In work W = F*D, what does D represent?

  • #1
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Does D represent the distance through which the force is MAINTAINED, or the distance that the object moves given an instantaneous force?

Because if Work is integral of F dx, and we assume F is applied only at a single instant and not maintained, and also assume the D represents the distance through which force is maintained, then Work is essentially 0 Joules because the force was maintained only in a small instantaneous distance, right? Like for a single moment you have force, then for the rest of time you have no force.

So based on this, I'm inclined to believe D represents the distance the object moves as a result of instantaneous force.
 

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  • #2
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In my opinion, "D" represents the displacement of the action point on which the force act, rather than the displacement of the "center of the mass". In the case of walking, the work done by the static friction is zero although the displacement of center of mass of man is not zero. Therefore, there is no energy flowing from ground to our body through static friction.

Reference : Sherwood, B. A., & Bernard, W. H. (1984). Work and heat transfer in the presence of sliding friction. Am. J. Phys, 52(11), 1001-1007.
 
  • #3
sophiecentaur
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So based on this, I'm inclined to believe D represents the distance the object moves as a result of instantaneous force.
I can see where you're coming from here but work may also be involved in rotating the object. To find the Work Done On the object you have to use the F (vector) dot idx (also a vector), which will include any motion that is achieved on the object. F may be changing with time or distance, of course.
 
  • #4
jbriggs444
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So based on this, I'm inclined to believe D represents the distance the object moves as a result of instantaneous force.
To answer the question directly: No, D represents the distance the object moves while the force is applied, not the distance it moves after. The work done by an instantaneous, finite force is zero. The work done by a not-quite instantaneous force can be non-zero.

In the case of a non-rigid or rotating body, Philethan is correct that it is the motion of the [material at the] point of application that is relevant rather than the motion of the center of mass. However, it is often helpful to ignore the details of the point of application and consider "center of mass work". Introductory physics texts often pay little attention to the distinction (in my experience).
 
  • #5
Merlin3189
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..., or the distance that the object moves given an instantaneous force?
As an aside, could a stationary object move at all given only an instantaneous force?
Even a moving object would not change its velocity (neither speed nor direction) given only an instantaneous force.

If you observe a change in velocity, work must be done and momentum must change, implying a non-zero impulse and a force applied for a finite time, which since the object moves, means the force is also applied over a finite distance.

Edit: with apologies to Paul Dirac!
 
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  • #6
Dale
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Because if Work is integral of F dx, and we assume F is applied only at a single instant and not maintained, and also assume the D represents the distance through which force is maintained, then Work is essentially 0 Joules because the force was maintained only in a small instantaneous distance, right?
Right. If F is finite, but applied for only an instant then the displacement is 0 and the work is 0. The only way to have an acceleration or displacement for an instantaneous force is for F to be infinite. Then ##W=F\cdot D=\infty \cdot 0## is undefined and you have to use another method to determine the work.
 
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  • #7
sophiecentaur
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'Work Done On', is often not a very useful quantity to know yet it causes more aggro than most of the other quantities used in Mechanics. It doesn't actually tell you how the object being acted on will behave and it doesn't tell you how much Energy the agent of the Force has used. In collisions, you don't know the instantaneous forces applied so it's not a useful concept. There's the usual problem of using a lot of your energy, just keeping an object suspended in front of you and doing no work on it. In fact, Work Done is pretty much an abstraction in many cases. Avoid getting in a stew about it when it lets you down.
Conservation of Momentum is about the only thing you can really rely on - and even that can lead you into trouble when things are in contact with the Earth or a much larger object.
I have to say, the only way to get this topic sussed is to look at each situation and just worry at it, like a dog with a bone, until you come up with a valid way of approaching it, in which you have actual knowledge or control of all the relevant , usable quantities. There isn't a rule of thumb.
 
  • #8
A.T.
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In the case of a non-rigid or rotating body, Philethan is correct that it is the motion of the [material at the] point of application that is relevant rather than the motion of the center of mass. However, it is often helpful to ignore the details of the point of application and consider "center of mass work". Introductory physics texts often pay little attention to the distinction (in my experience).
If you move the reference point (used for displacement) from the point of application, then you also have to consider the angular work done by the torque of the force around that reference point. The net work (linear + angular) should be the same in both cases.
 
  • #9
jbriggs444
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If you move the reference point (used for displacement) from the point of application, then you also have to consider the angular work done by the torque of the force around that reference point. The net work (linear + angular) should be the same in both cases.
I agree that for rotation of a rigid body, work measured by force cross displacement of the point of application can be accounted for equally well as force cross displacement of the center of mass plus torque cross rotation angle. For non-rigid bodies, that accounting is problematic because "rotation angle" is not well defined.
 
  • #10
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Cool, so just to clarify, if I push an object X meters, and I apply a force F throughout that distance X, then I stop pushing (now I apply 0 Force) and the object goes an additional Y meters, the work I did W = FX, not W = F(X+Y), right?
 
  • #11
jbriggs444
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Cool, so just to clarify, if I push an object X meters, and I apply a force F throughout that distance X, then I stop pushing (now I apply 0 Force) and the object goes an additional Y meters, the work I did W = FX, not W = F(X+Y), right?
Right!
 

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