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I know that $\text{Force} \times \text{Distance = Work}$.
But, what would be the physical meaning of $\text Force \times \text Area?$

Is such a quantity used in physics?

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An equation by itself is meaningless. You need context to be able to assign physical meaning. – ja72 Oct 23 '12 at 13:16
up vote 3 down vote accepted

No, not really. The reason why force x distance is a useful quantity is because work is typically defined on a moving particle: force x distance is really a special case of $W = \int F \,dx$. In other words, the particle has to have moved for there to be work.

Because movement is only one dimensional (distance) and not 2D (area), there is no clear interpretation of what force*area is. It is obvious what I mean when I say "a particle has moved 3 cm," but it is nonsensical to say that "a particle has moved 4 square miles."

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You have not address the question of $\int \vec{F}\cdot \vec{n} {\rm d}A $. A summed force through an area may have usefulness somewhere, I just don't know where. – ja72 Oct 23 '12 at 13:19

Well lets start with something similar that would be meaningful to the general physics population: Pressure = Force / area. As you may know this is a common measurement. It is easily calculated, and is a concept that many in the physics world will understand and recognize. The concept of Work is out of date. If you move something in the vertical direction and then let it go in a gravitational field, it will fall until it hits a surface that doesn't allow it to fall any farther, such as the ground or surface of the earth. Then when the "Work" is calculated, the answer becomes zero, because the distance that the object have moved is a net of zero. The "work" function assumes a net result of a change of distance for the object. Energy expenditure is different in my opinion, as it took energy to move the object to a height, K.E., then object had P.E. while at the height, then when released, the P.E. was turned into K.E., via gravity, and then the K.E. was converted into an Impulse when it hit the ground. Work = 0. Energy = depending on when and where the object was at the time.

Now lets look at your question: Force * Area =?

Force = mass * acceleration ; acceleration = distance /(time * time) area = distance * distance. Putting the three concepts together gets you : (mass * distance * distance * distance) / (time * time).

Now we can combine the units into useful concepts that a common physics person might understand. Mass is the concept of a unit-vector with ‘punch’. Mass is a tangible abstract concept. Mass is what gives, matter and energy, reality. Mass is a fundamental property of matter, energy, momentum, inertia, and gravity. Mass is not “inertia”. Mass has been given a scalar physical unit of measure, in SI units, which is the amount of matter in a Platinum / Iridium cylinder with the label of “kilogram”, and is the ‘K’ in MKS. Mass is a concept that has misleading terminology. For the purpose of weights and measures this was needed to standardize measurement so that trade could be conducted without major confrontations as were going on before it was standardized. Once accepted, the world around, it became the standard and a kilogram in Europe was the same in Australia, South America, and in the USA.

The concept of “mass”, as I use it here, is not a scalar quantity, nor a static-past quantity, instead it is considered as the Causative Vector, as it causes the future to be the present, and holds onto the past. Mass is an ‘at Cause’ unit-vector that forms physical, structure producing, relationships between the various primary concepts of dimension by extension and time. Mass is that property that gives solidness to matter. Mass is what gives, and takes, the effort behind the action of change. Mass is the ‘Woof’ Thread of reality. Mass is what causes stuff to be.

Then we have (distance*distance*distance) which can easily be seen as the concept of space or the units of the concept of volume. Given that we also have 1/(time*time) one can use various math transformational properties:

(distance/time) * (distance*time) = velocity^2 and if you remember the speed of light 'c' is a velocity, then you can say 'c^2'. So now we have mc^2 * distance or Energy * distance. This formula of units may have a concept attached to it already with some kind of name for that concept, but I think that it could be a "Practical Work Function" as energy expended in any direction, even in two directions opposite each other, resulting in a net displacement of zero, still has a positive non-zero value.

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</randomgarbage> – Rody Oldenhuis Oct 23 '12 at 19:09

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