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Take the typical phyiscs / classical mechanics example of a parachuter descending at constant speed. The two forces acting on the parachuter are weight, which has direction towards the ground, and air resistance, which has direction upwards. In classical mechanics, for this case, we say that the net force – the force of air resistance minus the force of weight – is zero, because the parachuter is descending at constant speed, and so acceleration is zero.

Even though I understand the argument, and I understand this concept in a "memorization" type of way, I can tell that I do not actually understand the concept/physics in a proper / sufficiently deep way. For instance, we say that the "net force" is zero, which means that the force of air resistance and the force of weight cancel out to equal zero; but why should zero net force mean that velocity is constant, rather than an object (in this case, the parachuter) being stationary (that is, being stationary in the middle of the atmosphere once air resistance force equals weight force)? Now, I understand that this is going back to Aristotle's theory of motion (the idea that all motion was centered upon the object trying to reach its "natural resting position") and Newton's first law of motion / the law of inertia – an object at rest stays at rest, and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force – but I still do not feel logically comfortable with this idea that, even though the net force is zero, that does not mean that the parachuter will become stationary in the atmosphere, but, rather, that velocity just becomes constant.

I suspect that I am misunderstanding/misinterpreting the concept of "net force," and so am expecting the fact that the air resistance force vector and the weight force vector cancel out to mean that the object must become stationary, rather than velocity to just be constant; the object becoming stationary just seems logical, given the mathematics of how these forces/vectors interact, rather than velocity becoming constant. Why does "net force" being zero not mean that the parachuter/object becomes stationary in the atmosphere, but, rather, that they just descend at a constant speed (acceleration becomes zero)?

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  • $\begingroup$ You may have a hidden assumption that was common before Newton's time. In the distant past, people believed that a constant velocity required a constant force to keep an object moving. That is definitely an incorrect assumption. $\endgroup$ Mar 25, 2022 at 17:16
  • $\begingroup$ @DavidWhite That's not it; it's moreso an interpretation of the mathematics/vectors associated with the forces and the concept of "net force" that is leading me to this conclusion. $\endgroup$ Mar 25, 2022 at 17:19
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    $\begingroup$ I can't directly clear up your confusion, but I will state that physics is NOT math. The language of physics is math, but the concepts drive the math rather than the math driving the concepts. $\endgroup$ Mar 25, 2022 at 17:24
  • $\begingroup$ @DavidWhite I understand that too. But in this context of Newtonian/classical mechanics, this is not the issue, since I'm dealing with basic, well-established physics that is known to be correct. So, unless I'm doing the mathematics incorrectly, which I don't think I am, then the issue must be in misunderstanding/misinterpreting the mathematics, with (false) hidden assumption in my mind, which then leads to faulty reasoning. $\endgroup$ Mar 25, 2022 at 17:29

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By $F=ma$, a net force of zero ($F=0$) implies that there is no acceleration ($a=0$), assuming the object has mass. This simply means that the velocity is not changing, but makes no statement about what the velocity actually is. It could be positive, negative, or zero, so long as there is no change in velocity, there is no acceleration and therefore zero net force. A skydiver falling at a constant rate is not accelerating, but $a=0$ does not imply that $v=0$.

Also note that there is nothing "special" about the ground reference frame. Any inertial reference frame is equally valid, you could measure the skydiver's velocity with respect to the ground, or with respect to a balloon rising at a constant rate, or with respect to another skydiver falling at the same constant rate. There is no reason why the velocity should be zero with respect to the ground rather than the balloon or the other skydiver - there is no such thing as absolute velocity, it depends entirely on what you measure it with respect to. From the ground reference frame, a skydiver at terminal velocity has a constant, non-zero velocity. But according to another skydiver falling at the same rate, they have constant zero velocity. Neither view is any more "correct" than the other - although they disagree about the velocity of the skydiver due to their different reference frames, they both agree that the skydiver is not accelerating. The skydiver's velocity can take different values depending on the choice of equally valid inertial reference frames, but they all see that the net force is zero and the acceleration is zero.

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  • $\begingroup$ I think this answer is closest to what I was looking for. I'll have to ponder this further. $\endgroup$ Mar 25, 2022 at 18:06
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Net force zero implies zero acceleration, hence constant velocity. It does not imply stationary position.

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The acceleration vector and the velocity vector at any instant are completely independent of one another. The magnitude and the direction of the acceleration vector can be chosen independtly of the magnitude and direction of the velocity vector. They can point in the same direction and both be large; the can point in opposite directions and both be small; they can point at right angles and you can have a large acceleration and small velocity; etc.

Newton's second law says that the net force vector is parallel to the acceleration vector, and the magnitude is related by $F_{net}=ma$. So the force vector will tell you the acceleration vector. Since the velocity vector is independent of the acceleration vector, and since the net force vector is parallel to the acceleration vector, then it must be the case that the net force vector is independent of the velocity vector. That is indeed true, in general. (Although, there are special cases like uniform circular motion where the net force does depends on the velocity and the acceleration and velocity are not independent)

This is a very common source of confusion when first learning physics. It is not intuitive, because in the world we experience there is friction, and so it does require application of a constant force opposing friction to keep an object moving at a constant velocity on Earth. So at first sight, Newton's second law seems obviously in conflict with experience. However, this is just an artifact of the environment we find ourselves in. Properly accounting for all the forces, Newton's second law holds. And, we now have the technology to go to space, where we can get rid of the Earth which just serves as a distraction. In space, Newton's second law is more obvious; if you throw a baseball in space, it will keep moving at the speed you threw it at in the direction you threw it at forever (ignoring gravity).

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  • $\begingroup$ "Newton's second law says that the net force vector is parallel to the acceleration vector, and the magnitude is related by $F_{net}=ma$." That's a good point. $\endgroup$ Mar 26, 2022 at 18:09
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If you think it mathematically, $force= dp/dt$ where $p$ is the momentum in the direction of the falling parachutist. If the sum of forces is zero, it means that the momentum is constant. If it is larger than zero it means that the momentum is increasing.

As $momentum=mv$ where $v$ is the velocity zero force means constant velocity, not zero velocity.

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Perhaps it would help to remember that the only reason why there is an upward force counterbalancing the weight of the parachutist is because the parachutist is moving. If the parachutist became stationary there would be no air resistance to stop them dropping.

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You may like to consider the state where the parachute is open and the jumper is falling, but has not reached the stable velocity.

The drag force on the parachute increases as the velocity through the air increases. But in the early stages of the jump that drag force is LESS than the weight of the jumper, so he accelerates. As he accelerates the drag force increases until it is EQUAL to the weight. At that stage he has a given velocity, and there is no net force to change that. So the velocity remains constant. In order for him to DE-celerate there would need to be a net UPWARDS force to reduce the velocity to zero.

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  • $\begingroup$ This answer is a good addition. Thanks. $\endgroup$ Mar 25, 2022 at 22:58
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Air drag force $F_D \propto v^2$, i.e. is proportional to air flow speed around object (parachute). So at first time moment $t_0$ when parachuter jumps out of plane,- drag is zero and gravity commands them all. Then as human weight produces acceleration due to gravity,- parachuter speed increases and so does - air drag force. At some point drag force becomes so big that it reaches gravity force in magnitude, thus due to reached equilibrium, object aquires terminal speed,- maximum speed which can be reached until drag stops gravity effect. This terminal speed can be calculated by equating drag force and weight and solving for $v$ :

$$ \,{\tfrac {1}{2}}\,\rho \,v^{2}\,C_{D}\, = mg$$

If at the time of jump $t_0 = 0~s$ two opposing forces would be instantly at equilibrium,- then as you say object would stay at rest, not even starting to move at all down from plane. Similarly as we are sitting on chair - two opposing forces gravity and normal force are constantly at equilibrium, so you never start to move either up or down, so simply stay at rest with $v=0$ speed in a vertical axis.

Btw, Newton second law $$ F_{net} = m \frac {dv}{dt} $$ implies that when net force is zero - speed change must be zero, because mathematically derivate of constant is zero : $$ \dot {const} = 0 $$. And 0 is same constant as any other numeric speed value. Whatever exact constant of speed body will have,- it will depend on initial conditions of system, how/when force equilibrium is reached, etc. That's the main beauty of Physics.

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