Clapeyron's Theorem: Explaining the 1/2 and Why It Matters

[Zouatine]

I am new here, so hi for all ^-^
I have a problem in the Clapeyron theorem, the work of a force by definition is equal to the force multiplied by the displacement, and in this theorem it says that the work = (1/2) * force * displacement. why there is 1/2 ?

I also saw Maxwell-Betti's theorem
it's says: we have a corp with two forces F1 and F2 and the displacement d1 of the force F1 and d2 of the force F2 so:
W = W1 + W2 + W '
W = [(1/2) * (F1) * (d1)] + [(1/2) * (F2) * (d2)] + [(F1) * (d ')]
d = displacement of the application point F1 under the effect of the force F2.
why there is no 1/2 in W '?

Thank you 'excuse my English is not good'

 

[mjc123]

Because the force is not constant. Work = ∫Fdx = ∫kxdx = 1/2 kd²
This assumes equilibrium static displacement. If you apply a constant force F, the work done will be Fd. What happens to the extra energy?

 

[Zouatine]

Thanx Sir for your answer .I understand. But what is the physical sens of à force is not constant?

 

[Zouatine]

I mean what's the différence between a constant force and à force is not constante un the displacement of the beam.

 

[mjc123]

Suppose you extend a spring by applying just enough force to extend it a little bit, then a little bit more, and so on, so that it is in equilibrium throughout the process. Strictly speaking this would take infinitely long, but imagine we approximate this by applying a force dF so that it extends by dx, where dF = kdx; then increase the force to 2dF and the extension to 2dx, and so on. At each step we apply a force F = kx and it moves through a distance dx, so the work done is Fdx = kxdx. If you integrate this from 0 to d, the work is 1/2 kd² = 1/2 F_dd.
Now instead suppose you apply a constant force F = kd to the spring (e.g. by attaching a mass to it and letting go.) When it has extended to the equilibrium extension d the work done is Fd. But the energy stored in the spring is 1/2 Fd, just as in the first case. The remainder of the energy is the kinetic energy of the mass, which oscillates about the equilibrium displacement d.
Clapeyron is considering the first case.

 

[Zouatine]

I get it , thanks for your help Sir.