Reaction force of and its relation to normal force and friction

[fog37]

TL;DR Summary

Reaction force, friction, normal force. Friction and normal force seem the components of the reaction force and not forces on their own standing...

Hello,

When we consider a block sitting on a surface, the gravitational force $W$ and the normal force $F_N$ are applied to the block. Both equal i magnitude and opposite in direction. We call the normal force the reaction force exerted by the surface on the block.

Now we consider the block on an inclined plane. The block is not moving. Gravitational force $W$, normal force $F_N$, and static friction $f_s$ act on the block to produce static equilibrium.

Question: both the normal force and static friction are produced by the same entity, i.e. the surface. In general, by Newton's 3rd law, an entity (surface) produces a force on another entity (block) and vice versa, and not multiple forces...That said, when normal force and friction are both present, is it more correct to consider friction and normal force as the parallel and perpendicular components of the single reaction force $F_r$ or as two separate forces both generated by the surface itself? It feels more correct to interpret normal force and friction as components with the reaction force pointing at some non perpendicular angle to the surface.

Thank you!

 

[Dale]

You are not wrong in principle. However, the functional form of the normal and frictional components are different. So it also makes sense to treat them separately, as individual forces.

 

[kuruman]

I agree with you and for other reasons that just in principle. A foolproof way to account for the forces acting on a system in a free body diagram (FBD) is to count all the entities outside the system that can exert a force on it. Then draw arrows representing the forces, as you have done. I would draw the FBD for the block at rest on the incline exactly as you have done, one "arrow" for the force exerted by the Earth and one equal and opposite arrow for the force exerted by the incline. I may have to resolve these vectors into components with respect to some convenient coordinate system in order to answer a question, but that's a whole 'nother issue. The fact remains that, to satisfy Newton, when the block is in equilibrium with only two pieces of the Universe acting on it, the arrows representing the two forces have equal magnitudes and opposite directions because that's the only way to add two vectors and get zero.

That said, it is useful to give names to frequently encountered components of forces in order to signify their direction. I differ from @Dale about treating them separately. We have directed components,
Normal = the component of the surface force that is perpendicular to the surface.
Friction = the component of the surface force that is parallel to the surface.
Centripetal = the component of any force that points towards the center of a circle described by the object.

Clearly, a friction force can also be centripetal as in the case of a car going around in a circle on level ground. Thinking of friction separately could be confusing to beginners who might ask "how can friction be centripetal?"

Finally, I commend you on your FBD drawn to scale.

 

[Dale]

I differ from @Dale about treating them separately.

How can you avoid treating them separately? There is no one equation that can be used to treat them together. Particularly for static friction.

 

[jbriggs444]

How can you avoid treating them separately? There is no one equation that can be used to treat them together. Particularly for static friction.

I do not see any particular difficulty. Vector algebra works. $\sum \vec{F} = m\vec{a}$. We need not split this into two (or three) equations involving real values. Perhaps I am missing your point badly.

For static friction, there is no equality that relates the normal component to the tangential component. One is free to treat the contact force as a vector with no pre-defined direction.

There is, of course, an inequality. If one wishes to incorporate the coefficient of static friction, it can be embodied in a restriction that the total contact force is restricted to angles which lie within a particular cone about the normal, otherwise slipping will be initiated.

For kinetic friction, the angle of the contact force is fixed based on the coefficient of kinetic friction to lie on the boundary of a (possibly different) cone about the normal.

 

[kuruman]

How can you avoid treating them separately? There is no one equation that can be used to treat them together. Particularly for static friction.

I guess I didn't make myself clear. Of course I would treat them separately the same way $x$ and $y$ components are treated separately when adding vectors algebraically. This is my point, We all agree that when we write $$\mathbf{F}=F_x~\mathbf{\hat x}+F_y~\mathbf{\hat y}$$symbols $F_x$ and $F_y$ are not separate forces and we don't think of them as such. Likewise, when we write parallel and perpendicular components $$\mathbf{F}=F_{\parallel}~\mathbf{\hat r}_{\parallel}+F_{\perp}~\mathbf{\hat r}_{\perp},$$ symbols $F_{\parallel}$ and $F_{\perp}$ are not separate forces and we don't think of them as such. Therefore, when we call $F_{\parallel}$ "friction" and $F_{\perp}$ "normal", it would be consistent not to think of them as separate forces. It's a minor point, but I think it clarifies the issue of special names given to components for beginners who think that named forces indicate separate entities generating them.

 

[hutchphd]

Summary: Reaction force, friction, normal force. Friction and normal force seem the components of the reaction force and not forces on their own standing...

That said, when normal force and friction are both present, is it more correct to consider friction and normal force as the parallel and perpendicular components of the single reaction force Fr or as two separate forces both generated by the surface itself?

There is a vector force. We can always represent any vector as the sum of (many) other vectors however we please, so long as we use the rules for vector addition correctly. The question is what is the most useful way to solve the problem at hand. Clearly if you believe that the parallel component is proportional to the perpedicular component, then these would likely be a useful representation of the vector force. It is a utilitarian choice.

 

[Dale]

when we write parallel and perpendicular components $$\mathbf{F}=F_{\parallel}~\mathbf{\hat r}_{\parallel}+F_{\perp}~\mathbf{\hat r}_{\perp},$$ symbols $F_{\parallel}$ and $F_{\perp}$ are not separate forces and we don't think of them as such. Therefore, when we call $F_{\parallel}$ "friction" and $F_{\perp}$ "normal", it would be consistent not to think of them as separate forces.

But the symbols $F_{\parallel}~{\hat r}_{\parallel}$ and $F_{\perp}~{\hat r}_{\perp}$ are each individually a valid force and we can and do think of them as separate whenever it is convenient to do so.

So the question is, when is it convenient to think of components of a force as separate forces in their own right?

As I mentioned before, the key is the functional form of the force law. For example, the Coulomb force law is $$\vec F =k \frac{q_1 q_2}{r^2}\hat r =k \frac{q_1 q_2}{r^2}\hat r_x+k \frac{q_1 q_2}{r^2}\hat r_y+k \frac{q_1 q_2}{r^2}\hat r_z$$ This one vector law gives the force $\vec F$ and the different components all behave the same, so it is convenient to think of this a single force.

This is not the case with the contact force where $$\vec F = F_n ~\hat r_\perp + F_f ~\hat r_\parallel$$ Here there is no vector law for writing the whole force that doesn’t explicitly use this decomposition. $F_n$ doesn’t even have a functional form, it is a constraint force taking whatever value is needed to satisfy the constraint. And $F_f= F_n~\mu$ requires knowing the other component for kinetic friction, and for static friction $F_f$ is also a constraint force taking whatever value is needed to constrain a different degree of freedom. So it is usually convenient to think of them as different forces.

This occurs elsewhere too. For aircraft lift and drag are both parts of the combined force of the air on the wing. But it is useful to treat them separately because the functional form of the lift law is different from the functional form of the drag law. More abstractly, in thermodynamics convection, conduction, and radiation are all just parts of one overall heat transfer, but they are useful to consider separately because they have different functional forms.

So the idea of categorizing physical phenomena according to the math that you use to quantify the interaction is reasonable, convenient, and ubiquitous. It isn’t wrong to mention them as a combined interaction either, but there is a good reason that they are usually treated separately.

I do not see any particular difficulty. Vector algebra works. $\sum \vec{F} = m\vec{a}$. We need not split this into two (or three) equations involving real values. Perhaps I am missing your point badly.

The point is that there is usually no other way to write the overall interaction force other than as a sum of the normal and frictional forces, each of which has a different function describing it. There is no equivalent of Coulomb's law.

 

[fog37]

Thank you everyone for the helpful comments.

What bothered me when I created the thread was:

• A FBD included ALL the forces acting on the body of interest (aka the system). Each force is supposedly due to a specific single agent in the environment. The fact that there would be two forces both due to the surface of the inclined was a point of confusion for me (it seems to clash with Newton's 3rd law because the pair block/surface should have a pair of force but, in this other view, the block exerts a force on the surface and the surface exerts two forces on the block).
• It flows better, for me at least, to view the normal force and the static friction (in my specific example) as the components of a single force, the reaction force of the surface.
• I see Dale's point that normal force and static friction are obtained using very different equations.
• I agree that a force and its components are the same thing. But the components are coordinate system dependent while the FBD is coordinate system independent.
• All physics books would have the normal force and the static friction as separate vector arrows/forces on the FBD though. That seems to be the conventional approach...

 

[Dale]

But the components are coordinate system dependent while the FBD is coordinate system independent.

Note that the components we are talking about here are not coordinate system dependent. Splitting a force into coordinate basis components is coordinate system dependent, but splitting a contact force into normal and frictional components is coordinate system independent. This has to be the case since the force laws have different forms, so if you couldn't express them in a coordinate independent manner then you would get contradictory results in different frames.

The fact that there would be two forces both due to the surface of the inclined was a point of confusion for me

I think the best case is to be flexible and aware so that you can use the one-force view when it is convenient and use the two-force view when that is convenient. You even can switch back and forth within a problem since they are equivalent. Maybe you find it convenient to use the one-force view to keep your FBD less cluttered but a two force view for some computations in the same problem.

 

[gmax137]

I think the best case is to be flexible and aware so that you can use the one-force view when it is convenient and use the two-force view when that is convenient. You even can switch back and forth within a problem since they are equivalent. Maybe you find it convenient to use the one-force view to keep your FBD less cluttered but a two force view for some computations in the same problem.

I think this is the key take-away here.

 

[fog37]

Note that the components we are talking about here are not coordinate system dependent. Splitting a force into coordinate basis components is coordinate system dependent, but splitting a contact force into normal and frictional components is coordinate system independent. This has to be the case since the force laws have different forms, so if you couldn't express them in a coordinate independent manner then you would get contradictory results in different frames.

I think the best case is to be flexible and aware so that you can use the one-force view when it is convenient and use the two-force view when that is convenient. You even can switch back and forth within a problem since they are equivalent. Maybe you find it convenient to use the one-force view to keep your FBD less cluttered but a two force view for some computations in the same problem.

I see. Thanks.

I didn't know about the distinction of coordinate basis components and non-coordinate basis components. To me, components have always been equivalent to scaled unit vectors along the chosen coordinate axes.

Using the local TNB triad as coordinate system, centered on a point along the trajectory, we can split the velocity vector into a component that is tangent and one normal to the trajectory at that point. Is that what we are talking about in the case of non coordinate basis dependent components of a vector?

 

[fog37]

I think I may understand what Dale is saying: a vector can always be expressed as the sum of two other vectors of a certain magnitude and at a certain angle w.r.t. each other. In our case, the vector is the reaction force $R$ and the two composing vectors are the normal force #F_N# and friction $f$. This has nothing to do with projecting any of those three vectors on the axes of a chosen coordinate system which we can choose as we like...

Thanks!

 

[Dale]

we can split the velocity vector into a component that is tangent and one normal to the trajectory at that point. Is that what we are talking about in the case of non coordinate basis dependent components of a vector?

Yes, that is one example. Perhaps your coordinates axes are north, west, and up. If you have a car moving up hill on a north west road then the unit tangent vector would be a valid basis vector even though it doesn’t align with any of the coordinate basis vectors.

 

[Mister T]

It feels more correct to interpret normal force and friction as components with the reaction force pointing at some non perpendicular angle to the surface.

The block exerts a force on the surface and the surface exerts an equal and opposite force on the block, in accordance with Law III. Either one may be called the action and the other the reaction, because they are symmetrical in every way.

Consider the force the surface exerts on the block. It is customary to consider two components of that force. One is parallel to the surface and is usually called the friction force, the other is normal to the surface and is called the normal force. Although it is conventional (and convenient) to refer to the friction force and the normal force as separate forces, they are indeed components of the same force. For lack of a better term we could call it the contact force exerted on the block by the surface.

 

[fog37]

The block exerts a force on the surface and the surface exerts an equal and opposite force on the block, in accordance with Law III. Either one may be called the action and the other the reaction, because they are symmetrical in every way.

Consider the force the surface exerts on the block. It is customary to consider two components of that force. One is parallel to the surface and is usually called the friction force, the other is normal to the surface and is called the normal force. Although it is conventional (and convenient) to refer to the friction force and the normal force as separate forces, they are indeed components of the same force. For lack of a better term we could call it the contact force exerted on the block by the surface.

Agreed. The (my) dilemma was really: do we put two forces on the FBD or a single force, the reaction/contact force? It seems that, for problem solving purposes (this is what all physics textbooks do), the two forces, friction and normal force, are considered separately...

 

[jbriggs444]

Agreed. The (my) dilemma was really: do we put two forces on the FBD or a single force, the reaction/contact force? It seems that, for problem solving purposes (this is what all physics textbooks do), the two forces, friction and normal force, are considered separately...

The free body diagram works either way -- either with the contact force drawn as friction and normal force separately or as a single combined force. Almost universally, the preferred choice is to draw them separately.

If friction is a concern or if slipping is a question then one will need to consider the two components separately eventually and one should draw them as such.

If friction is not a concern and one is sure that no slipping will occur (e.g. a floor lamp beginning to topple over) we may want to model the device as an inverted pendulum and consider that the contact force acts along the axis of the lamp. [Yes, for this to work accurately, the moment of inertia of the lamp + stand about its center of mass would need to be negligible]. In such a case, there would be no need to consider the normal and frictional components separately and we would produce a free body diagram with a single force.

You do whatever makes it most convenient to solve the problem at hand.

 

[Mister T]

The (my) dilemma was really: do we put two forces on the FBD or a single force, the reaction/contact force?

Makes no difference. Same is true of all forces. Take for example a block on an inclined plane. There is a downward weight force $\vec{w}$ that we can draw on our FBD. Alternatively we could instead draw the components $w_x$ and $w_y$.

 

[Dale]

Makes no difference. Same is true of all forces.

Indeed, it makes no difference in terms of the final answer. But it can make a bit of difference how easy the math is.

 

[PeterO]

TL;DR Summary: Reaction force, friction, normal force. Friction and normal force seem the components of the reaction force and not forces on their own standing...

Hello,

When we consider a block sitting on a surface, the gravitational force $W$ and the normal force $F_N$ are applied to the block. Both equal i magnitude and opposite in direction. We call the normal force the reaction force exerted by the surface on the block.

Now we consider the block on an inclined plane. The block is not moving. Gravitational force $W$, normal force $F_N$, and static friction $f_s$ act on the block to produce static equilibrium.

Question: both the normal force and static friction are produced by the same entity, i.e. the surface. In general, by Newton's 3rd law, an entity (surface) produces a force on another entity (block) and vice versa, and not multiple forces...That said, when normal force and friction are both present, is it more correct to consider friction and normal force as the parallel and perpendicular components of the single reaction force $F_r$ or as two separate forces both generated by the surface itself? It feels more correct to interpret normal force and friction as components with the reaction force pointing at some non perpendicular angle to the surface.

View attachment 315142

Thank you!

The way I would consider it would be to imagine if the Friction Force was NOT big enough, so the block slips (accelerates) down the slope.
The Normal Reaction Force is still the same size (and direction of course - perpendicular to the surface)
I would thus always consider the Normal Reaction Force to NOT be a component of any force but being a Reaction Force, is just a Force which is as strong as it needs to be.
Your first example, where the surface is horizontal, you even mentioned two forces equal in magnitude and opposite in direction - seeming to want then to be a Newtons 3rd Law pair.
The sloping surface is the simplest indication that they are NOT a Newtons pair. They are not even opposite in direction.
The other way to show they are not a Newtons pair is it place a 20 cent coin on top of the block.
The mass of the block has not changed, so the force of gravity acting on the block has not changed, but NOW the normal Reaction Force from the surface equals the weight of the block PLUS the weight of the coin. So opposite in direction, but not equal in magnitude.
Peter

 

[nasu]

I see. Thanks.
Using the local TNB triad as coordinate system, centered on a point along the trajectory, we can split the velocity vector into a component that is tangent and one normal to the trajectory at that point.

The velocity vector has no component normal to the trajectory. If there is a component of velocity normal to some curve, that curve is not the trajectory.
The acceleration may have a normal component though.