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So my textbook says that the reaction forces between two surfaces in contact can be resolved into the normal reaction, which is normal to the surface and friction, which is parallel to the the surface. However, if friction is a reaction force, then I don't understand how frictionless surfaces can exist. On a previous question I asked about a car using friction to move, one of the answers claimed that static friction was a reaction force. Furthermore, this indicates that in frictionless surfaces, the parallel component doesn't exist. This means that the normal force will be equal to the vector sum of the reaction forces and I cannot understand how a vector sum will always yield a force normal to the surface, which means there's a problem here somewhere.

Recently, I needed to solve a problem in which I needed to list all the forces. Upon looking at the solution, I suspected a lot of forces were either mislabeled or left out. The answer to this question would help me confirm/refute my suspicions.

The title is too broad, my apologies. However, you see that my doubts were too many and too interlinked to be put into a single line and warrant a discussion as such.

Edit: Link to the car question where it has been claimed that static friction is a reaction force: Why does frictional force cause a car to move? Also, is friction a reaction force?

Edit: While viewing the answer, also read through associated comments.

lee_dong-eun
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Well all forces exist in pairs, which is the original and which is the reaction is just a matter of naming really. But maybe this example will help a little, when you have a box on an incline, as you increase the inclination the friction force increases to match the component of the weight down the incline; so you can't really calculate friction independently until you reach maximum friction (whether limiting equillibrium or the object is moving). This happens because the irregularities of both surfaces on the molecular level interlocks, thus it is in a sense due to the summation of all the reaction forces between the irregularities.

I think your problem with the reaction force having one component is that you assume that an equillibrium is always possible which isn't true. If theoretically a frictionless surface exist and you apply a force that has a horizontal component, the reaction will be cancel out the force in the vertical direction only, and thus the object will have an acceleration in the horizontal direction.

Peter
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So my textbook says that the reaction forces between two surfaces in contact can be resolved into the normal reaction, which is normal to the surface and friction, which is parallel to the the surface.

This statement is true only if there is friction between the surfaces and if there is a component of an externally applied force that acts parallel to the surface. Otherwise the only reaction force will be the force normal to the surface. You didn't put your book statement in context, but it sounds like it is referring to an object on an inclined plane with friction. That example does not apply to a horizontal surface where there is no external horizontal force applied to the object.

However, if friction is a reaction force, then I don't understand how frictionless surfaces can exist. On a previous question I asked about a car using friction to move, one of the answers claimed that static friction was a reaction force.

Static friction force is a reaction force, but it only exists in opposition to an externally applied force parallel to the surface. In the case of the car the static friction force exists due to the torque applied to the wheel. Without any force applied by the wheel to the surface there would be no static friction force.

If an object rests on a horizontal surface with or without friction, and there is no externally applied horizontal force on the object, the only reaction reaction force is the force normal to the surface in opposition to the weight of the object. There is no friction force. The character of the surface is irrelevant since there is no external horizontal force to be opposed by friction.

Furthermore, this indicates that in frictionless surfaces, the parallel component doesn't exist. This means that the normal force will be equal to the vector sum of the reaction forces and I cannot understand how a vector sum will always yield a force normal to the surface, which means there's a problem here somewhere.

I'm not quite sure of what you are saying when you refer to the "parallel component", but again it sounds like you may be thinking that a parallel friction force will always exist as a component of the reaction force on a surface. As I've shown above, it is not.

Perhaps the free body diagrams below will help. The top diagram shows an object and a frictionless surface. The bottom diagram shows the same but with a surface with friction. The diagrams show action-reaction pairs per Newton's 3rd law. There is no friction reaction force in the top diagram because there is no friction between the surfaces. The only reaction force at the surface is the normal force.

Note that a friction reaction force exists in the bottom diagram in opposition to the applied force $F$ due to friction between the surfaces. Here the static friction force equals the externally applied force so the net force is zero. However, if the externally applied force $F$ were removed from the bottom diagram, the friction reaction force no longer exists to oppose it, even though the surfaces are rough. The net force will still be zero and the only reaction force at the surface will be the normal reaction force.

if a force is at an oblique angle to the surface, there will be a component that is parallel to the surface and normal to it as forces can be resolved into two components. However if it is a frictionless surface, you say it will not exist. Why not? You said in case of a frictionless surface, the only reaction is the normal force. In this case, the vector sum can always be resolved into one component, normal to the surface. I cannot gain intuition regarding this

I have added two more diagrams to show an oblique applied force $F$, and labeled all the diagrams Figures 1-4.

Figure 3 is for a frictionless surface. It shows an oblique force $F$ applied to the mass $M$. The force $F$ is resolved into its x- and y- force components as shown. The y- component of the force adds to the downward weight of the mass and equals the normal reaction at the surface, for a net force of zero and no vertical acceleration.

There is no friction force so there is no reaction force at the surface. The reaction force to the x- component of the applied force is directed at the source of the applied force (source not shown). This satisfies Newton's 3rd law. But since there is no friction opposing the x- component of the force applied to the mass $M$, there is a net force in the x- direction to the left and per Newton's 2nd law and the block $M$ accelerates to the left.

Figure 4 shows the same situation as Figure 3, but now there is a static friction force opposes and equals the x- component of the applied force $F$ as long as $F_x$ does not exceed the maximum static friction force of $μ_{s}Mg$. So there is now a net force of zero in the x direction and the block does not accelerate. Note, however, that there is still the action-reaction pair of forces between $M$ and the source of $F$ per Newton's 3rd law. That doesn't matter if there is friction or not.

I'm afraid I don't have any more time to put into this, so I hope this finally helps.

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Bob D
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  • Yes a force at an oblique angle has a parallel component. But if its a frictionless surface there is no friction reaction. The mass itself has an equal an opposite reaction to the parallel component of the oblique applied force. Just like the mass M has a reaction to the applied force F in both diagrams per Newtons 3rd law. The difference is if there is no friction then there is a net horizontal force to the left due to the parallel component of the applied force and the mass will accelerate to the left per Newtons 2nd law. You need to differentiate between the two laws. – Bob D Nov 03 '19 at 13:05
  • if a force is at an oblique angle to the surface, there will be a component that is parallel to the surface and normal to it as forces can be resolved into two components. However if it is a frictionless surface, you say it will not exist. Why not? You said in case of a frictionless surface, the only reaction is the normal force. In this case, the vector sum can always be resolved into one component, normal to the surface. I cannot gain intuition regarding this – lee_dong-eun Nov 03 '19 at 13:11
  • What mass were you referring to? – lee_dong-eun Nov 03 '19 at 13:18
  • @kimi The only one in the diagrams, $M$ – Bob D Nov 03 '19 at 13:50
  • My doubt was something like this. Bunch on action forces on a surface will warrant a reaction force from the surface, which can be split into normal and tangential components in 2-D. What I realise is that without friction, the tangential component of the action force , though applied, doesn't act on the frictionless surface. With a bit of intuition, I can understand that in case of a surface with friction, there are tiny "bumps" the tangential component of the action force can act upon? So this is not the case in a frictionless surface. – lee_dong-eun Nov 03 '19 at 13:58
  • Thus the only forces that act on the surface are the ones that are normal to it. So there is only a normal reaction. However, this normal reaction is NOT a vector sum of all the forces that are applied to the surface as some, like the tangential component don't act on the surface at all. It is only a vector sum of the component of action forces that are normal to the surface. This is what I wanted cleared up. Do I have it right? – lee_dong-eun Nov 03 '19 at 14:01
  • The reason I asked this is because in a question I came across, the normal reaction of the frictionless force was taken as the vector sum of all the forces applied to the frictionless forces, some of which had components along the surface, which clearly wouldn't have an effect on it or warrant a reaction force. – lee_dong-eun Nov 03 '19 at 14:05
  • I am accepting your answer. However, the comments helped me to come to the conclusion I did. Please tell me whether it is a correct conclusion or not. – lee_dong-eun Nov 03 '19 at 14:06
  • @kimi Thanks for accepting. But I see my latest revision to hopefully end your doubts. – Bob D Nov 03 '19 at 14:08
  • Thank you so much. This was helpful. – lee_dong-eun Nov 03 '19 at 14:09