have acceleration due to a force acting on a body.
In school, I was taught that the first cause of acceleration was also due to a force, called gravity. This line of thinking works in the everyday world. It's
accurate enough, much of the time, but conceptually it is misleading.
Thank you for your though-provoking question.
radiolistener:
--- Quote from: bostonman on May 14, 2021, 03:13:53 am ---One thing we can all agree on is mass remains constant, it can't change throughout the universe.
--- End quote ---
I'm not sure about that. The quantity of matter in the body will be the same at any point of universe, but if we're talking about mass in context of interaction force with other matter in the universe, it may depends.
Nominal Animal:
--- Quote from: bostonman on May 14, 2021, 03:13:53 am ---From my initial assumption, it seemed a special case existed where ma = mg because we are on Earth and/or a connection existed with 9.8 to make ma = mg.
Basically, they can equal each other anywhere in the universe providing the accelerating object accelerates at the same rate as the gravity as to where you are.
--- End quote ---
Another way to realize the same thing, is to consider how one can measure weight as opposed to mass.
To measure weight, we use a spring, a counterweight (for comparative analysis with known masses), or a strain gauge. They all measure the combined forces (gravity, buoyancy due to surrounding fluid – atmosphere –, and so on). So, "weight" is just an indication of the force pulling the object and the planet closer together.
To measure mass, we typically use rotating contraptions, since angular momentum is not dependent on gravity, assuming good enough bearings. By measuring the angular momentum caused by a known force at a known distance from the center of rotation, say a spring, we can determine the actual mass.
Which also means that while in microgravity, say on a space station, you are weightless, you are most definitely not massless. If you push a little too hard when you careen down an access tube, and catch yourself awkwardly with one hand, it is exactly as easy to break your hand there as it would be here on Earth, given the same velocity of your body at the point of impact.
In particular, trying to move or slow down a large mass, is exactly as hard in microgravity as it is on Earth on frictionless bearings – and air bearings are close to frictionless, which is why those air tracks are often used in freshman physics classes.
Also, going up to a low Earth orbit, does not mean you get much "further out" of Earth's gravity well. In fact, at the International Space Station, at about 420 km above sea level, the gravity exerted by Earth is still about 8.6 m/s2; or about 87% of that at the sea level. It's just that due to being in orbit, ISS is in free fall; it basically simply misses ever hitting the Earth.
--- Quote from: bostonman on May 14, 2021, 03:13:53 am ---If I place a 100kg block on a table, is the block a Weight with the table acting as a Force in the upward direction, or is the block also a Force? During this discussion, I began realizing, I'm confused as to when to use a Weight and when to use a Force (unless we are talking about say a car driving into an object, then that's a Force).
--- End quote ---
I'd say it would be worth your time and effort to understand the concept of force, potential, and work.
Let's consider gravity a field, a field of gravitational potential energy, but ignore all spacetime-bending and relativistic effects, keeping things very straightforward. We cannot measure it directly, but its units would be the units of energy. (It would also be a simple scalar field, and not say a vector field with a direction at each point, like magnetic fields have.)
Gravitational potential energy is not something we can tap into, either. Gravitational potential energy is just a way to express the kinetic energy loss/gain between points at different gravitational potentials. In a real sense, it does not "exist"; but, unless you stay at the exact same potential, you either gain or spend, kinetic energy. Total energy is the sum of kinetic and potential energy; in the same way one might consider the total amount of funds they have the sum of cash at hand, and the amount of funds available in bank accounts (although there is really no physical guarantee you can access those).
(Zero-point energy is a completely different concept, and involves quantum mechanics.)
In a pendulum, the kinetic energy is maximum at the equilibrium position (where potential energy is zero); the kinetic energy is zero and potential energy maximum at the extremum positions. Ignoring losses due to friction etc., the total energy stays constant, and describes the system (not at any specific point in time, but how it behaves). So, potential energy is a very useful notion when dealing with anything moving. In a gravitational potential field, the "zero" is at infinite distance.
The force exerted by a potential field – be it gravity or anything else – is exactly \$F = -\nabla U\$, where \$U\$ is the field at that particular point, \$F\$ is the force exerted, and \$\nabla\$ is the gradient; in Cartesian coordinates,
$$\left\lbrace \begin{aligned}
F_x &= \frac{\partial \, U(x, y, z)}{\partial \, x} \\
F_y &= \frac{\partial \, U(x, y, z)}{\partial \, y} \\
F_z &= \frac{\partial \, U(x, y, z)}{\partial \, z} \\
\end{aligned} \right.$$
i.e., the 3D analog of tangent of the field. The minus sign is there because the force is always "downhill". Without the minus sign, the force would repel.
This is, by the way, exactly how molecular dynamics simulators simulate atom motion. Classical models define potentials (force fields in chemistry terms); quantum mechanical or ab initio methods model electron charges (and atomic nuclei as point-like positive charges) as fields, but using quantum mechanics to model the way those fields interact. Because of the heavy calculation involved, we're limited to thousands or at most tens of thousands electrons, and the boundaries of the simulation must also repeat (cannot be in "empty space").
For two point-like objects, we have \$U = -G\frac{M \, m}{r}\$, where \$G\$ is the gravitational constant, \$M\$ and \$m\$ are their masses, and \$r\$ is the distance between the two. Note that the divisor really is the distance; this the gravitational potential fields reach far indeed. If you calculate its derivative, it is the familiar \$F = -G\frac{M \, m}{r^2}\$ (as a vector quantity, in the direction to the center of mass of the system; this applies to each point separately). So, this is just another way to describe Newtonian physics.
Work is the amount of force exerted over a specific displacement, say from \$a\$ to \$b\$. If the force is constant, \$W = (b - a)F\$; if the force varies, \$W = \int_a^b F(p) \, d \, p\$. With respect to the gravitational potential field, it is exactly the difference in gravitational potential energy between the final and initial points. If you move in a gravity field, but stay at the same potential, you do no work; only the change in the potential matters.
Orbits – and even driving a car in hilly countryside – work very much like pendulums, since the Earths gravitational field is not really uniform, and besides is affected by the Moon anyway: because you don't stay in the same potential, you trade potential energy with kinetic energy. When you go "downhill", you gain speed (trade gravitational potential energy to kinetic energy); when you go uphill, you slow down (trade kinetic energy to gravitational potential energy). The International Space Station occasionally needs additional speed boosts, but this is because even at 400 km up, there is still a bit left of Earth's atmosphere, slowing the ISS down. (Note that ISS orbital speed is also about 7.66 km/s ≃ 27600 km/h ≃ 17100 mph. Currently, atmospheric drag causes the ISS to slow down by about 0.00000000066 km/s per second.)
If we go back to the object on a table, there is a force pulling the object and the center of gravity closer together; a small force due to fluid displacement which is either in the same direction (if denser than air) or opposite direction (if lighter than air). These are countered by the "static" force of the chair and ground and surrounding material involved. (Such systems are examined in the Mechanics subfield of physics – classical mechanics in this particular context. The forces that exist due to materials requiring energy to change shape, and the overall forces being less than that, are often called "static" in this context, because they are opposing dynamic forces – forces that try to change things. In other words, "static" and "dynamic" in this context are just labels that help conceptualize the situation, just like "acceleration" and "deceleration" are just two words describing change in speed or velocity.)
It is often easier/shorter to just say that it's the table causing the equal but opposite force, but it's actually all material supporting the object not changing, that contribute to it. At the atomic level, it's due to Pauli exclusion principle; because electrons repel each other, you need to spend a lot of energy to smush atoms closer together. (We call those kinds of forces pressure, by the way.)
Since it's me writing this, there may have crept inaccuracies or poor analogs into this post. If you notice them, let me know; definitely do point them out so we can fix them. The last thing I want is to lead others astray in their basic grasp of physics!
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