# Why is there no action-reaction pair for normal force?

jlyu002@ucr.e
Here is an example in which the normal force would not be counted as an action-reaction pair. A 100g block is sitting at rest on a horizontal table. According to Newton's third law, which of the following indicates the correct action-reaction pair of the two forces? So I was stuck between C. The weight of the block and the normal force exerted by the table on the block and D. the weight of the block and the gravitational force exerted by the block on Earth. The correct answer in my text book was D. It's pretty weird. Why is there no action reaction pair for the normal force?

this is the action reaction pair of normal force:the block applies a normal force on the table, which is equal to its weight, in response to which the table exerts a normal force on the block of the same magnitude.

Sourabh N
Hi jlyu002@ucr.e! Welcome to PF!

I find that strange too. When the block is falling, say, there is no weight but only the gravitational force. I don't think one part of an action-reaction pair can stay alone.

EDIT: I misunderstood. Doc Al below has the right idea.

jlyu002@ucr.e
By the way, this is not homework or course work. It is a general question that has been nagging at my mind for months. The example I used is to further supplement my question by guiding the pundit to the right answer.

Mentor
Here is an example in which the normal force would not be counted as an action-reaction pair. A 100g block is sitting at rest on a horizontal table. According to Newton's third law, which of the following indicates the correct action-reaction pair of the two forces? So I was stuck between C. The weight of the block and the normal force exerted by the table on the block and D. the weight of the block and the gravitational force exerted by the block on Earth. The correct answer in my text book was D. It's pretty weird. Why is there no action reaction pair for the normal force?
The normal force certainly is part of an 'action/reaction' pair. It's just that that choice wasn't given.

The action/reaction pairs in this situation are:
-the downward gravitational pull of the earth on the block (its weight) and the equal and opposite upward gravitational pull of the block on the earth
-the upward push of the table on the block (the normal force) and the equal and opposite downward push of the block on the table

The big mistake that most make is thinking that the weight and the normal force are action reaction pairs. They are not. If you remove the table (toss the block in the air), the weight remains but the normal force disappears.

jlyu002@ucr.e
Hey Doc Al,
The first action-reaction pair made plenty of sense to me. The second part, not so much. This is probably due to my lack of comprehension of the subject. Is this "push," something other than the weight of the block?

jlyu002@ucr.e
I think I see the problem. The force that the table exerts on the book does not equal the force of the Earth on the book. The force of the Earth on the book is the weight of the book. The forces in an action-reaction pair must act on different object. Hmmm. Yeah, I still cannot understand the second action reaction pair you addressed Doc Al - the upward push of the table on the block (the normal force) and the equal and opposite downward push of the block on the table.

Mentor
Hey Doc Al,
The first action-reaction pair made plenty of sense to me. The second part, not so much. This is probably due to my lack of comprehension of the subject. Is this "push," something other than the weight of the block?
Remember that 'weight' is the gravitational force exerted by the earth on the box. The normal force, on the other hand, is the contact force (electromagnetic) between two objects that push against each other.

Under the usual zero-acceleration conditions, the upward normal force on the box will equal the downward weight of the box. But that doesn't make them action/reaction pairs. (For one thing, action/reaction pairs act on different bodies!)

Mentor
I think I see the problem. The force that the table exerts on the book does not equal the force of the Earth on the book.
They may well be equal and opposite, but that doesn't make them Newton's 3rd law pairs.
The force of the Earth on the book is the weight of the book.
Right.
The forces in an action-reaction pair must act on different object.
Right!
Hmmm. Yeah, I still cannot understand the second action reaction pair you addressed Doc Al - the upward push of the table on the block (the normal force) and the equal and opposite downward push of the block on the table.
See my last post.

But realize that whenever body A pushes on body B, that body B must push back on body A with an equal and opposite force.

So the box and the table push on each other with equal and opposite forces.

Gold Member
Perhaps it would help to recognize that the "purpose" of action/reaction pairs is not to create a force balance on a given object within a system (as Doc Al said, action/reaction forces don't cancel on the pieces of a system, they only cancel on a whole system that includes all the interacting objects), their purpose is to ensure overall conservation of momentum of the whole system. If the Earth could exert a force of gravity on a book, but the book did not exert the same force on the Earth, then when a book falls, the Earth+book system would not conserve its momentum. Also, when the book hits the table and experiences a normal force, if the book did not also push on the table, then the downward fall of the book would be stopped, while the "upward fall" of the Earth would not be stopped-- so again momentum of the Earth+book system would not be conserved. Maybe that will help.

jlyu002@ucr.e
Sweet! Thanks you guys. I can sleep better now. 8)