• B
Sundown444
I have a question about action-reaction forces, and it mostly has to do with contact action and reaction forces, but it can include action-at-a-distance forces, too. So, can there be multiple action-reaction forces on the same two objects, like a chair and the floor, for example? By multiple action-reaction forces, I mean those of normal force, applied force, friction force, drag, etc.

Mentor
I have a question about action-reaction forces, and it mostly has to do with contact action and reaction forces, but it can include action-at-a-distance forces, too. So, can there be multiple action-reaction forces on the same two objects, like a chair and the floor, for example? By multiple action-reaction forces, I mean those of normal force, applied force, friction force, drag, etc.
Yes, every force acting on an object will separately have its own 3rd law counterpart on a different object. So if a chair has the normal force from the floor, its weight from gravity, and a contact force with someone's butt then there will be three third law forces. One will act on the floor, one will act on the earth, and one will act on the butt.

• vanhees71 and Sundown444
Sundown444
Yes, every force acting on an object will separately have its own 3rd law counterpart on a different object. So if a chair has the normal force from the floor, its weight from gravity, and a contact force with someone's butt then there will be three third law forces. One will act on the floor, one will act on the earth, and one will act on the butt.

I see. I probably misspoke here, but what I meant would there be more action-reaction forces acting between just the chair and the floor? Like friction, normal force and applied force all acting as separate action-reaction forces between the chair and the floor? This does not include a person sitting on the chair, by the way.

Mentor
Like friction, normal force and applied force all acting as separate action-reaction forces between the chair and the floor?
Sure, you could have that. Suppose that the chair is magnetic and the floor is metal. Suppose further that the chair is being pushed from the side. Then you would have three forces from the floor acting on the chair: the magnetic force, the normal force, and the friction force*. Each would have its own 3rd law counterpart acting on the floor.

*In principle you could say that the normal force and the friction force are just different components of the same contact interaction so they together really form one force. However, you still have two forces with the combined contact force and the magnetic force.

• Sundown444
Homework Helper
Gold Member
I see. I probably misspoke here, but what I meant would there be more action-reaction forces acting between just the chair and the floor? Like friction, normal force and applied force all acting as separate action-reaction forces between the chair and the floor? This does not include a person sitting on the chair, by the way.
Normally, one would add vectorially all the forces exerted by one entity on another and call that the action force. The reaction force exerted by the other on the first would be equal in magnitude and opposite in direction to the first force. Now this doesn't mean that the single force exerted by the entity cannot be broken down into components. For example, if the chair is dragged across the floor, there is a single force exerted by the floor on the chair. That force is usually resolved into two components: a component perpendicular to the floor which given the name "normal force" and a component parallel to the floor which is called "force of kinetic friction". The point here is, one entity - one contact force with another entity. Obviously, you can resolve that one contact force into as many components as you like and give them separate names, but this does not make them separate forces. If nobody is sitting on the chair and the chair is not dragged across the floor, then the contact force exerted by the floor on the chair is pependicular to the floor, i.e. there is no parallel component.

• Dale
Homework Helper
Normally, one would add vectorially all the forces exerted by one entity on another and call that the action force. The reaction force exerted by the other on the first would be equal in magnitude and opposite in direction to the first force.
I would resist the urge to call one sum the "action force" and the other sum the "reaction force". There is nothing special about an "action force" that distinguishes it from a "reaction force". Neither is cause. Neither is effect. They simply coexist. Forces come in pairs.

I would also resist the urge to lump the force from the floor onto the three legs of a three legged stool into one force and then combine that with any magnetic force that might also exist. It might be useful to summarize those forces. Or it might not. Choosing to label the sum as the "action force" would be irrelevant either way.

• sophiecentaur
Homework Helper
Gold Member
I would resist the urge to call one sum the "action force" and the other sum the "reaction force". There is nothing special about an "action force" that distinguishes it from a "reaction force". Neither is cause. Neither is effect. They simply coexist. Forces come in pairs.
Right. However, I think that what distinguishes the "action" from the "reaction" forces is the choice of system in a FBD. The action forces are the ones that are drawn in it, act on the system and are added to give the net force that causes the acceleration of the system. By contrast, their reaction counterparts are out of sight but not out of mind. The inherent symmetry that you mention is broken when one chooses a system.
I would also resist the urge to lump the force from the floor onto the three legs of a three legged stool into one force and then combine that with any magnetic force that might also exist. It might be useful to summarize those forces. Or it might not. Choosing to label the sum as the "action force" would be irrelevant either way.
Of course. I oversimplified and I should have said "one contact force per leg". I also did not mention that the point of application of these forces is relevant because there might be torques involved.

• jbriggs444