Is Newton's 3rd Law of Motion always in equilibrium on inclined planes?

[Bashyboy]

Okay, I am currently reading about motion on inclined planes and how the normal force acts directly perpendicular to the surface of the plane and how the gravitational force, obviously, acts downwards. Since the two forces are not inflicted in opposing directions, we split the gravitational force into two vectors: the first vector acts perpendicularly against the surface and the second vector acts in parallel with the surface. So now where I am a sort of confliction is when it on as follows..." The perpendicular component of the force of gravity is directed opposite the normal force and as such balances the normal force." Do they mean there is an equilibrium, that the forces cancel each other out in accordance with Newton's Law, "For every action there is an equal and opposite reaction." And so the normal force is accounted?

Also would this be a good way to define friction, "The force of gravity acts directly downward on an object on a plane and the reason why two vectors are formed is because the force of gravity is pushing it into the plane, but since there is like charged atoms in the plane and the object it does not actually push into it, which is the reason why it slides and friction is formed because it digs into the surface."

I know it might not possibly be coherent, per se; but, by chance, do you see where I am going? Or is it just completely wrong conjecture?

Thank you.

 

[Doc Al]

So now where I am a sort of confliction is when it on as follows..." The perpendicular component of the force of gravity is directed opposite the normal force and as such balances the normal force." Do they mean there is an equilibrium, that the forces cancel each other out in accordance with Newton's Law, "For every action there is an equal and opposite reaction."

Yes, there is 'equilibrium' perpendicular to the plane, but that's due to Newton's 2nd law, not the 3rd. Action/reaction pairs never 'cancel', since they act on different bodies. The two forces on the body--gravity and the normal force--are not 'action/reaction' pairs.

 

[gsal]

Yes, for every action there is a reaction and so, the component of the weight normal to the plane pushes against the plane and the plane produces a reaction of equal magnitude...in this sense the object remains in equilibrium (in that direction) in the sense that it does not penetrate into the place...if the plane weren't able to produce a reaction of the same magnitude, then the object sinks.

The component of the weight parallel to the plane tends to move the object down the plane...if there was no friction, the object would simply slides down without rotating, even it was a ball. With friction, you have two parallel forces acting on the object that attempt to make the object rotate, depending on the shape of the object and the value of friction, it may slide down or may roll down.

my 2 cents

 

[Doc Al]

Yes, for every action there is a reaction and so, the component of the weight normal to the plane pushes against the plane and the plane produces a reaction of equal magnitude

The weight acts on the object, not on the plane. The object and plane exert equal and opposite forces on each other, but neither of those forces is the weight of the object. (Although they are equal in magnitude to the normal component of the weight.) The weight and the normal force are not Newton's third law pairs.

...in this sense the object remains in equilibrium (in that direction) in the sense that it does not penetrate into the place...if the plane weren't able to produce a reaction of the same magnitude, then the object sinks.

Yes, the object is in equilibrium (normal to the plane) due to the combined effect of the weight and the normal force. This is Newton's 2nd law in action.

 

[gsal]

Doc:

With all due respect, allow me to argue back...

When I say " for every action there is a reaction.." and I am referring to the component of the weight normal to the plane and the reaction from the plane ...I think I am correct.

Observe the choice of words...weight is already a force, which is a result of mass under the influence of acceleration...acceleration is not a force, weight is ...and I did not use the word mass, either.

and for as long as the normal from the plane is simply going to match the component of the weight normal to the plane...well, there you have action and reaction...

Also, you say that weight acts on the object, I say acceleration acts upon the mass of the object and makes it weigh...If I am the plane, I see (feel, really) that an object is pushing on me (may not even care how it manages to do that) and if I want to stay put, I need to respond with equal force.

 

[Doc Al]

When I say " for every action there is a reaction.." and I am referring to the component of the weight normal to the plane and the reaction from the plane ...I think I am correct.

When you say "for every action there is a reaction.." I assume you are talking about Newton's 3rd law. The object's weight is the gravitational force of the Earth pulling down on the object; the 'reaction' to that force--in the Newton's 3rd law sense--is the gravitational force of the object pulling up on the earth.

The plane exerts a contact force--the normal force--on the object; at the same time, the object exerts an equal and opposite contact force on the plane.

There is a bit of a semantics issue here, since in common parlance if there's a rock pressing down on you we often say the weight of the rock is pushing you down. That's fine in general terms, but not in a physics class where 'weight' is usually (but not always) taken as the gravitational force. Note that if you and the rock were falling, there would no longer be a force pushing you down (at least from the rock) but the rock is still being pulled down by gravity.

 

[gsal]

I see.

When I say " for every reaction..." , of course you assume correct I am referring to Newton's 3rd Law, but you assumed wrong thinking I was talking about the gravitational attraction between the object and the earth...

But now that you brought it up...I had never thought of the attraction between two masses as being action and reaction; for me, this event happens when something ' acts ' and something else ' re-acts ' to that...like me leaning on the wall and the wall exerting just as much force back, but the initial force of me on the wall is independent of the wall...

...when it comes to two objects attracting each other, the attraction force depends on the mass of both of them...I hardly think of one of them as reacting to the other one...in fact I don't think of one of them as reacting at all...they are both acting for the same purpose ...to attract each other...sure, a free body diagram may show the two forces in opposite direction, but I still fail to see it as a action-reaction pair.

...then again, I am just a practical engineer and tend to see thing that best way it serves me.

If you could offer another explanation that helps me see gravitational attract as a action-reaction, please do.

thanks.

 

[Doc Al]

Part of the problem is with the archaic 'action/reaction' terminology, which implies that first one object acts on another, then, some time later, the other object 'reacts' back on the first. That is not how the Newtonian model works. Nowadays we talk in terms of 'third law pairs', not 'action/reaction'.

The two forces involved in Newton's 3rd law are best thought of as two aspects of a single interaction. You cannot press against the wall without the wall simultaneously pressing back on you with equal force. It's not first one, then the other.

Take gravity as an example: Gravity is an interaction between two masses, in this case the object and the earth. They exert equal and opposite forces on each other, which are 'third law pairs'. The Earth pulls the object down and the object pulls the Earth up.