Author Topic: Physics Question - ma = mg  (Read 7659 times)

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Offline bostonman

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Re: Physics Question - ma = mg
« Reply #25 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?
 

Online IanB

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Re: Physics Question - ma = mg
« Reply #26 on: May 14, 2021, 04:25:59 am »
Okay.... is it safe to just say Mass remains the same for purposes of discussion to answer the question in hand?

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.
I'm not an EE--what am I doing here?
 

Offline thermistor-guy

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Re: Physics Question - ma = mg
« Reply #27 on: May 14, 2021, 05:21:43 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).
...

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.
 

Offline radiolistener

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Re: Physics Question - ma = mg
« Reply #28 on: May 16, 2021, 01:13:56 pm »
One thing we can all agree on is mass remains constant, it can't change throughout the universe.

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.
 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #29 on: May 17, 2021, 09:19:57 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.
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.

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).
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!
« Last Edit: May 17, 2021, 09:54:07 am by Nominal Animal »
 

Offline bostonman

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Re: Physics Question - ma = mg
« Reply #30 on: June 15, 2021, 03:54:30 am »
I guess the more one reads about physics, the more questions arise.

The episode of The Big Bang Theory where Sheldon teaches "a little physics" to Penny was on last week. Since the initial start of this message, I somewhat learned little 'g' is the gravity on Earth. Now if it's the gravity on Earth, what happens to W=mg if we moved to Mars and blew up Earth?

Since little 'g' is gravity on Earth, and Earth ceased to exist, then how can that formula exist?

This all goes back to not understanding why ma = mg.

As you discussed, an object on the table, but in more detail. Speaking purely of ideal numbers and not the many other forces, if a 100kg box is on the floor, and I care to lift it, the force on it is F=ma. This means I need to exert a force of 980N.

One thing that occurred to me the other day is that a = 9.8m/s^2, but to lift it (ignoring initial force to overcome gravity), aren't we technically applying greater than 9.8m/s^2, else, if we didn't, the box would remain still because we are only applying an equal upwards force that is being applied in the downward force by Earth's gravity?

 

Offline T3sl4co1l

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Re: Physics Question - ma = mg
« Reply #31 on: June 15, 2021, 06:05:52 am »
Well g is just a matter of convention.  Without Earth, but with g still being an accepted mean value around 9.8m/s^2, it would probably just drop out of favor.  Or we'd adopt the analogous g for Mars mean surface acceleration.  Maybe over time the subscript would drop off and we'd be back to g anyway.

The transient force to move a box vertically, is the sum of its weight and inertial force, i.e. depending on how fast you move it.  Nothing could be simpler!

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Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #32 on: June 15, 2021, 06:17:59 am »
As you discussed, an object on the table, but in more detail. Speaking purely of ideal numbers and not the many other forces, if a 100kg box is on the floor, and I care to lift it, the force on it is F=ma. This means I need to exert a force of 980N.
No.  The box exerts a force of 980N to the floor.  If you exert a force of 980N on the box upwards, that just means the box doesn't move, you are just holding it in place – as if someone else were supposed to slip say a sheet of paper under the box.

To lift the box, you need to do work.  If the force against which we do the work is constant, then the work W = F h.

Because acceleration due to gravity drops inversely to the distance squared, and sea level is about 6,371,000 m from the center of gravity, the difference between the force due to gravity at sea level and 1,000 m above sea level is less than 0.0314%.  Even at 420 km altitude (International Space Station in low Earth orbit), the force due to gravity is 88% of that at sea level.  So, for any kind of lifting by hand or by crane, or even aeroplane, we can approximate the force due to gravity being constant.

Work has units of energy; this is the minimum amount of energy needed to displace something against the force, as well as the maximum amount of energy one can extract by letting something be displaced due to the force.  So, for lifting on any planet or whatever with gravity g, we do work W = m g h, where the m g part is due to the force F exerted by gravity on something having mass m.

Note, this mass m is NOT WEIGHT.  Mass is not dependent on gravity; weight is.  If you have a frictionless surface with an object with mass m, you need force F to accelerate it by a (velocity per time unit, i.e. distance unit per time unit per time unit), F = m a, no matter whether you were on Earth, free fall, Moon, or anywhere else.

Weight is what we call the force due to gravity.  Typical measurement devices nowadays are springs and strain gauges.  Essentially, we take something that deforms preferably linearly depending on the force exerted on it, and then measure the deformation somehow.

Now, let's say you have a box on the ground that exerts a downward force of 980 N due to gravity (and we'll exclude buoyancy due to air etc.).
If you exert a force of 981 N upwards on it, the forces almost balance out, leaving a sum total 1 N upwards force.
This force causes an acceleration a, a = F/m, upwards. I we assume g = 9.80 m/s2, then the mass m = 100 kg, because 1 N = 1 kg m s-2.  Therefore, the acceleration would be about 0.01 m/s2.  Let's say you kept that up for ten seconds, so the velocity would be 0.01 m/s2 × 10 s = 0.1 m/s.  If you then reduce your upwards force to 980 N, the acceleration drops to zero, but the velocity remains.  If you reduce your upwards force to 979 N, the acceleration reverses, becoming 0.01 m/s2 downwards.

It does not matter how you apply the force; it does not need to be a wasteful rocket engine.  Stuff like a solid four-by-four plank of wood can easily handle a kiloNewton of compressive force; lots more in the direction of the grain.  Lots of lifting mechanisms use levers, hoists, pulleys etc. to maximize the energy efficiency, so that to lift a 100 kg by one meter upwards in standard Earth gravity (which takes a theoretical minimum of 100 kg × 9.80 m/s2 × 1 m = 980 kg m2 s-2 = 980 N m = 980 J = 980 W s) you need less than say 2 kJ of energy, typically electrical energy – say, 500 W for four seconds.

Note that this also means that if we ignore electrical motor efficiency differences (including internal friction), a 1 kW winch that lifts that 100 kg in two seconds, uses exactly the same amount of energy than a 250 W winch that takes eight seconds to lift the same weight.
However, the difference in force exerted is huge.  Because of the mechanical leverage involved, the instant acceleration at startup can be pretty huge – in physics, the change of acceleration is called jerk – so examining the forces to understand how things happen isn't as important as examining the work done to achieve a specific outcome.

(You've watched Dave's videos.  The energy needed to do specific work is what we examine in physics, to do the back-of-the-envelope calculations; we don't examine the forces involved.)

The forces come into play when we need to worry if the materials we use for the mechanism can take the "load"; whether the stress exerted by the forces exceed the material properties.

We can, however, estimate the minimum forces involved in events, if we are interested in.

If you have a friend who can lift a 100 kg box one meter high in one second, we can immediately estimate the minimum force they exerted on the box upwards to achieve that.   Because the total acceleration is linearly dependent on the force, a = F/m, we first need to estimate the minimum acceleration needed to start from standstill and reach one meter in one second; this happens to be constant acceleration of a = 2 m/s2; and because the friend had to overcome the force from gravity as well, they exerted a force of F = m g + m a = 100 kg (9.80 m/s2 + 2 m/s2) = 1180 kg m/s2 = 1180 N.

Due to human physiology, the initial jerk was probably much higher, and the force exerted at the end somewhat less, but the above is the minimum instantaneous force the friend lifting the box did exert.  To find out exactly how much, we'd need to track for example the height of the center of mass of the box as a function of time.

(When humans do height jump, they exploit that, by the way: by arching their body in a suitable pattern, they can go over a boundary, while keeping the center of mass below the top of the boundary at all times.  Nifty trick, that.)
 

Offline bostonman

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Re: Physics Question - ma = mg
« Reply #33 on: June 17, 2021, 03:46:30 am »
Quote
Let's say you kept that up for ten seconds, so the velocity would be 0.01 m/s2 × 10 s = 0.1 m/s.  If you then reduce your upwards force to 980 N, the acceleration drops to zero, but the velocity remains.

If my upward force would be 980 N, equal to the downward force (100kg mass and Earth being 9.8m/s^2, why would velocity remain the same?

Unless you're talking about no air resistance.
 

Offline Circlotron

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Re: Physics Question - ma = mg
« Reply #34 on: June 17, 2021, 06:14:38 am »
My 'mass' (using 75kg) is the same throughout the universe and can't change (unless I eat lots of food while venturing around space)
To be pedantic, your mass increases proportionally to your velocity. If you at 75kg are moving at 100 km/h or 27.777 m/s you will have 28,935 Joules of kinetic energy. That amount of energy is equivalent to 0.0000000000003215 kg  or 0.3215 nanograms of matter. That is how much mass you will gain, so I am led to believe.

Edit -> because your mass is now slightly more than 75kg then presumably your kinetic energy is also slightly more than 28,935 Joules.
« Last Edit: June 17, 2021, 06:17:18 am by Circlotron »
 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #35 on: June 17, 2021, 07:21:39 am »
Unless you're talking about no air resistance.
That.

It is important to realize that having some force exist, does not mean energy is being spent/transferred.

In the case of a box sitting on the ground, the forces are static, and no energy is transferred.
Similarly, when the total forces acting on an object – including air resistance, gravity, everything – are equal, the object is in free fall, and retains its velocity.

For energy to be transferred via some force, the force has to do work.
 

Offline thinkfat

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Re: Physics Question - ma = mg
« Reply #36 on: June 17, 2021, 07:46:49 am »
I don't think you need to dig deeply into Relativity to understand this equation. It's just a statement, an important one, though, expressed as a mathematical equation.

ma = mg just says that  mass behaves equally in a gravitational 'field' as when being accelerated by any force. The equation states that gravity can be understood as a force, causing a mass to accelerate.
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Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #37 on: June 17, 2021, 08:31:06 am »
Edit -> because your mass is now slightly more than 75kg then presumably your kinetic energy is also slightly more than 28,935 Joules.
Kinda, but also no (because you end up in a cyclical forever increasing mass and velocity argument if you follow that).

Kinetic energy isn't \$E_K = \frac{1}{2} m v^2\$, it is actually \$E_K = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}}\$ for a rigid body.
The two agree for relatively small velocities \$v\$, but start to differ when velocity \$v\$ gets closer to the speed of light in vacuum \$c\$.
Here, \$m\$ is the rest mass, that is invariant of velocity.  This is based on the linear momentum as described by Einstein's theory of special relativity.

In practice, it means that given a specific kinetic energy, your velocity is smaller than the classical kinetic energy (\$E_K = \frac{1}{2} m v^2\$) would indicate; and given a specific velocity, your kinetic energy is higher than the classical formula suggests.

Special relativity describes two related concepts: invariant mass (the mass at rest, described by \$m\$ above), and relativistic mass which bends space-time.  It is easier to think of relativistic mass as relativistic energy, though, because us humans so closely associate "mass = weight"; and it really is the total energy, E, that bends space-time.  As an example, even though photons have no rest mass (\$m = 0\$), they do have linear momentum (as described in special relativity) and kinetic energy; and they do bend space-time.
« Last Edit: June 17, 2021, 08:43:31 am by Nominal Animal »
 

Offline bostonman

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Re: Physics Question - ma = mg
« Reply #38 on: June 17, 2021, 02:24:12 pm »
Quote
I don't think you need to dig deeply into Relativity to understand this equation. It's just a statement, an important one, though, expressed as a mathematical equation.

ma = mg just says that  mass behaves equally in a gravitational 'field' as when being accelerated by any force. The equation states that gravity can be understood as a force, causing a mass to accelerate.

I think I dug too deeply initially. When I saw this on TBBT, I initially thought it meant the Weight of (as an example) a box at rest on a table, and the table has an upward Force exerting a F=ma equal to the weight of the box W=mg.

After digging into it more, I began reading that they are equal because we are on Earth. This didn't make sense to me because a can be any object with acceleration that just so happens to be what the gravity is of that planet (or a point in space).

What I make of ma=mg is that an object sitting on the ground is experiencing a force F=ma where 'a' is Earth's gravity, and its weight is W=mg where 'g' is gravity on Earth; therefore ma = mg.

If we were on the moon, ma would also equal mg in the example above.
 

Offline Ground_Loop

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Re: Physics Question - ma = mg
« Reply #39 on: June 17, 2021, 07:10:21 pm »
Or, for the case originally cited, a = g. Something that given enough time, even Penny would have realized.
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Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #40 on: June 18, 2021, 08:06:13 am »
What I make of ma=mg is that an object sitting on the ground is experiencing a force F=ma where 'a' is Earth's gravity, and its weight is W=mg where 'g' is gravity on Earth; therefore ma = mg.
This is the correct intuitive understanding, I believe.



Mass is the property of matter that does not depend on gravity.
We call the force due to gravity 'weight', and that does depend on gravity; it is that force that weighing scales etc. measure.

On other planets and moons, 'g' differs, so objects have different weights there, despite having the same mass.

On Earth, it takes 0.45 seconds to drop 1 meter from standstill, because g = 9.80 m/s².
On the Moon, it takes 1.11 seconds to drop 1 meter from standstill, because there g = 1.62 m/s².

A one liter bottle of water weighs 1.00 kg on Earth, but only 0.165kg on the moon.
Its mass m=1.00 kg everywhere, because to accelerate it at 1 m per second per second, 1 m/s², you need a force of 1 N, if we ignore possible drag and friction, on Earth as well as the Moon.
It is only the downwards force, 'weight', that varies due to different gravities.

Physicists use an air cushion, like an air hockey rail, to create a very nearly frictionless surface to slide tiny carriages on.
This is a very illustrative device, giving the correct intuition in this matter in a practical way.
No matter in what kind of gravity we are in (as long as the air cushion rail still works), the same mass needs the same force to accelerate at the same rate.
It is only the 'weight' and not 'mass' that is affected by gravity.
 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #41 on: June 18, 2021, 08:24:45 am »
In case someone is wondering whether 'mass' itself could actually be split into two different quantities based on what I wrote above, wonder no longer: Even Einstein contemplated that.

The two "different" aspects of mass are called gravitational mass ("slow mass") and inertial mass ("fast mass").  One is the quantity related to gravity (the 'm' when dealing with weight); and the other is the one related to movement and inertia (the 'm' when dealing with actual acceleration).

Experiments on Earth and in Earth orbit (microgravity) have shown that the two must be equal to at least 12 significant decimal digits: that even if they were different, their relative difference must be less than 1:1,000,000,000,000.  So, for all intents and purposes in the human scale, there is just one 'mass' for each object.

A core starting concept in Einstein's theory of general relativity is the equivalence principle: that gravitational mass and inertial mass are equal and indistinguishable.
It actually goes even further, as the strong equivalence principle states that locally, acceleration and gravity are indistinguishable.  (That in a small closed box, given any scientific instruments and experiments you want, it is impossible to distinguish whether the box is standing still on a planet, or in constant acceleration.)
All experiments we have managed to devise thus far indicates these are exactly, precisely true.
 

Offline bostonman

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Re: Physics Question - ma = mg
« Reply #42 on: June 18, 2021, 02:16:04 pm »
Do we ever really care about Weight in equations unless we want to know based on what an object is on planets?

Another words, if a box is resting on a table, it's just all based on Forces, correct? It would be the force in the upward direction the table exerts, and the force in the downward direction the box exerts. Tilt the table 45 degrees, and now there is a X and Y axis force (the force the box has in the X direction and friction in the X direction opposite the box), and a Y, with a Fnet.

Do we care what the Weight is in at all?

 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #43 on: June 18, 2021, 06:20:51 pm »
Do we ever really care about Weight in equations unless we want to know based on what an object is on planets?
We derive friction from weight, but that's about it.

Another words, if a box is resting on a table, it's just all based on Forces, correct?
Correct.

Tilt the table 45 degrees, and now there is a X and Y axis force (the force the box has in the X direction and friction in the X direction opposite the box), and a Y, with a Fnet.
An illustration should be in order here.  The three forces shown are vector quantities, with direction and magnitude.


Red Fw is the force due to gravity the box exerts on the tilted surface.
Blue Ft is the static force the surface exerts on the box, because it resists deformation.
Green Ff is the force due to friction.

If the velocity of the box does not change, then Fw + Ft + Ff = 0.
If the velocity of the box does change, then Fw + Ft + Ff = m a, where m is the mass of the box, and a is the acceleration vector.  Its direction is opposite to friction, Ff.

(This assumes perfectly smooth but not frictionless surface, and assumes the friction and tilt angle are such that the box does not start tumbling.  If we'd add angular momentum, we'd see that the friction causes torque, which causes the leading edge of the box to exert larger force than the trailing edge of the box, and even a minor imperfection in the surface can catch the leading edge, increasing the forces such that the box starts rolling.  Also, if the center of mass is not supported by the surface the box is on, it will tumble.)

(If we add momentum and angular momentum, we have a pretty accurate simple physics engine at hand.)

So, if we are trying to model the movement of an object, we care about weight, because it relates to friction and the stress on materials holding it in place (like the table surface above).  To calculate acceleration, we use mass.

Static friction is basically always larger than dynamic friction.  In other words, if the object is not moving with respect to the surface it is on, the friction is much greater than when the object is moving with respect to the surface.  This is why objects on tilted surfaces often only need a small nudge, but then keep moving, often accelerating.  The initial nudge overcomes the static friction, and the dynamic friction is not large enough to stop the movement.
Aside from this, it is important to realize that forces cause acceleration; without forces, velocity stays constant.

If we push the abovementioned box to some speed on an inclined surface, and the friction is larger than the other forces combined, then the box slows down: it decelerates (decelerates == acceleration vector opposite to its velocity vector).  When the box stops, the friction is exactly the other two forces combined.  Because static friction is greater than dynamic friction (consider it a curve that drops quickly to a near constant, as velocity increases), there is often a bit of a jerk at the end.
If the box "bounces" backwards a bit, that is due to temporary deformations relaxing; essentially spring-back.
« Last Edit: June 18, 2021, 06:31:16 pm by Nominal Animal »
 

Offline RJHayward

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Re: Physics Question - ma = mg
« Reply #44 on: June 20, 2021, 03:12:12 am »
Ok, uh BRUMBY: (thank you)
   Is it alright, if my HEAD HURTS now ??  Jeez, but you make the best summary...I agree but interesting to read.
Question about 'g' is that it's weird to call it acceleration directly. Perhaps it just has exact same units, being in meters per second, per second. So an object, on a scale, is not moving, per SE, but rather it just calculates out in a very similar math process.
   How did Newton prove (or did he?)  a derivation ?
I did college level 'classic' physics. Did Newton use integration, using another formula first ?
  I find I can't clearly address that...
 

Offline bob91343

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Re: Physics Question - ma = mg
« Reply #45 on: June 20, 2021, 03:39:37 am »
Yes, mass changes with velocity.  So the proper equation is f = d(mv).  Or f = m(dv) x v(dm).  Or something like that.
 

Offline RJHayward

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Re: Physics Question - ma = mg
« Reply #46 on: June 20, 2021, 07:59:23 am »
 What I mean is, you have to start with integration, of the conventional little bits of matter, within a sphere (Earth). That's calculus, and I believe you can do this with a 'point mass', like a basketball for example.
   So you integrate that: The 3-D sphere with the point mass, to get the total force. The actual force equation has m1 x m2 and divided by r cubed (to 3rd pwr). And some sort of constant.
   After integrating that mess, you've got a force but no motion. Of course, any free mass will begin to 'fall' or accelerate downward. AND, any mass supported by the springs of a scale is going to exert that force as static.
At that point you have, simply, Velocity = Accel X Time
To me, it always looks like either stationary, with a static force, or it's going to be moving, in which case you've got Position / Velocity / and Acceleration.
  Non-relativistic masses, or slightly relativistic situations are, well, EXACTLY THAT. Although likely sensible to view mass as warping space - time, you need a lot of effect, to get, well, a lot of effect.
No swirls around a 'hyper-big' mass as we blast thru hyperspace.....
 

Offline bostonman

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Re: Physics Question - ma = mg
« Reply #47 on: June 23, 2021, 02:03:13 pm »
I've been away and haven't had a chance to reply.

Quote
To lift the box, you need to do work.  If the force against which we do the work is constant, then the work W = F h.

To divert a bit, I know (or believe to be true) that if I lift a box fast or slow, the same work is still done. If W = F * h, and F = ma, why would picking up the box faster not use more Work?

Assuming no other forces except Earth gravity and the Force of the box, if I pick up the box 1m in 1s, my acceleration is faster than if I pick up the box 1m in 10s; thus making the Work larger.
 

Offline T3sl4co1l

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Re: Physics Question - ma = mg
« Reply #48 on: June 23, 2021, 03:00:00 pm »
The power is proportionally larger, the work is the same.

Mind that biological effort is irrelevant, for example you must spend effort just holding the box out in front of you (while doing no work on it).  This doesn't need to be the case, as you could rest it on your knee or shoulders or whatever and not expend any effort beyond keeping it balanced.  Or, y'know, rest it on the floor where it's obviously not doing any work.

Electrical analogs include a lot of actuators, which incur idle power to hold position.  Stepper motors, servos, solenoids, etc.  The only reason these dissipate power is wire resistance.  If they were wound with lossless wire (superconductor), the voltage drop would be zero while the holding force is proportional to applied current (which can circulate without decay).

Tim
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Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #49 on: June 23, 2021, 03:39:03 pm »
I've been away and haven't had a chance to reply.

Quote
To lift the box, you need to do work.  If the force against which we do the work is constant, then the work W = F h.

To divert a bit, I know (or believe to be true) that if I lift a box fast or slow, the same work is still done. If W = F * h, and F = ma, why would picking up the box faster not use more Work?

Assuming no other forces except Earth gravity and the Force of the box, if I pick up the box 1m in 1s, my acceleration is faster than if I pick up the box 1m in 10s; thus making the Work larger.

The key to understanding the conundrum is recognizing that you accelerate and decelerate.  Doing both faster requires more power but the opposite signs mean that extra power doesn't translate into work, Only the net change in potential energy of the box is work. 
 


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