Author Topic: Physics Question - ma = mg  (Read 27348 times)

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Offline SiliconWizard

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Re: Physics Question - ma = mg
« Reply #125 on: July 03, 2021, 12:12:07 am »
An accelerating force is that in Newton's First Law, usually stated now as "an object at rest will stay at rest, and an object in motion will stay in motion unless acted on by a net external force".

I just do not like this term "accelerating force" much, hence my remark.

But as to what I said about weight and freefall, and that was also mentioned later on in the thread, I think it's an important, yet not always fully understood point. And not so trivial after all.

It's actually the thought of weight having virtually no effet in freefall that led Einstein to formulate the equivalence principle (that, true, was something already more or less "known" before, but in a more restricted way), and eventually led to the theory of general relativity. So, even though it's in itself nothing extremely special when you know Newton's laws, the idea is still key in "recent" physics history.

What Einstein noticed thinking about weight in freefall is that, in short, the effect of gravity, for an object in freefall, would sort of cancel gravity. That was seminal enough to have inspired the theory that followed...
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #126 on: July 03, 2021, 03:33:02 am »
Should a force effect an acceleration, pursuant to Newton's Second Law, what should one call it?
If a strong wind damages your property, it is called a "damaging wind", etc.
« Last Edit: July 03, 2021, 01:53:49 pm by TimFox »
 

Online bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #127 on: July 03, 2021, 02:27:16 pm »
Quote
I just do not like this term "accelerating force" much, hence my remark


Certain terminology is confusing, especially when you've spent your entire life having a specific naming convention. As an example, Weight has always been known to me as weight, but it's technically Mass. Then you say you're picking up X kg of weight, but you're really using Newtons. Personally it's hard for me to comprehend that a car hitting a wall is X Newtons, but it's easier to think of some logical comparison like it's the equivalent of bench pressing 1000lbs.

Anyway, just some input about how spending a lifetime using wrong terminology can get confusing when you need to learn the correct way.

Acceleration can be confusing. This entire time we've talked about picking up stuff, so "acceleration" is 9.8 due to gravity. Now what if I'm pushing an object (leaving out anomalies like friction)? I need to accelerate the object in order for it to move, but in this case, am I applying Kinetic energy to perform Work instead of using an acceleration formula?

Another Work related question, if I walk on a treadmill at a normal pace, I believe it will read approximately 200 calories are burned (not that important) - obviously this is the computer doing a theoretical calculation based off distance. If I decide to burn calories by picking up a 25kg object several times at a distance of 0.5m (let's forget about Work exerted by putting down the object), that is 245N, which is 122J of Work.

Due to laziness, I used an online conversion calculator and got 29 calories burned just by picking up that object once (although it seems high to me).

Does this mean I can save myself an hour on the treadmill by picking up that item seven times?



 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #128 on: July 03, 2021, 02:41:09 pm »
I must refer you to a good freshman physics textbook to keep the technical terms straight.
For example, you don't "apply kinetic energy"  to accelerate an object, you apply a force to accelerate the object, which thereafter has kinetic energy.
Similarly, weight is not mass.  There is confusion in the way that weight is used in commerce (at the deli counter, for example) for what is really mass.
If I take 1 kg of sausage from the deli and carry it to the moon, it no longer weighs 9.8 N.
Questions about calories (technically, kilocalories) burned during exercise are not Newtonian physics, but physiology:  complicated biological processes.
In your example of lifting a weight, the (Newtonian physical) energy for one lift is the force (245 N) times the distance of the lift (0.5 m), and your body does 122 J of work.
If you lift the weight very slowly, so the acceleration is negligible, then after one lift, the object has 122 J of potential energy, and zero kinetic energy.  If you then drop it, it will accelerate downwards and will have 122 J of kinetic energy after dropping 0.5 m, illustrating the conservation of energy.
 

Offline CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #129 on: July 03, 2021, 08:06:24 pm »
I agree with TomFox that you need time with Resnick and Halliday or other good introductory physics course.

While improper nomenclature is hurting you, improper concepts are your bigger problem.  It will be worthwhile for you to do the experiments described in the text and work the end of chapter problems.  In general if you find the problems hard it is because you didn't understand the presented concepts. Go back and reread and redo the experiments.

Think of your body as a very inefficient car.  It burns something like 1000-1500 calories a day sitting in a chair doing nothing.  The work perform lifting a weight or walking a treadmill is almost negligible.
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #130 on: July 03, 2021, 08:23:35 pm »
From my younger days, I remember Resnick and Halliday as having very well written explanations on the basic topics for freshmen who could do elementary calculus.
 

Offline CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #131 on: July 03, 2021, 10:16:56 pm »
From my younger days, I remember Resnick and Halliday as having very well written explanations on the basic topics for freshmen who could do elementary calculus.

The first several chapters don't even require calculus.  The text is written assuming that the course is concurrent with first year calculus.
 

Online bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #132 on: July 04, 2021, 02:59:50 pm »
Quote
Think of your body as a very inefficient car.  It burns something like 1000-1500 calories a day sitting in a chair doing nothing.  The work perform lifting a weight or walking a treadmill is almost negligible.

Ignoring for a moment that apparently Newtonian work is different than biological, I'm uncertain I agree with this statement. If someone has a desk job, and the best they can burn is 1500 calories, ninety-minutes on the treadmill burning 300 plus calories is 20% or more additional burnt calories. That could mean the difference of burning off your lunch, or not.

As for thinking I could use lifting an object to calculate calories burned, guess I need to research this.
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #133 on: July 04, 2021, 03:41:41 pm »
Warm-blooded mammals need to consume energy in order to maintain their body temperature and vital processes, hence the energy consumption at rest and that one can starve to death while totally sedentary.
Again, another misnomer in units:  when dietitians say "calories", they mean "kilocalories".  Neither are SI units, since heat energy and other energy forms are equivalent and are measured in Joules.
The "calorie" is defined as the amount of heat energy that will raise the temperature of one gram of water by one Kelvin.  Similarly, the "kilocalorie" or "kilogram-calorie" or "large calorie" is the energy required to raise the temperature of one kilogram of water by one Kelvin.  These units are convenient when doing wet chemistry.  Confusion between small-c and large-C calories persists amongst non-physicists and is enough reason to use Joules for serious work.  At least the BTU is well-defined.
 

Offline ucanel

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Re: Physics Question - ma = mg
« Reply #134 on: July 04, 2021, 03:44:24 pm »
Great video that discusses the subject:
(in a funny way also)
Science Ashlyum
Why did i take a scale on an elevator?
https://youtu.be/c6KliNGReTQ
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #135 on: July 04, 2021, 03:48:14 pm »
My favorite descriptions of Einstein's equivalence between a uniform gravitational field and a uniformly-accelerated reference frame take place in elevators.  This is a good way to show the deflection of a light beam by gravitation, since the entrance and exit of the beam on the two walls of the elevator depends on the acceleration.
 

Offline RJSV

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Re: Physics Question - ma = mg
« Reply #136 on: July 04, 2021, 05:34:56 pm »
...A good thread going on, so please don't misunderstand, my mild distress, faced with recent BULK (simply stated), I tend to skim past. My reasons being medically based. Various ailments having associated 'Chronic Fatique' limits on time / energy.

So, my take on movements and expending 'WORK' in formal defined terms includes the non-linear real world of friction. In casual terms, I learned in basic physics, about how friction expresses stronger incremental increases, as you go up the scale of speed. Maybe not increasing as the SQUARE of applied speed; certainly not linearly, usually somewhere in between. It's not a clean business, for example the parachute jump, with all the attendant air resistance and limiting velocity of the jumper (some 120 mph I believe).
   But I don't dedicate all needed time, on friction or WORK debates, and there is a reason:
   The blog posts, on eevblog generally, I've been using as I speculate on improving personal JOB positioning.
And so it makes reasonable sense, to skip over some details (here),...especially while having time / energy limits.
Heck, all creatures operate within limits, huh ?
Thank you. - Rick
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #137 on: July 04, 2021, 05:46:05 pm »
Friction is complicated.  In freshman physics, it is normally approximated as a constant value of force, independent of speed, opposite in direction to the velocity, proportional to the "normal" component of force between the object and base.  (Here, "normal" means perpendicular to the surface.)  A block sliding on an inclined plane is a good homework exercise, where you need to find the normal force as a component of the gravitational force, and then subtract the frictional force from the gravitational force component in the direction of motion.  One can find tables of friction co-efficients, including "waxed hickory on snow".  To be careful, one distinguishes static friction (before the sliding starts) and dynamic friction (while the object moves).
Later, one learns of viscous friction, also opposite in direction to the velocity, whose value is roughly proportional to the velocity.  The "dashpot" (piston moving through a fluid, like an automobile's shock absorber) is used to provide damping in the basic model of forced, damped harmonic oscillator (mass, spring, dashpot, and possibly gravity), which is the model for the R-L-C electrical resonant circuit.
« Last Edit: July 04, 2021, 06:56:45 pm by TimFox »
 

Offline CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #138 on: July 04, 2021, 07:22:04 pm »
Friction is complicated.  In freshman physics, it is normally approximated as a constant value of force, independent of speed, opposite in direction to the velocity, proportional to the "normal" component of force between the object and base.  (Here, "normal" means perpendicular to the surface.)  A block sliding on an inclined plane is a good homework exercise, where you need to find the normal force as a component of the gravitational force, and then subtract the frictional force from the gravitational force component in the direction of motion.  One can find tables of friction co-efficients, including "waxed hickory on snow".  To be careful, one distinguishes static friction (before the sliding starts) and dynamic friction (while the object moves).
Later, one learns of viscous friction, also opposite in direction to the velocity, whose value is roughly proportional to the velocity.  The "dashpot" (piston moving through a fluid, like an automobile's shock absorber) is used to provide damping in the basic model of forced, damped harmonic oscillator (mass, spring, dashpot, and possibly gravity), which is the model for the R-L-C electrical resonant circuit.

Friction IS complicated.  Some folks refer to pressure drag as a form of viscous friction, with some justification.  But this type of friction is "sort of" proportional to the square of velocity.  With "sort of" meaning over a limited range of speed, pressure and a few other variables.

When examining the force required to move and keep a car in motion you can observe at least four classes of friction.  A breakaway force, a constant force at low speeds, a force linearly increasing with speed and one proportional to the square of speed.  If the dynamic friction is large and the drag is large the viscous term may not be observed without very careful measurements as the pressure drag may become large before viscous drag causes significant growth in the total drag.

The simple concepts and models of physics are extremely useful.  But many real world problems require a great deal of observation and thought to determine which simple models apply and how they are used.
 

Offline Nominal Animal

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Re: Physics Question - ma = mg
« Reply #139 on: July 04, 2021, 08:03:29 pm »
(I'll try to be brief!)
friction expresses stronger incremental increases, as you go up the scale of speed
Yes.

Friction itself can be described as a force opposite to velocity.
Usually, we choose a coordinate system where our object is stationary, and use the velocity of the surrounding fluid (for drag equations) or the velocity on the surface the object is rubbing against.

Velocity is not the only thing involved in determining its magnitude.  Pressure of the fluid affects its drag coefficient, and the force pushing the object to the surface it slides on affects the coefficient of friction.  As mentioned above by Tim Fox, CatalinaWOW and others, its magnitude can be proportional to the square of the velocity, to the magnitude of the velocity, or something in between or only slightly different, depending on the situation.  The proportionality is useful for us humans to estimate the effect, but arises from shapes and flows so there is no clear formulae, except for idealised shapes and fluids and surfaces.

The 'work' done by those forces describes exactly the amount of energy transferred due to drag and friction.
That energy is usually lost in heat and deformation.

If you have ever done lathe work or milling, "chip breaking" is important because the tools rely on the removed particles (chips) to carry off the extra heat generated by friction.  With good tools and proper setup – surface speed of the cutting bit and the depth of each cut –, your work piece can remain relatively cool, while the metal shavings flying off are so hot they rapidly oxidize and change color.  (I love machining videos.  Knowing the funky physics that occur there just makes them even more enjoyable.)

In his 1980 Science Fiction book Sundiver, David Brin describes a craft skimming in the chromosphere of our Sun, that uses a high-tech version of that to remain cool: a very, very powerful refrigeration laser.  The "chips" are then photons instead of matter, with energies corresponding to crazy high temperatures, but the core principle is the same.  As with lathe and mill chip breaking, to work, those chips have to be hotter than the ambient and hotter than the object you are trying to keep cool(er).
« Last Edit: July 04, 2021, 08:08:36 pm by Nominal Animal »
 

Offline CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #140 on: July 04, 2021, 09:37:11 pm »
Quote
Think of your body as a very inefficient car.  It burns something like 1000-1500 calories a day sitting in a chair doing nothing.  The work perform lifting a weight or walking a treadmill is almost negligible.

Ignoring for a moment that apparently Newtonian work is different than biological, I'm uncertain I agree with this statement. If someone has a desk job, and the best they can burn is 1500 calories, ninety-minutes on the treadmill burning 300 plus calories is 20% or more additional burnt calories. That could mean the difference of burning off your lunch, or not.

As for thinking I could use lifting an object to calculate calories burned, guess I need to research this.

I was flippant when I said almost negligible.  If 90 minutes on a treadmill had no impact there would be no market for gyms.  But the dominant use of the fuel you dump into your body is keeping the fires burning, powering your brain (the single largest consumer of energy in most humans) and doing other basic functions.  Running, jumping and lifting and doing the kind of work described in physics makes relatively minor variations in the fuel consumption.  Even for top athletes running ultra marathons or world strongest man champions the work output is far smaller than the housekeeping consumption.

In principal you can relate lifting weights to calories but the relationship will be very imprecise.  Your body speeds up the background consumption during and after exercise so calories will be burned that aren't directed toward the lifting of the weight.  Similarly you lift and drop parts of your body while lifting your weight set, complicating the calculation.  That is why you will find the kind of empirical numbers you are referring to on your treadmill.  Scientists have basically put people into a calorimeter while performing various types of exercise and work and measured the output.  Your mileage will vary.  Different individuals will get different results so while one person may burn 300 calories on that treadmill others might burn 200 to 600 (made up numbers, I know there is a range, but haven't studied it).  I am quite sure that the marketing people for gyms and exercise equipment bias their quotes to the higher results.
« Last Edit: July 04, 2021, 09:38:55 pm by CatalinaWOW »
 

Online bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #141 on: July 09, 2021, 02:52:11 am »
It's unfortunate calories isn't more precise.

Some of these questions do revolve around the gym. For years I wondered how many say bench press reps would equal X minutes on a treadmill.

 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #142 on: July 09, 2021, 01:25:59 pm »
In physics, these terms are precise, but due to historical usage, the popular language confuses them.
Precisely:  a calorie is the amount of heat that increases the temperature of 1 gram of water by 1 K.
A kilocalorie is the heat that increases the temperature of 1 kg of water by 1 K.
A BTU is the heat that increases the temperature of 1 lb av of water by 1 Fo, which brings us back to the definition of the pound:  mass or weight.
 

Online bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #143 on: July 09, 2021, 01:30:51 pm »
Quote
A kilocalorie is the heat that increases the temperature of 1 kg of water by 1 K.

Is this the reason when I first mentioned "calorie" burning at the gym, someone said it's really a kilocarlorie?

In other words, people who say they burned "calories" at the gym are really using it incorrectly?
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #144 on: July 09, 2021, 02:18:56 pm »
That was me.
Of course, if you burn calories you certainly burn kilocalories, but a different rate.
The SI uses only Joules for mechanical, electrical, and thermal energy.  1 kcal = 4184 J.
At the gym, if you lift 1 kg up by 1 m, than is 9.8 J of mechanical work and potential energy.  Therefore, lifting a 50 kg dumbbell by 1 m is 490 J, or only 0.117 kcal.
Once you leave the quiet rational space of the SI, you venture into a dense thicket of other units at your own risk.
 

Online bostonmanTopic starter

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Re: Physics Question - ma = mg
« Reply #145 on: July 09, 2021, 02:35:42 pm »
Quote
herefore, lifting a 50 kg dumbbell by 1 m is 490 J, or only 0.117 kcal.

Just tell me if I'm going over the deep end or just reverting to "calories" being incorrectly used. So if I eat a Snicker's Bar at whatever.... 100 calories, does this theoretically mean I have to lift that dumbbell 855 times (100 divided by 0.117) to burn off the snack?

Or are calories on nutrition labels using the wrong units?
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #146 on: July 09, 2021, 03:28:59 pm »
The question about the conversion of food energy to mechanical energy through biological processes is far more complicated than the Newtonian physics behind lifting a mass in a gravitational field.
Since your body is very inefficient, it takes more food energy to do a given amount of physical work, therefore it takes less physical work to burn off a given amount of food energy.
Nicolas Léonard Sadi Carnot​ in 1824 originated the careful consideration of "heat engines", that convert thermal energy into mechanical energy, just as steam engines (invented in the previous century) were becoming important.
(A distinguished family.  His father Lazare Nicolas Marguerite, Count Carnot was a French revolutionary and did fundamental work on engineering mechanics, and his nephew Marie François Sadi Carnot was President of France at the end of the 19th century.)
Compared to a Diesel engine, the human body is not an energy-efficient engine (but it suffices for our normal purposes).
As I stated, what dietitians and candy-bar manufacturers call a "calorie" is technically a "kilocalorie" in physics, and sometimes called a "large calorie" to distinguish it from the "small calorie" as defined properly.
 

Offline CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #147 on: July 09, 2021, 03:35:17 pm »
Just to pound the point in calories are not imprecise, but calculations of their consumption in the human body are.  There are a great number of processes going on, many variables associated with each process all of which are difficult to measure and many of which are either not simple enough or well enough understood to write an input-otput equation.
 

Offline TimFox

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Re: Physics Question - ma = mg
« Reply #148 on: July 09, 2021, 03:41:23 pm »
Of course, nutritional scientists work hard to develop legitimate quantitative models and calculations of the human body's usage of and needs for food energy.  It is a far more complex problem than the Newtonian mechanics of apples dropping from trees.  Since calories are popular units for chemists, the nutritional guys have tended to use them.  Not my field, man.
 
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Offline CatalinaWOW

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Re: Physics Question - ma = mg
« Reply #149 on: July 09, 2021, 06:21:19 pm »
Not only do they work hard, but they do good work.  But one size definitely does not fit all and there is a huge population demanding simple answers.  Bostonman is on his way to understanding the complexity.  Most aren't interested or don't have the capacity.
 


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