Definition for Mechanical Work

[kgm2s-2]

We all know that the definition for work is
W = ∫F.ds
Why can't it be
W = ∫s.dF?

 

[meldraft]

Because it doesn't have a physical meaning.

if a body moves over a distance, the work done by the force is defined as the force component in the direction of the motion times the distance traveled:

For a differential distance, the differential work is the dot product of the force vector times the displacement vector:

dW=\vec F\cdot d\vec s

and by integrating over the trajectory of the motion you get the total work of the force.

If the force is not constant at every point, and therefore a function of s, you would need the total differential:

dW=\frac{\partial W}{\partial F}dF+\frac{\partial W}{\partial s}ds

 

[chrisbaird]

Think of an integral as an idealization of adding up the contribution of many small increments: ∫f(x) dx = ΣΔx f(x) = (x₂-x₁)f(x₁) + (x₃-x₂)f(x₂) + ...

So W = ∫s dF (I've made it one-dimensional to keep things simple) would be:
W = (F₂-F₁)s(F₁) + (F₃-F₂)s(F₂) + ...

Do you see the problem? What is s(F₁) supposed to mean physically? The displacement applied at the certain force strength F₁? This doesn't mean anything. Forces causes displacements to do work. Displacements don't cause the forces.

 

[bcrowell]

We all know that the definition for work is
W = ∫F.ds
Why can't it be
W = ∫s.dF?

Well, I don't consider that to be the definition of work. I'd define work as the transfer of mechanical energy. The equation W = ∫F.ds isn't even valid in all cases.

In a purely one-dimensional context, you could actually take W=\int F dx and apply integration of parts to get W=Fx-\int x dF.

Do you see the problem? What is s(F₁) supposed to mean physically? The displacement applied at the certain force strength F₁?

Yes, actually that's what it would mean if you did the integration by parts thing.

This doesn't mean anything. Forces causes displacements to do work. Displacements don't cause the forces.

I don't see how the cause-and-effect relationship is relevant. And in any case, it is often possible to view displacements as causing forces. For example, if I have a mass on a spring and I displace it, that causes a different force to exist.

 

[AlephZero]

Why can't it be
W = ∫s.dF?

That definition would mean that if F was constant, dF would be 0 and W would be 0 for any value of s.

That isn't how "mechanical work" is defined.

 

[256bits]

W = ∫s.dF would be the co-energy of the system.

See
http://en.wikipedia.org/wiki/Coenergy

And that's all I know about it.

 

[bcrowell]

That definition would mean that if F was constant, dF would be 0 and W would be 0 for any value of s.

But W=Fx-\int x dF (derived using integration by parts, evaluated at the beginning and end of the displacement) doesn't have this problem.

That isn't how "mechanical work" is defined.

That doesn't answer the OP's question. The OP wants to know *why* work is defined a certain way.

 

[kgm2s-2]

It seems like it is because ∫s.dF doesn't have a physical meaning so we can't use it to find work done?

So for a force vs displacement graph, ∫F.ds is talking about the area under the curve and the x-axis while ∫s.dF is talking about the area bounded by the curve and the y-axis. Hence ∫s.dF does not have a physical meaning?

 

[atyy]

It seems like it is because ∫s.dF doesn't have a physical meaning so we can't use it to find work done?

So for a force vs displacement graph, ∫F.ds is talking about the area under the curve and the x-axis while ∫s.dF is talking about the area bounded by the curve and the y-axis. Hence ∫s.dF does not have a physical meaning?

Work isn't really an area, except in 1D. The idea is that it's a path integral, where the path can be in 3D space.

The concept of work is useful, because the work done causes a change in kinetic energy.

There are special sorts of forces, called "conservative" forces, in which total energy = kinetic energy plus potential energy is conserved. In general, the work done by a force on a particle depends on the path taken, but if the force is conservative, then the work done by the force is depends only on the start and end points of the path.

Gravity is an example of a conservative force.

 

[Studiot]

∫s.dF is talking about the area bounded by the curve and the y-axis. Hence ∫s.dF does not have a physical meaning?

This is called the 'complementary' energy in Structural Engineering and appears in some theorems.