# Mechanical Work

1. Mar 6, 2008

### Mahsud

Hi, I am facing difficulty in trying to prove how the mathematical relation for WORK

W = force X distance

it might sound strange to a few but I was wondering how it was decided upon that force has to be multiplied by distance and not added or something else! Its understandable that to make the units (N-m) or Joules the only logical way is to multiply the two together. If I was the person discovering this formula then, after pushing a vehicle, logically I would think that I applied force and displaced the vehicle and this is work, but what would make me think that I need to multiply the applied force with the displacement of the vehicle to make it equal to something called 'work'….I mean HOW do you come up with that…why not do some other operation on it.

I would really appreciate it if you could explain this to me in very simple words or by plots.

2. Mar 6, 2008

### tiny-tim

Conservation laws

Hi Mahsud! Welcome to PF!

There's several ways of looking at this. Here's one:

Ultimately, we can only use conservation laws.

But force doesn't appear in any conservation law.

Our conservation laws are for momentum and energy.

Force is mass x acceleration, = mass x length /time/time.

We write that ML/TT for short

Momentum is ML/T. Energy (non-relativistic) is MLL/TT.

So we multiply force (ML/TT) by a length (L) to get something we can put in an energy equation (ML/TT)!

For example, we can put work on the left, and make it equal to increase in kinetic energy, or gravitational potential energy, on the right!

3. Mar 6, 2008

### Math Jeans

Another way to think of it is by considering that the units are very useful.

Mainly, the combination of force and displacement creates a useful type of measurement.

When you get into things such as thermodynamics, you will find that there are many different equations for work, and the units all end up being the same type.

4. Mar 6, 2008

### pmb_phy

That is an excellant question. W = Fx is a definition and as such cannot be proved. The real questions are (1) what is the motivation for defining work and (2) why as W = F*x. The work-energy theorem will come in handy at this point. It can be shown that the change in kinetic energy of a particle equals the definite integral of F(x)dx between the start and end points. Note that if we define W = integral F(x)dx then this will be consistent with the idea of "work" as used when one thinks of "working" as in pushing a car or lifting a brick etc. Think of some of the things we can note about our ideas of work. We intuitively think doing "work" if we have to exert a force on something to move it. One might think that work is done if we simply exert a force on something like resting leaning against a wall. When we lean against a wall we don't consider this as doing work since nothing actually gets done. We could just as easily rest a ladder against a wall and we wouldn't think of that as doing "work." But if we need to push something along the floor when friction is acting then we'd agree that we'd actually have to do some "work." We thus want to keep these things in mind when we define "work." The problem of adding force and distance is that we can't add two physical quantities which have different units since it has no physical meaning or interpretation. We would also like there to be a linear relationship between work and distance and work and force. It would be most efficient to define work dW as the product of F(x)dx, i.e. dW = F(x)dx. If we do this then the work-energy theorem will have a simpler result.
I disagree. The conservation of momentum for a single particle is defined as F = 0.
Dimensional analysis will not give you the full answer. One can easily ask why W = 2*force*distance instead of W = force*distance.

Best wishes

Pete

Last edited: Mar 6, 2008
5. Mar 6, 2008

### mikelepore

We would expect to find variables being multiplied if we expect the answer that we are getting out of a formula to be directly proportional to each of those variables.

If you were pushing an object, and the amount of friction that you had to oversome suddenly became six times as much, requiring to apply six times as much force to overcome it, then you would expect that you would have to do six times as much work.

If you were going to push that object a given distance, and then I told you that you had to push it a distance five times as long, then you would have to do five times as much work.

So the work is proportional to the force, and the work is also proportional to the distance. That means those two things have to be multiplied in the formula for work.

A case of being inversely proportional would be different. If some property were *inversely* proportional to some variable, if some property were to go down at the same rate as some variable goes up, then we would find the formula containing a division instead of a multiplication.

6. Mar 7, 2008

### Mahsud

mikelepore what you are saying sounds absolutely logical! W α F, W α S, but then again digesting W=F.S is a little difficult....let me try to explain WHY! lets compare the logic with this! "A car that is accelerating on a road is increasing speed/velocity & covering distance/displacement at the same time, so if i say "Acceleration is directly proportional to Speed/Velocity, and Acceleration is directly proportional to distance/displacement then why not write Acc=Vel.Disp with units (m²/s)... or call it the rate at which Area changes as Area has units m² and m/s*m = m²/s....

What Pete mentioned earlier is something that is logical! ." The problem of adding force and distance is that we can't add two physical quantities which have different units since it has no physical meaning or interpretation...but cant we come up with some thing that is analogous to the area of a rectangle Area=Length*Width (A = L X W)or as logical as like the (3X4) row and column of asteriks '*' shown below the total number of asterisks can be given in the form (1+1+1+1+1+1+1+1+1+1+1+1) or alternatively as {1+1+1+1=4}X{1+1+1=3} = 12 or [(L=4) X (W=3)=12]

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If force and distance/displacement are plotted on the X-Y table and the area under the curve is found out it gives you 'work' just like the velocity time graph gives you displacement S=V X T....so if i start living from here on with this belief that the area under the curve for the asteriks above gives me the total number of asteriks 4X3=12 asteriks...and the area under the curve for a force and displacement graph gives me this thing that is called work or N's-m's = joules and the velocity time graph gives m/sec's X sec's = metres then i wonder what the displacement time graph gives me....metres X sec's and i shall call it mahsud ;) I think i have driven every one mad here i cant seem to put it behind me and live with it which i guess i will have to! Thanx all
P.S...since there is this invisible cos θ because of the dot product of these two vectors then may be now i should let go of the rectangular analogies and start researching in to the dot product of vectors to find logic.....

Last edited: Mar 7, 2008
7. Mar 7, 2008

### Staff: Mentor

It doesn't sound to me like you've been exposed to calculus yet, because what you just described sounds a lot like the point of integral calculus.

8. Mar 7, 2008

### mikelepore

Hi, Mahsud,

That sounds funny to me, where you said acceleration proportional to speed and also acceleration proportional to distance. I wonder if there is some confusion about vocabuary words. Remember that a constant acceleration will have the speed rising linearly with time. If, at times 3,6,9,12, seconds ... the speed is increasing like 5,10,15,20,... meters/second, that's a constant acceleration, not an increasing acceleration. In that example, a graph of acceleration versus time would be a horizontal line; a graph of speed versus time would be a diagonal line; a graph of distance versus time would rise while bending upward like a parabola.

The reason for the dot product is that some of your force might not be able to accomplish anything, depending on the direction. If a train is on a straight track pointed east, and you try to push it northeast, cos θ represents some of your efforts that are able to get the job done, the train can go that way, east, but sin θ represents some of your efforts just trying to push north, pushing sideways against the rails, and those rails just push back at you, achieving nothing.

9. Mar 7, 2008

### TVP45

It's not an arbitrary thing at all. Remember that Huygens had discovered the conservation of mv^2 by the late 1600s and this was reinforced by Leibniz. And, of course, they had the notion of a force causing a change in motion. So the question was this: How does the application of a force relate to a change in this mv^2 quantity? The answer, of course, turned out to be 2Fx. I don't know who actually derived that, but it was a foregone conclusion that someone would once you had mv^2.

Last edited: Mar 7, 2008