Are frictional forces action forces or reaction forces?

[lemonxx]

TL;DR

are frictional forces action forces or reaction forces?

if a box is sitting on a flat surface and I apply 10 N on it, then the box would apply 10 N on my hands, but the static friction (assuming maximum static friction is greater than 10 N), would oppose the applied force with 10 N thereby keeping the box at rest. what I do not understand is this: is this static force an action or a reaction force in accordance with newton's 3rd law? if it is action, what is the reaction? if it is reaction, what is the action?
if the applied force exceeds maximum static friction, it begins to move, and the box exerts an action force on the ground in the direction of motion, so the ground exerts a reaction force opposing the motion, which is kinetic friction, is that correct?

 

[anuttarasammyak]

There are three bodies in your story, the hand, the box and the floor.
The hand push the box, the box push the hand. If you choose that the former is action, the latter is reaction. If you choose that the former is reaction, the latter is action.
The box push the floor, the floor push the box. If you choose that the former is action, the latter is reaction. If you choose that the former is reaction, the latter is action. In these namings whether the box is moving or not does not matter.
The hand and the floor has no action-reaction relation directly.

 

[jbriggs444]

There is no distinction between an "action" and a "reaction" force. Neither is cause. Neither is effect. Neither is primary. Neither is secondary. Neither comes first. Neither comes after. They are part and parcel of a single interaction, always existing together.

 

[lemonxx]

this might be a silly question, but if I slide my hand on the floor thereby exerting force on the floor, then by newtons 3rd law, the floor should exert an equal and opposite force back on my hand, which is kinetic friction, but if there is a fixed coefficient of kinetic friction (kinetic friction is constant), how is it equal to the force I apply? i could apply a strong force once and then a weaker force the second time but the kinetic friction on both would be the same, yet newtons 3rd law would tell me that this reaction force is equal to the force I apply on the floor each time

 

[anuttarasammyak]

There are three bodies in your story, the hand, the box and the floor. The box is between the hand and the floor. The box gets force from the hand H and the floor F. The equation of motion is
H+F=ma
where m,a are mass, acceleration of the box.
The hand gets force -H from the box.
The floor gets force -F from the box.

 

[PeroK]

this might be a silly question, but if I slide my hand on the floor thereby exerting force on the floor, then by newtons 3rd law, the floor should exert an equal and opposite force back on my hand, which is kinetic friction, but if there is a fixed coefficient of kinetic friction (kinetic friction is constant), how is it equal to the force I apply? i could apply a strong force once and then a weaker force the second time but the kinetic friction on both would be the same, yet newtons 3rd law would tell me that this reaction force is equal to the force I apply on the floor each time

The force of your hand on the floor is a vector and will have both vertical and horizontal components. You have control over the vertical component but kinetic friction determines the horizontal component. If the floor were frictionless, then you could still push down with whatever force you can, but your hand would slide along the floor without horizontal resistance.

 

[Lnewqban]

this might be a silly question, but if I slide my hand on the floor thereby exerting force on the floor, then by newtons 3rd law, the floor should exert an equal and opposite force back on my hand, which is kinetic friction,

The magnitude of the resulting kinetic friction is only certain percentage of the force exerted by your hand on the floor.
That percentage is what we call kinetic μ.

but if there is a fixed coefficient of kinetic friction (kinetic friction is constant), how is it equal to the force I apply?

The value of kinetic μ is constant, but again, the amount of kinetic friction force is not, and also smaller than the force you apply.

i could apply a strong force once and then a weaker force the second time but the kinetic friction on both would be the same, yet newtons 3rd law would tell me that this reaction force is equal to the force I apply on the floor each time

What you are writing is equivalent to say that, in a motor vehicle, your foot could apply a strong force on a brake pedal once, and then a weaker force the second time. but the braking effect on both would be the same.

Common braking practice contradicts that statement and confirms Newton's 3rd law, as the magnitudes of the reaction forces between brake pads and discs, as well as between contact patches of tires and pavement, are proportional to the magnitude of the force you apply on the brake pedal each time.

 

[lemonxx]

The magnitude of the resulting kinetic friction is only certain percentage of the force exerted by your hand on the floor.
That percentage is what we call kinetic μ.


The value of kinetic μ is constant, but again, the amount of kinetic friction force is not, and also smaller than the force you apply.


What you are writing is equivalent to say that, in a motor vehicle, your foot could apply a strong force on a brake pedal once, and then a weaker force the second time. but the braking effect on both would be the same.

Common braking practice contradicts that statement and confirms Newton's 3rd law, as the magnitudes of the reaction forces between brake pads and discs, as well as between contact patches of tires and pavement, are proportional to the magnitude of the force you apply on the brake pedal each time.

i’m still confused. my hand exerts an applied force on the ground and the ground exerts a force back, but my hand also exerts friction on the ground and the ground exerts friction back on my hand?

 

[A.T.]

i’m still confused. my hand exerts an applied force on the ground and the ground exerts a force back, but my hand also exerts friction on the ground and the ground exerts friction back on my hand?

Yes, as these are only components (normal and tangential) of the total contact force vector. You can decompose any force vector into components, and say it exerts this component and also that component.

 

[Lnewqban]

i’m still confused. my hand exerts an applied force on the ground and the ground exerts a force back, but my hand also exerts friction on the ground and the ground exerts friction back on my hand?

Yes.

Let's assume that your hand exerts an applied vertical force of 10 Newtons on the ground and the ground exerts a vertical force of 10 Newtons back.

Then, let's assume that the coefficient of kinetic friction between your hand and the ground is μ_k=0.80.

Then, your hand also exerts a horizontal friction force of 8 Newtons on the ground and the ground exerts a horizontal friction force of 8 Newtons back on your hand, which resists its horizontal movement on the ground.

 

[hutchphd]

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Then, your hand also exerts a horizontal friction force of 8 Newtons on the ground and the ground exerts a horizontal friction force of 8 Newtons back on your hand, which resists its horizontal movement on the ground.

This statement seems ill considered. The hand touches only the block (not the ground). The ground touches only the block(not the hand) Perhaps you should restate this more carefully?

 

[jbriggs444]

If you put your hand on an icy horizontal surface and press down with 8 Newtons of force, that works fine. The upward force of the ice on your hand will also be 8 Newtons.

If you now try to exert 100 Newtons of horizontal force on the ice, you will fail to do so. Your hand will slip away and only a miniscule force (in the ballpark of 0.4 Newtons) will be seen. You will be exerting 0.4 Newtons of frictional force on the ice. The ice will be exerting 0.4 Newtons of frictional force on your hand.

Your arm may be attempting to impose a greater force, but the result will be a large horizontal acceleration as your hand slips rapidly across the ice.

 

[Lnewqban]

This statement seems ill considered. The hand touches only the block (not the ground). The ground touches only the block (not the hand)
Perhaps you should restate this more carefully?

What block?

Please, see:
Post #4:

this might be a silly question, but if I slide my hand on the floor thereby exerting force on the floor, ...
yet newtons 3rd law would tell me that this reaction force is equal to the force I apply on the floor each time

Post #8:

i’m still confused. my hand exerts an applied force on the ground and the ground exerts a force back, but my hand also exerts friction on the ground and the ground exerts friction back on my hand?

 

[russ_watters]

i could apply a strong force once and then a weaker force the second time but the kinetic friction on both would be the same...

No, it isn't. No, you can't. Driving a car, take the same corner in the sun and in the rain, and you'll see that you(the car) don't get to decide if the friction is the same. The road decides.

 

[hutchphd]

What block?

My apologies, box
Please see #1.


But my objection remains. We need to answer to answer one question at a time, which has been done

There are three bodies in your story, the hand, the box and the floor. The box is between the hand and the floor. The box gets force from the hand H and the floor F. The equation of motion is
H+F=ma
where m,a are mass, acceleration of the box.
The hand gets force -H from the box.
The floor gets force -F from the box.

Confusion will otherwise result.
Define the object.
Consider the forces on the object using a free body diagram.
Solve for the motion of the object.
If you wish to remove the box from the problem, let the OP restart the process.