# What is the action-reaction pair of static friction?

First of all sorry for this basic question.

Newton's third law states that for every force there is a force, which is equal in magnitude and opposite in direction to those of given force.So by that logic friction should also have a reaction force.

Now my questions are, what is the reaction force for static friction ?

For instance in the picture, what is the direction of reaction force of static friction and on which object will that force act ?

There is no reaction force on this free body diagram. That is because this is the diagram for the object only.

Draw a free body diagram of the Earth, and there you have your reaction force. You will see that the same static friction pulls the opposite way in the Earth.

Free-body force diagrams tell a story about one object (or system) and are thus useful for e.g. Newton's 2nd and 1st laws. But the action/reaction force couples that come from Newton's third law appear between objects, or you could say that they appear on two different objects. Not on the same objects. Therefor they might not both be included within a free-body diagram - if they are both included (when you define your system as including both objects), then they will always exactly cancel out and will thus make no different for Newton's 2nd and 1st laws, and thus we ignore them. You will thus never see an action/reaction force pair both appearing on a free-body diagram.

To explain your specific scenario: The static friction force is trying to prevent sliding from happening. So it pulls leftwards in the box in order to try to prevent it from moving, and it pulls rightwards in the Earth to try to make the Earth follow along with the box, so that there is no sliding.

• Of course i know that there is no reaction force on that diagram because i did not draw those. I just showed a situation for somebody to show me the direction of reaction force of friction. Thanks for the answer. I will edit the question to make it a little clear. $\ddot \smile$
– user115480
Aug 30, 2016 at 19:23

If the object stays stationary and no force acting on it, there is no frictional force on it (static or kinetic). However when the force is applied from zero, until the object just starts moving, the static frictional force is acting on the object and increasing in magnitude. The direction is equal and opposite to the applied force until critical point is reached (where it starts moving) hence that obeys newton's third law.

• On which body it acts ? is what i asked ? :)
– user115480
Aug 30, 2016 at 19:19
• This answer is an example of Newton's second law, not the third law. Aug 30, 2016 at 19:24
• static frictional force is "equal and opposite"; reflects third law Aug 30, 2016 at 19:35
• "equal and opposite" also applies to the second law when the object is at rest. The forces that you (correctly) say are equal and opposite are two different forces that happen to be equal and opposite. They are not the two forces associated with a single interaction. They are not an action/reaction pair. Aug 30, 2016 at 19:38

The red arrow is presumably the force on the block due to the table. It is a friction force.

By Newton's third law, there is a force on the table due to the block. It points to the right. It is a friction force.

You drew your diagram with a bit of an ambiguity. You drew the red arrow right on the interface between the block and the table, so it is not clear which object that force is acting on. To be clearer, you should draw the red arrow unambiguously connected to the block. The reaction force would be drawn as an arrow pointing to the right unambiguously attached to the table.

I am just giving anology for the problem. Frictional force arise due to surface roughness so both surface create frictional forces in action-reaction pair, surface1 for surface2 and vice versa. You can imagine dragging two sandpapers onto each other to figure out direction of frictional forces in pair. Therefore ground is acting on box in left direction and box is acting on ground in right direction.