Second Law of Motion (F = ma)

THE SECOND LAW
OF MOTION

.

F = ma?

The above scan is a page from the standard basic physics book (Basic Physics - Ramsay, Eubank, McMilan Canada, 1963)

The page states Newton's second law of motion as mass and acceleration having a relationship of inverse proportionality. I have decided to show to whom ever may be interested:

• that this law is down right wrong and unjustified
• where has the originator failed to correctly comprehend and interpret
  the experimental evidence
• what is the correct relationship between force, mass and
  acceleration.

I have designated the textbook "F" as M1 (mass 1), and the textbook "M" as M2 (mass 2) and I will follow this designation from now on.

The M1 (stated as "F" in the textbook figure 5:6) is first of all mass. Its weight (force) is used to pull (accelerate) another mass M2 (stated as "M" in the same figure). M1 hangs over a pulley on a string attached to a trolley containing M2. While the gravitational force is pulling (accelerating) M1 down, M1 is pulling (accelerating) M2 horizontally.

"If we double the mass of the body and apply the same force, we obtain half the acceleration. That is, the acceleration is inversely proportional to the mass, if the force is kept constant"

A Small Mistake

I have to postulate here that mass of M1 = mass of M2 through out this treatise. Once the mass of M2 is increased, it is presented as set of masses M2, i.e., [2 x M2], [3 x M2] etc.

When we look at the figure 5:6, the downward weight of M1 (F) is obviously considered to be a source of constant force. Then it follows that only the mass M2 (M) is meant as being doubled. Both masses, M1 as well as M2, are being accelerated in this experiment as the earth gravitation acts against the inertia of M1 as well as M2. The setup in the figure 5:6 already doubles mass M1 (which would be otherwise under free fall conditions) by M2 added into the horizontally dragged (assumably mass less) trolley.

When only M2 is further doubled from the depicted (fig. 5:6) [1 x M1] + [1 x M2] = [2 x M] setup to [1 x M1] + [2 x M2] = [3 x M] setup, the inertial mass opposing the earth gravitation acting on M1 alone is increased only by one half of the depicted total [1 x M1] + [1 x M2]. The correct interpretation of the [1 x M1] + [2 x M2] setup would have been: "If the inertial mass is increased by one half, the acceleration drops to one half of its original value."

If the originators of this experiment gave it a good thought, they might have realized that the first step in this experimentation should have been the measurement of the free fall rate of M1 alone. Adding horizontally dragged M2 of equal mass into the Fletcher's trolley connected to M1 by a line would really double the mass of the experiment.

THE BIG MISTAKE

Lets say that the textbook figure 5:6 represents M1 (F) at m = 1 kg and M2 (M) at m = 1 kg. Lets say that adding M2 in a trolley to M1 slows what would have been free fall of M1 down to 1/2, that is 4.9 m/s². (4.9 is wrong, but that does not really matter at this point).

"Everybody" knows that the acceleration in the free fall is a balancing act between inertia of a mass and gravitation. If there is no acceleration of a body toward ground, the body weights the full equivalent of its mass, i.e. its full weight, in our case 1 kg. When the body is accelerating at the free fall rate toward ground, it is weightless. When a body is hindered in its free fall by a force smaller than the gravitational force and opposing the gravitational force, its weight is in some proportion to the strength of the opposing force. In other words, if we should place 1 kg weight onto a scale and the scale into an elevator mounted on a slip brake, which would allow the elevator to fall only at 4.9 m/s², the 1 kg mass of the weight would show substantially lighter on the scale than 1 kg.

Therefore, 1 kg M1 has its weight at 1 kg before the Fletcher's trolley experiment is executed but, M1 weights only a portion of its weight at rest while the experiment is in progress.

The weight of a body in a limited accelerated fall changes with the degree of limitation imposed on its free fall. It means that while this kind of experiment is in progress, the force applied by ("constant mass") M1 on different sets of M2 masses varies with the rate of acceleration, because the weight of M1 varies with different rates of limited fall acceleration.

Therefore, it is wrong to assume that different M2 sets like [1 x M2], or [3 x M2] etc. may be dragged into acceleration by a constant mass M1 with a constant weight, therefore constant force F.

The key word in the interpretation of these experiments is "WEIGHTLESS". A free falling mass of 1 kg (M1) is weightless and it can't exert any force onto anything. The same 1 kg mass M1 supported at a constant height, lets say by a desk top, applies force equivalent to its weight against the desk top. These two are the extremes. One extreme is "no weight = no force opposing gravitation", the other is "full weight = force equal to gravitation opposing gravitation".

When we insert a spring (or a rubber band) into the line connecting M1 and M2, we can readily observe that the spring (or rubber band) contracts as soon as the experiment is released into motion. The contraction lasts as long as long is M1 accelerating M2. This proves beyond any reasonable doubt that the force stretching the flexible portion of the connecting line between M1 and M2 diminishes while the experiment is in progress. We should realize that the weight of M1 in a limited (not so free) fall is not equivalent to 1 kg weight. It is less, depending on the strength of the force opposing gravitation pulling M1 down. (By the same token, a body accelerated upwards weights more than its mass equivalent at rest.)

The force between M1 and an M2 set in accelerating motion is greater, the lower is the rate of acceleration of M1 fall. Vice versa, the force between M1 and M2 set is smaller, the greater is the rate of acceleration of M1 fall. This means the greater is the ratio between mass M1 and M2 set in favor of M2 set, the greater is the force between M1 and M2 set and the greater is the weight of M1 during the experiment and vice versa.

The direct and inversely direct proportion between force, mass and the rate of acceleration "proven" by this experimentation is fundamentally faulty. On the contrary. This experimentation proves that relationship wrong, once properly understood.

• The formula F = ma is wrong.
• The Second Law of Motion is wrong.
• Any physics formula and, or set of calculations containing, or based on F = ma is fundamentally wrong.
•  

THE CORRECT RELATIONSHIP

One readily available way to come to the realization of the relationship between magnitude of force and the rate of acceleration of mass comes from the simple behavior of liquids, namely the initial acceleration of a “fountain” jet of liquid.

Observation as well as experimentation shows us that a jet of liquid squirting from an orifice pointing upwards at the bottom level of a tank will almost achieve the height of the head (surface) of the liquid in the tank. It never goes as high as the head. It goes up approximately only 85% of the head height. But, there is friction within the liquid before and while it passes the orifice, as well as friction between the liquid and the orifice, plus the air friction once the fountain exited the orifice to name just a few. These just about provide for the difference. (A bit of that difference comes also from the fact that a conical column of liquid in the tank accelerates downward and therefore looses some of its weight, therefore static pressure drops some too within this stream. Never the less, if the vessel is rather large in comparison to the orifice size, this loss is negligible for our purpose.)

We know that an object dropped from a certain height will attain a certain impact speed (velocity if you insist) by the time it hits the ground. If we accelerate it to the same speed up at the ground level, it will achieve the height from which it was originally dropped.

A pendulum shows the same effect, swinging just about as high as from where it was released. By the same token, the liquid of a fountain starts off at the speed equivalent to the speed it would have reached at the bottom level of a tank if poured from the level of the liquid head. (Theoretically, friction etc. disregarded)

Below is the chart of head heights and the fountain initial velocities, equivalent to free fall final velocities corresponding to the head height(s). I have fudged the standard value of “a = 9.8 m/s²” to a = 10 m/s². It is irrelevant to the principle, but it allows us to readily recognize the pattern of relationship between the force (static pressure at the fountain orifice with a particular head height) and the initial velocity of the upward liquid stream, i.e. acceleration. I would also use water at 4C in order to keep the liquid mass at a wholesome 1, therefore unnecessary to take into account. This also makes it easy to derive the static pressure from the head, because water column 10 m high (10 m head) exerts 1 Atm (1 Atmosphere = 1 kg pressure per square centimeter) of pressure on the bottom, because 1 cm x 1 cm x 10 m of water equals 10cm x 10cm x 10 cm = 1 dm³ = 1 liter of water, which weights 1 kg. (Good, original and easy metric system)

Rate of acceleration at 1 g fudged to a = 10 m/s²
mass = 1, t1 = 1 sec, t2 = 2 seconds etc.
head = top of the water column in meters

time	free fall vertical distance	impact velocity	liquid head	static pressure  (force)	fountain initial velocity
t1	5 m	10 m/s	5 m	0.5 Atm	10m/s
t2	20 m	20 m/s	20 m	2   Atm	20 m/s
t3	45 m	30 m/s	45 m	4.5 Atm	30 m/s
t4	80 m	40 m/s	80 m	8  Atm	40 m/s
t5	125 m	50 m/s	125 m	12.5Atm	50 m/s
t6	180 m	60 m/s	180 m	18    Atm	60 m/s
t7	245 m	70 m/s	245 m	24.5 Atm	70 m/s
t8	320 m	80 m/s	320 m	32    Atm	80 m/s
t9	405 m	90 m/s	405 m	40.5 Atm	90 m/s
t10	500 m	100 m/s	500 m	50    Atm	100 m/s
t11	605 m	110 m/s	605 m	60.5 Atm	110 m/s

This table shows in clear terms that the rate of increase in the potential energy of the static pressure is not matched by the rate of increase in the initial velocity of the fountain in a direct proportion. When we compare t1 and t10 in particular, we can immediately recognize that while the water column at t10 is 100 x taller than the water column at t1, the initial velocity of the upward liquid fountain at t10 is only 10 x greater than the initial velocity at t1.

In other words, while the static pressure of 0.5 Atm produces twenty times its arithmetic value in acceleration of water at the orifice of a fountain, the static pressure of 50 Atm produces only 2 times its arithmetic value in acceleration of water at the orifice.

Please note that pressure of 12.5 Atm at t5 (right down the middle as per “t”) produces only 4 times its arithmetic value in acceleration. This is not anywhere near half way between 20 at t1 and 2 at t10. The 24.5 Atm at t7, just about the middle of the head heights, produces not quite 3 times its arithmetic value in acceleration, which again is nowhere near half way between 20 and 2.

When we produce a table with an imaginary liquid whose specific mass is 10, that is 10 kg/dm³ we may realize that as long as the rate of acceleration is kept constant, increasing the mass 10 times requires increasing the force 10 times. In general, keeping mass and force in a direct proportion produces a particular rate of acceleration. The same can be derived from the Fletcher's trolley experiments, which is hardly surprising.

Never the less, the theoretical acceleration of a body as per applied force can be only calculated using an altered energy formula E = 1/2mv², which is this F = 1/2ma². (it would be F = ma² /200 when calculated in the units I used.)

In other words: Energy, Time and Force are mathematically more or less on equal footing regards the relationship between mass, acceleration and velocity.