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Physics Question - ma = mg

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bostonman:
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?

T3sl4co1l:
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!

Tim

Nominal Animal:

--- Quote from: bostonman on June 15, 2021, 03:54:30 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.

--- End quote ---
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.)

bostonman:

--- 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.
--- End quote ---

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.

Circlotron:

--- Quote from: bostonman on May 13, 2021, 03:38:12 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)

--- End quote ---
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.

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