Misconception of Newton's laws?

easwar2641993

A bird flies by pushing downwards the air below its wings with a force. According to Newton's third law, the air pushes the wings upwards. As a result, the bird is lifted.

Also consider the case of a rocket. The fuel it ejects exerts an upward force and as a result, the rocket moves upwards.

The second law states that vector sum of the all the forces acting on the particle is given by the product of its mass and acceleration.

In the above cases, should they have acceleration as the action force shouldn't be included in the "net " force "acting" on them?Is it the reaction force causes them to accelerate?

Another question. I am standing on the ground. I am exerting a force on the ground and ground of course by the 3rd law exerts a normal force.If I consider only normal force as the force acting on me, then according to 2nd law,I should have an acceleration right?

I am bit confused whether I take gravitational force as it is the one which causes us to exert a force on the ground.

0xDEADBEEF

I guess that the answer to this confusing question ("was is an action force?") is yes. The force of the bird wing on the air accelerates the air and not the bird. The counter-pressure by the air lifts the bird (in a very simplified picture).
There is no difference between an "action force" and a "reaction force". This only matters in the way how you decide to discuss the problem. The fact that the wing can produce a force on the air, or that the air can produce a force on the win comes from muscle tension and fluid dynamics. The bird accelerates due to the force of the air on the wing, if it is larger or smaller than the force of gravity on the bird.
And if you consider that there is no air, you will suffocate... It is simply not true that only the normal force acts on you.
There are 4 forces. 1 Gravitational pull of the earth on you, 2 Gravitational pull of you on the earth, 3 normal force preventing the earth from entering your body, 4 normal force preventing your body from entering the earth. I hope that this makes it a bit more clear.

Philip Wood

Consider carefully which body's acceleration you wish to find. If you want the rocket's acceleration you need the net (resultant) force on the rocket. Neglecting air resistance, this is the upward force exerted by the rocket's escaping gases minus the downward pull of gravity on the rocket. You don't include the downward force of the rocket on the gases, because this is a force acting not on the rocket but on the gases.

The terms 'action' and 'reaction' are notorious for confusing students, and a lot of teachers never use them - they're not needed. Instead, they speak of the force that A exerts on B, and the force that B exerts on A, in which A and B are the relevant objects.

Penultimate paragraph: yes the normal contact force exerted by the ground will cause you to slow down and stop when you land on the ground. The peak value of the force will be greater than the pull of gravity on you - much greater if you've dropped from a large height. When you're standing still and not in the process of springing up or sinking to your knees, i.e. not accelerating, the normal contact force on you and the pull of gravity on you are equal and opposite. [Don't fall into the trap that these forces being equal and opposite in this special circumstance has anything to do with Newton's third law.]

DaleSpam

The second law doesn't have anything to do with action and reaction forces, that is the third law. For the second law you add up ALL forces acting on one single object to get the object's acceleration.

There are two forces acting on you, the normal force and gravity. You just add them up (and divide by your mass) to get your acceleration. The fact that they each come in 3rd law action/reaction pairs is not important for the 2nd law.

WannabeNewton

I wanted to emphasize this. It seems to be a very common trap to think, for example in this case, that the normal force is the 3rd law reaction force of gravity. The normal force when standing on the contact surface is NOT a 3rd law reaction force of gravity. It's cause is much more complicated than that.

Chestermiller

We learn to use free body diagrams in freshman physics, but then, for some reason, people stop using them, and then the trouble begins. The importance of using free body diagrams cannot be overemphasized.

vanhees71

Well, for me forces and free-body diagrams are the nuiscance. Of course you have to learn about them and also teach them to freshmen, but one shouldn't overemphasize their importance. The true fundamental laws, including Newtonian mechanics, are all best represented by Hamilton's principle of least action and a careful analysis of the underlying symmetry assumptions of the fundamental theories.

Chestermiller

If the OP had used free body diagrams, he wouldn't have encountered the conceptual difficulties he displayed. Anyway, regarding free body diagrams and forces vs Hamilton's principle, potatoes - potahtoes.

WannabeNewton

I agree completely Chet. I remember there's a quote in Kleppner about the importance of free - body diagrams but I can't seem to find it
To be honest, I enjoyed this kind of mechanics much more than more advanced physics because there was just so much more actual physics. The "physics" that came later on just felt like nothing but mathematics with little to no actual physics (e.g. the QM in Ballentine's book).

Fredrik

What do you mean? One of the forces acting on you is gravity. You are not accelerating. Therefore, the sum of all the forces is zero. So we just choose to call the force that cancels the effect of gravity the "normal force".

Are you talking about when the surface is a slope? Then we simply write the force that cancels the effect of gravity as a sum of a vector perpendicular to the surface and a vector that's parallel to the surface, and we call the former "normal force" and the latter "friction".

WannabeNewton

The normal force isn't a 3rd law reaction pair of gravity; this isn't in conflict with what you said however. When there is a gravitational pull on you due to the Earth's field, the 3rd law reaction force is simply your gravitational pull back on the earth, not the normal force. For a body resting on a surface, the normal force is equal and opposite to the resultant of all other forces acting on the body that are orthogonal to the surface; in the situation talked about here we simply have only the gravitational force. The cause of the normal force comes from pauli exclusion etc.

Fredrik

Ah, now I understand. I agree. What I said has nothing to do with the third law. The third law says that when A exerts a force F_AB on B, then the force F_BA that B exerts on A is equal to -F_AB. F_BA is a force on another object, not a second force on the same object.

D H

You are accelerating. The Earth rotates once per sidereal day, and thus you undergo circular motion. Since the Earth has an atmosphere, you are also buoyed upward a tiny bit by the buoyant force. The normal force is not equal but opposite to gravity. There is an equal but opposite counterpart, the force your body exerts locally on the surface of the Earth. (Strictly speaking, it's only the component that is normal to the surface. The transverse component is friction, not the normal force.)

The normal force is a constraint force. It's directed normal (upward) to the surface and the magnitude is however much is needed to keep you from sinking into the surface. Suppose you grab a chin up bar and pull down on it, but not with your full weight. The normal force from the ground is still acting on you, but now reduced in magnitude. Gravity however hasn't changed a bit.

stevendaryl

There are two different "equal and opposite" forces in elementary physics, and people often mix them up. First, if object A exerts a force on object B, then object B exerts an equal and opposite force on object A. This is always true (at least in Newtonian physics).

Second, in the special case in which B is at rest (static case), it must be that there is some other force on B that is equal to and opposite the force due to A. This is not Newton's 3rd law, it's Newton's 2nd law: if the acceleration is zero, then the net force must be zero.

So in the case of a person standing on the floor, we have gravity pulling down on the person, and we have the floor pressing up on the person. Because the person is not accelerating, these two forces have to be equal and opposite. But they are not the equal and opposite forces from Newton's 3rd law.

The more complete accounting of forces is:

1. [itex]\boldsymbol{F_{e,p}} = [/itex] the downward gravitational force of the Earth on the person.
2. [itex]\boldsymbol{F_{p,e}} = [/itex] the upward gravitational force of the person on the Earth.
3. [itex]\boldsymbol{F_{f,p}} = [/itex] the upward normal force of the floor on the person.
4. [itex]\boldsymbol{F_{p,f}} = [/itex] the downward normal force of the person on the floor.
5. [itex]\boldsymbol{F_{e,f}} = [/itex] the upward force of the earth on the floor.
6. [itex]\boldsymbol{F_{f,e}} = [/itex] the downward force of the floor on the earth.

[itex]\boldsymbol{F_{e,p}}[/itex] and [itex]\boldsymbol{F_{p,e}}[/itex] are Newton 3rd law pairs of equal and opposite forces.

[itex]\boldsymbol{F_{e,p}}[/itex] and [itex]\boldsymbol{F_{f,p}}[/itex] are equal and opposite, but they are not Newton 3rd law pairs.

[Edit: I see that in the time it took for me to explain this, it's already been resolved. So nevermind.]

stevendaryl

I don't see how the principle of least action, or the symmetry of fundamental theories, helps you solve static force problems involving ropes and pulleys and ladders and friction, and so forth. Force diagrams are intended for solving those type of problems.

WannabeNewton

Yeah exactly. Btw, you might find these old discussions interesting, I found them back in AP Physics C when my teacher didn't give an adequate description of the normal force: http://physics.stackexchange.com/que...e-normal-force
http://physics.stackexchange.com/que...m-or-and-pauli