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How is the work done by the internal forces acting in a rigid body zero?

Actually I read in a book an example for the same.

Let me present that example here.

Consider a rigid body having two particles $A$ and $B$. Suppose, the particles move in such a way that the line $AB$ translates parallel to itself. The displacement $d\textbf{r}_A$ of the particle $A$ is equal to the displacement $d\textbf{r}_B$ of the particle $B$ in any short interval of time. The net work done by the internal forces ,i.e, the force that $A$ exerts on $B$ and the force that $B$ exerts on $A$, is zero.

How can it be analysed mathematically that the work done is zero?

This example is very unclear to me.

UKH
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ADITI SAINI
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3 Answers3

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Your confusion might be coming from not clearly understanding that you need to define a system, and then all of your quantities are referenced to the system you have defined.

If your system is A and B, then the force that A applies to B is by definition an internal force, and no work is done by that force on the system. But you can choose your system however you would like. Nature doesn't care where you draw your boundary. If you take the system to be A, the the force of B on A is external, and work can be done on the system. Note that it's not the same system as the first case!

What's internal and what's external is a matter of bookkeeping. The setup of the equations will be different, but the final answer will always be the same.

garyp
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I disagree with both other answers.

  • It's not enough that the sum of forces is zero
  • This is not a vocabulary problem

The work-energy theorem applies also to systems of interacting bodies, where the total work of all forces (internal and external) equals the variation of the sum of the kinetic energies of all bodies.

Now to answer the question.

Newton's third law gives you $\mathbf F_{A→B}=-\mathbf F_{B→A}$ (without condition, in any FoR). In your example, $\mathrm d\mathbf r_A=\mathrm d\mathbf r_B$, so you get $δW_{A→B}=\mathbf F_{A→B}·\mathrm d\mathbf r_B=-\mathbf F_{B→A}·\mathrm d\mathbf r_A=-δW_{B→A}$, hence the net work $δW_{A→B}+δW_{B→A}=0$.

L. Levrel
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  • You've shown that the internal work done on the system A and B is zero. You've also shown that if you choose the system as B, the work done on that system is $-W$, and if you choose the system as A the work done on that system is $W$. The choice of system is essential in setting this up. – garyp Apr 29 '16 at 23:18
  • "You've shown that the internal work done on the system A and B is zero." Yes (although I'd say "work done in the system"), and that's precisely the question: "How is the work done by the internal forces acting in a rigid body zero?" For the A←→B forces to be internal, the system must be A+B. (Also, for a rigid body, it would be very weird to consider only part of it as the system.) – L. Levrel Apr 30 '16 at 09:20
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In a rigid body, the particles remain at their positions irrespective of the body's motion. So Newton's third law is applicable here. Here the two particles exert equal and opposite forces on each other (which you call the internal forces). So the resultant force acting along the line joining the two particles is zero. Since the net force is zero, there will be no work.

The same thing you can imagine for a mass at the center of earth. At the center of earth, one may imagine the acceleration due to gravity is so high and so the force of gravity on it is very high. But that's not the case. Due to spherical symmetry, there will be equal and opposite forces acting in all directions with the masses present around it which makes the resultant force zero.

UKH
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