Understanding Newton's Third Law and the Net Force on Objects

[dainceptionman_02]

In the physics text for F(net) = ma, the text expresses how
"F(net) must be the vector sum of all the forces that act on the body, and that only forces that act on that body are to be included in the vector sum, not forces acting on other bodies that might be involved in the given situation." "It does not include any push or pull on another object from you."
As an example, you are in an isolated frictionless system where 3 forces act on you giving you acceleration in a random direction, and a 4th force would be you pushing on an object giving you an acceleration in the opposite direction.
Am i just supposed to ignore this force when constructing Fnet or a free body diagram? Isn't the infamous normal force on free body diagrams a Third Law reactionary force?

 

[Dale]

Am i just supposed to ignore this force when constructing Fnet or a free body diagram?

In a free body diagram for a specified body you do not ignore any external force acting on the specified body. You do ignore any internal forces within the body. You also ignore any forces acting on any other body.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[berkeman]

After deleting a post trying to use AI as a reference, this thread is reopened provisionally.

 

[dainceptionman_02]

In a free body diagram for a specified body you do not ignore any external force acting on the specified body. You do ignore any internal forces within the body. You also ignore any forces acting on any other body.

would pushing an object and gaining an acceleration from Newton's third law be considered an external force then?
consider the normal force. on a free body diagram, the object experiences a gravitational force from the earth and a normal force from the surface. the normal force from the surface is supposed to be reactionary force from the gravitational force on the surface from the object. is it not? is this is contradiction in the text?

 

[kuruman]

As an example, you are in an isolated frictionless system where 3 forces act on you giving you acceleration in a random direction, and a 4th force would be you pushing on an object giving you an acceleration in the opposite direction.

You have not specified what the system is. The forces that are included in a free body diagram must be external to the system, i.e. originate in entities that are not part of the system. If in your example the system is "you", then you are not external to "you". Thus, the 4th force that you mentioned cannot be included. However, that 4th force has a Newton's 3rd law equal and opposite force outside you. That should be included.

Example: You are the system. You are sitting in a chair and you want to get up. If you grab your collar and lift up, you might tear your shirt but you are not going to rise. If you push down on the floor with your feet, the floor (external to you) pushes up on your feet with an equal and opposite force and you rise. Do you see how it works?

 

[Dale]

would pushing an object and gaining an acceleration from Newton's third law be considered an external force then?

If you push an object then Newton’s 3rd law says there are two forces. One force acts on you and one force acts on the object (and they are equal and opposite). The force acting on you would show up on a free body diagram for you. The force acting on the object would show up on a free body diagram for the object.

the normal force from the surface is supposed to be reactionary force from the gravitational force on the surface from the object. is it not?

The normal contact force on an object is not a 3rd law pair with a gravitational force. 3rd law pairs are always the same “kind” of force. The normal force on an object is a contact force, so its 3rd law pair is also a contact force. Specifically it is the contact force on the ground.

Consider what happens when you jump. Throughout a jump the gravitational force is constant. The normal force is much higher at first as you push up, and then goes to zero while you are in the air, and then is very high again as you land. It is not equal and opposite to gravity through most of the jump.

 

[kuruman]

One additional point to consider is that the normal is a contact force which, like static friction, adjusts itself to provide the observed acceleration. When you are crouching on the floor without moving, the floor flexes just enough under your weight to keep you at rest much like the compressed springs of your mattress support your body when it lies on it. When you push down as you unfold your legs and extend your body in preparation for the jump, your change in momentum is an additional downward force that flexes the floor some more. The normal force now is greater than your weight, as it should be, if you are to accelerate and jump up off the floor.

 

[Herman Trivilino]

the normal force from the surface is supposed to be reactionary force from the gravitational force on the surface from the object. is it not?

It is not.

The normal force exerted on an object by a surface and the normal force exerted on that surface by that object are Third Law pairs.

The gravitational force exerted on an object by planet Earth and the gravitational force exerted on planet Earth by that object are Third Law pairs.

 

[Meir Achuz]

Reading this thread, I see why students always get a third law question wrong.

 

[Herman Trivilino]

FWIW the term action-reaction can imply that the action has some kind of different status than the reaction. As if one precedes the other or is the cause of the other. The two forces are, however, perfectly symmetrical, neither having a different status than the other.

That is why I prefer the term Third Law pair, or Third Law partner.

To find the Third Law pairs you simply exchange the objects in your sentence. For example, table exerts force on book, book exerts force on table. Part of understanding this is understanding the terms "on" and "by". There is a force exerted on the book by the table so the Third Law requires that there is a force exerted by the book on the table.

It is surprising to learn the conceptual roadblocks that some students face.

Another example is the if-then clause. As in, do this if that happens. This gets understood as do this and that.

 

[sophiecentaur]

It is surprising to learn the conceptual roadblocks that some students face.

It's not helped by the fact that they learn about Equilibrium Forces at around the same time. The only difference between the two pairs of forces is that N3 applies always whilst the two (or more, of course) forces are equal and opposite only in Equilibrium. Deciding on which one of these applies to a situation can be confusing - and not only if you are thick as boots; 'thinking' too much about it can take you into confusionland too.

 

[A.T.]

The only difference between the two pairs of forces is that N3 applies always whilst the two (or more, of course) forces are equal and opposite only in Equilibrium.

That is not the only difference, and not even the key difference. For applying N2 and N3 correctly, the key difference to keep in mind, is that in N2 all the forces act on the same object, while in N3 they act at two different objects.

Also: "two (or more, of course) forces are equal and opposite" makes no sense. For example, how can three forces be "equal and opposite"?

 

[sophiecentaur]

how can three forces be "equal and opposite"?

In the examples we are dealing with, we assume one dimensional motion. If you want to introduce three dimensions then you can have vectors adding to zero. But why add problems at this stage by jumping through centuries?

 

[A.T.]

we assume one dimensional motion.

Are you using the royal 'we' here? Because I saw no reason to assume this, given your generally worded statment about Newton's Laws.

But even assuming one dimension only, how can three forces be "equal and opposite"? Can you give an example?

 

[Herman Trivilino]

In the examples we are dealing with, we assume one dimensional motion. If you want to introduce three dimensions then you can have vectors adding to zero. But why add problems at this stage by jumping through centuries?

Even in one dimension you can have three forces adding to zero. It wouldn't make sense to say that they are equal and opposite.

 

[sophiecentaur]

Are you using the royal 'we' here?

OK; bad style.

But even assuming one dimension only, how can three forces be "equal and opposite"? Can you give an example?

If the applied three forces act at a point then they could add to zero and that could be seen as two equal resultant forces acting in opposite directions. Really - what are you arguing about here? Could you explain to me why a summative statement about the situation can't be made by reducing multiple forces to opposing forces along some chosen axis? The simple statement of N1 is elegant and doesn't need to be taken further just to make a point. Be honest, when you state N1, do you consider multiple dimensions?

But even assuming one dimension only, how can three forces be "equal and opposite"?

Instead of trying to do this thing to death, why not try to be positive and avoid angels on pinheads. 'We' sometimes have to accept some overlap between reality and the simplest of ideal models. There's no practical one dimensional model - agreed - but we all frequently make the step in Physics. The nearest we can get to the point particle is to have multiple forces acting through the CM.

 

[A.T.]

If the applied three forces act at a point then they could add to zero and that could be seen as two equal resultant forces acting in opposite directions.

Why would you go though these contortions, just to force "equal and opposite" into Newton's 1st and 2nd Laws? What does it achieve, aside of creating confusion with the 3rd Law?

 

[Herman Trivilino]

If the applied three forces act at a point then they could add to zero and that could be seen as two equal resultant forces acting in opposite directions.

Suppose you have the following three forces, all measured in the same units: $7 \hat{i}, -4 \hat{i}, -3 \hat{i}$. They all lie along the same line, sum to zero, and are not equal but opposite.

 

[sophiecentaur]

Other examples are available lol.