Analysis of a deformable body and Newton's 3rd Law

[Adesh]

I'm posting the images of "Sommerfeld's Lectures on Theoretical Physics, Vol 2, Mechanics of Deformable Bodies"

I apologize for posting the images rather than writing out the whole thing. But I did it to so that nothing should be missed, as writing out the whole thing is subjected to errors.

I'm having problems in the argument which he has made (the blue box in second picture). How can we use Newton's 3rd Law when it is a consequence of deformation of bodies? A.P. French supports this statement by saying

The production of a force of reaction in response to an applied force always involves deformation to some extent.

When I have a spring attached at one end, and if I pull the other end we know that a restoring force will come into play and in equilibrium this restoring force will be equal to the applied force and hence we got Newton's 3rd Law , Action = Reaction ! But in the analysis above we have taken it to be true, why? I'm finding it hard to accept how "positive x-surface" will apply the same force on "negative x-surface" by the means of Newton's 3rd Law. Is there any other argument which can be used to say "force of negative x-surface will be same as it was on positive x-surface" ?

Thank you.

 

[Paul Colby]

Static masses are unaccelerated in some unaccelerated frame. In this frame the net force on them must be 0 otherwise they would accelerate. In a solid like the cube above, the net forces (and moments) on any sub portion of the cube must likewise sum to zero. So the traction forces across any surface must balance.

 

[Adesh]

Static masses are unaccelerated in some unaccelerated frame. In this frame the net force on them must be 0 otherwise they would accelerate. In a solid like the cube above, the net forces (and moments) on any sub portion of the cube must likewise sum to zero. So the traction forces across any surface must balance.

The forces on “positive x-surface” must be equal to zero as it is not moving and same should be true for “negative x-surface” but how to reason that they will experience the same forces. Although net forces on them is zero but how to prove that they are subjected to same forces?

 

[Paul Colby]

The positive x-surface is being pulled pushed or torqued by the negative x-surface in most solids?

 

[Adesh]

The positive x-surface is being pulled pushed or torqued by the negative x-surface in most solids?

Yes.

 

[Paul Colby]

Yes.

Okay, if the solid is stationary, then these forces, that supplied by +x must be equal and opposite to that supplied by -x. That doesn't imply that the forces supplied by +x are zero?

 

[Adesh]

Okay, if the solid is stationary, then these forces, that supplied by +x must be equal and opposite to that supplied by -x. That doesn't imply that the forces supplied by +x are zero?

I’m sorry but the wordings are quite confusing to me.

When the $-x$ pulled, pushed or torques the $+x$ then there must be a restoring force keeping $+x$ from moving. And when $+x$ applies force on $-x$ then the restoring forces on $-x$ must come into play for keeping $-x$ from moving. Am I right?

 

[Paul Colby]

Am I right?

Yes. When I sit on a chair my bottom pushes downward on the chair. The chair kindly reacts by pushing up on my bottom. Both the chair and I are somewhat deformed by these forces.

 

[Adesh]

Yes. When I sit on a chair my bottom pushes downward on the chair. The chair kindly reacts by pushing up on my bottom. Both the chair and I are somewhat deformed by these forces.

Yes. Then, how we should apply this concept in the problem above? What keeps $+x$ from moving? Is the restoring force which is caused by the left part of $+x$ (I mean the cube which the $+x$ is a face of) or due to the motion of $-x$ in opposite direction and hence causing $+x$ to remain as it was.

I mean if I take two blocks and push them towards each other then, of course, they will not move. Is this happening in the book? Or the spring system (the restoring force which causes the equilibrium) is keeping those parts from moving?

 

[Chestermiller]

In the figure, if you regard the surface between the right and left sides of the cut as a body of zero mass, then the forces on the two sides of the surface must always be equal, irrespective of whether the fluid or solid is deforming (by Newton’s 2nd law).

 

[Adesh]

In the figure, if you regard the surface between the right and left sides of the cut as a body of zero mass, then the forces on the two sides of the surface must always be equal, irrespective of whether the fluid or solid is deforming (by Newton’s 2nd law).

Will they going to be zero? $$ F = ma \\ F = 0 \cdot a \\ F=0$$

 

[Chestermiller]

Will they going to be zero?

F=maF=0⋅aF=0​

F=maF=0⋅aF=0

They have to be equal and opposite so that they sum to zero. The net force on the surface only has to be zero.

 

[Adesh]

They have to be equal and opposite so that they sum to zero.

Didn’t get you.

 

[Paul Colby]

The analysis we are discussing is called drawing a "free body diagram." It's a very common one and extremely useful. Basically one divides the system into a portion (free body) of the original problem and replaces the effect of the rest of the problem with exterior forces acting on the free body.

 

[Chestermiller]

Didn’t get you.

What part didn’t you get?

 

[Adesh]

What part didn’t you get?

How by Newton’s second law the forces on both sides need to be zero?

 

[Paul Colby]

What part didn’t you get?

The part where the forces on my bottom vanish, apparently.

 

[Adesh]

The analysis we are discussing is called drawing a "free body diagram." It's a very common one and extremely useful. Basically one divides the system into a portion (free body) of the original problem and replaces the effect of the rest of the problem with exterior forces acting on the free body.

Okay! So, we have drawn a free body diagram.

 

[Paul Colby]

Okay! So, we have drawn a free body diagram.

So we have a cube acted upon by exterior (non-zero) forces. The cube reacts to these with internal stresses which must exactly counter all exterior forces on a point-wise basis. Proof. Select a point x on the boundary of the cube. Excise a small portion of the body containing x and call this a free body. It is unaccelerated so the traction forces at x due to the small body must be equal and opposite those of the exterior applied forces. QED.

 

[Adesh]

So we have a cube acted upon by exterior (non-zero) forces. The cube reacts to these with internal stresses which must exactly counter all exterior forces on a point-wise basis. Proof. Select a point x on the boundary of the cube. Excise a small portion of the body containing x and call this a free body. It is unaccelerated so the traction forces at x due to the small body must be equal and opposite those of the exterior applied forces. QED.

Okay! Finally got you, means $x$ is not moving at all, and in our original figure we have pictured $x$ as two pictures and that’s why we (means I) were getting this conundrum, ha? But if $x$ is stationary then how does elongation going to occur?

P.S. :- First I though that when you wrote “traction”, I thought you missed “a” and “t” but then I saw in dictionary and found it means pulling of limbs.

 

[Lnewqban]

... When I have a spring attached at one end, and if I pull the other end we know that a restoring force will come into play and in equilibrium this restoring force will be equal to the applied force and hence we got Newton's 3rd Law , Action = Reaction ! But in the analysis above we have taken it to be true, why? I'm finding it hard to accept how "positive x-surface" will apply the same force on "negative x-surface" by the means of Newton's 3rd Law. Is there any other argument which can be used to say "force of negative x-surface will be same as it was on positive x-surface" ?

Could you see the the similarity with the statements of the Newton's law if you analyze the thing at molecular level?
On both sides of the imaginary plane X there are molecules that are very determined to keep a fixed distance with the neighbor molecules.

If an external force compresses both sides of the plane respect to each other, repulsion forces among molecules will fight the decreasing distances imposed by that force action-reaction induced by compression external loads.
If an external force pulls both sides of the plane respect to each other, attraction forces among molecules will fight the increasing distances imposed by that force: again, action-reaction induced by tension external loads.

Similar situation develops if external forces twist both sides of the plane respect to each other.
If the external force tends to bend the body, some pairs of facing molecules will develop repulsion forces and some pairs will develop attraction forces, according to their position in the plane X respect to an imaginary neutral line.

 

[Paul Colby]

But if xxx is stationary then how does elongation going to occur?

Simple in the linear limit, solve

$\ddot{u}_j = \nabla_i T_{ij} = 0$

and use the constitutive relations of the solid to extract the deformation.

 

[Paul Colby]

First I though that when you wrote “traction”, I thought you missed

Traction force is a technical term used in continuum mechanics. You should look it up.

 

[Adesh]

Simple in the linear limit, solve

$\ddot{u}_j = \nabla_i T_{ij} = 0$

and use the constitutive relations of the solid to extract the deformation.

I haven’t reached that far in my course, could please explain what does that mean?

 

[Paul Colby]

I haven’t reached that far in my course, could please explain what does that mean?

No, because I struggle with it myself. I find continuum mechanics is a complicated subject. My suggestion is drop the number of dimensions and consider a wire like a guitar string. Deformations of a wire obey a constitutive relation most call Hook's law.

 

[Adesh]

No, because I struggle with it myself. I find continuum mechanics is a complicated subject. My suggestion is drop the number of dimensions and consider a wire like a guitar string. Deformations of a wire obey a constitutive relation most call Hook's law.

Okay, but if we take a point $x$ on the string, then that point should be equilibrium otherwise the system will move. But for elongation $x$ must be displaced, isn’t it?

 

[Adesh]

Aha! I think it’s after the displacement that equilibrium will be achieved.

 

[Paul Colby]

Aha! I think it’s after the displacement that equilibrium will be achieved.

Bingo. Very same happens in 3d but with tensor fields and shear and a bunch of stuff that make the basics significantly less transparent. BTW tensors in continuum mechanics are quite interesting. Good luck with your studies.

 

[Adesh]

Bingo. Very same happens in 3d but with tensor fields and shear and a bunch of stuff that make the basics significantly less transparent. BTW tensors in continuum mechanics are quite interesting. Good luck with your studies.

I enjoyed talking (I mean learning) from you. Thanks :-)

 

[Chestermiller]

How by Newton’s second law the forces on both sides need to be zero?

Who said they need to be zero?

 

[Chestermiller]

The part where the forces on my bottom vanish, apparently.

The vertical surface doesn’t have a bottom?

 

[Paul Colby]

The vertical surface doesn’t have a bottom?

Who said I was slouching?