Does dW/dx = Fx and dW/dy = Fy?

[annamal]

Since dW = F dot product dr
dW/dr = F
So dW/dx = Fx
and dW/dx = Fy?

I am looking at conservative forces where dU/dx = Fx and dU/dy = Fy

 

[Chestermiller]

Since dW = F dot product dr
dW/dr = F
So dW/dx = Fx
and dW/dx = Fy?

I am looking at conservative forces where dU/dx = Fx and dU/dy = Fy

Those should be partial derivatives.

 

[Charles Link]

If $U$ stands for potential energy, there is a minus sign on each of the partial derivatives to give the component of the force.

 

[nasu]

The work is not a function of position. There is no "work at a specific point". I don't think that these derivatives make sense. This is why we introduce potential energy, which is a function of position.

 

[jack action]

dW/dr = F

That cannot be right. There are an infinite amount of vectors $\textbf{F}$ that can give the same amount of work $\partial W$ given the displacement $d\textbf{r}$. It is all a matter of what is the angle between the 2 vectors.

The dot product is more than a simple multiplication, it is an addition of products.

 

[Mister T]

$\vec{F} \cdot \mathrm{d} \vec{r}=F_x \ \mathrm{d}x + F_y\ \mathrm{d}y$

 

[annamal]

That cannot be right. There are an infinite amount of vectors $\textbf{F}$ that can give the same amount of work $\partial W$ given the displacement $d\textbf{r}$. It is all a matter of what is the angle between the 2 vectors.

The dot product is more than a simple multiplication, it is an addition of products.

Angle between F and dr is 0 for dW = F*dr

 

[annamal]

The work is not a function of position. There is no "work at a specific point". I don't think that these derivatives make sense. This is why we introduce potential energy, which is a function of position.

but if dW = F*dl, integrating work we must get a function of position because of dl and the F (force) function may have position variables as well.

 

[vanhees71]

Of course you can have a potential only for "conservative forces", i.e., the line integral between a fixed point $\vec{x}_0$ and the variable point $\vec{x}$ over the force, $\vec{F}(\vec{x})$ must be independent of the path.

In single connected regions of space that's the case if $\vec{\nabla} \times \vec{F}=0$. Then there exists, in this case, at least locally a potential $V$ such that $\vec{F}=-\vec{\nabla} V$, and then the work along an arbitrary path from $\vec{x}_1$ and $\vec{x}_2$ is given by $\Delta W=-[V(\vec{x}_2-V(\vec{x}_1)]$.

 

[nasu]

but if dW = F*dl, integrating work we must get a function of position because of dl and the F (force) function may have position variables as well.

Sorry, this does not make sense. At least, not to me. But what you call "dW" is not a differential in general.

 

[jack action]

Angle between F and dr is 0 for dW = F*dr

If this is your assumption then you are working with one dimension only ... so what are Fx and Fy in your OP?

 

[Mister T]

If this is your assumption then you are working with one dimension only ... so what are Fx and Fy in your OP?

You can have a one-dimensional problem where both $F_x$ and $F_y$ are nonzero.

 

[jack action]

You can have a one-dimensional problem where both $F_x$ and $F_y$ are nonzero.

if $x$ and $y$ represents dimensions, don't you have 2 dimensions?

I recall that the OP stated that both the force and the displacement are parallel in post #7. So if we assume $d\textbf{r}$ is along the $x$ direction, and that $\textbf{F}$ is parallel to $d\textbf{r}$, what could be the magnitude of $\textbf{F}_y$ but zero? What is the meaning of $y$ at all if there is no displacement or force along that axis?

 

[Mister T]

I recall that the OP stated that both the force and the displacement are parallel

Correct. And that direction, as you say, is chosen to be along the x-axis. But that is done out of convenience, in other words we use only one dimension because we have a one-dimensional problem.

If instead we had chosen the x-axis to be some other direction, then we could have a situation where both $F_x$ and $F_y$ are nonzero. But we still have the same one-dimensional problem.

 

[jack action]

Correct. And that direction, as you say, is chosen to be along the x-axis. But that is done out of convenience, in other words we use only one dimension because we have a one-dimensional problem.

If instead we had chosen the x-axis to be some other direction, then we could have a situation where both $F_x$ and $F_y$ are nonzero. But we still have the same one-dimensional problem.

I understand that, but if the resultant of $F_x$ and $F_y$ is guaranteed to be parallel to the displacement, it is completely useless to add a second reference axis (In the problem at hand, that is). You could still add the z-axis to include the third dimension of space and it wouldn't add more information to the problem either, just more confusion.

The dot product is used to ensure that the work is only composed of force components being multiplied by the corresponding displacement component, in the same direction. If the resultant force is known to always be parallel to the resultant displacement, all you need is a simple product (because there is only one dimension to consider), so why the need to split the force into two components?

If the OP wants to state that $\frac{dW}{dr} = F$, then he should replace $\frac{dW}{dx} = F_x$ by $\frac{dW_x}{dx} = F_x$ and $\frac{dW}{dy} = F_y$ by $\frac{dW_y}{dy} = F_y$, Where $dW = dW_x + dW_y$. Which is what you essentially stated in post #6. But I'm not sure what is gained by splitting everything into two components when the resultant product is already easily known.

 

[vanhees71]

Once more: You must not forget that this ONLY holds, if $\vec{F}$ is conservative, i.e., if there is a potential $V$ such that $\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x})$ or, equivalently at least locally, if $\vec{d} x \cdot \vec{F}$ is an exact differential one-form.

 

[Mister T]

I understand that, but if the resultant of Fx and Fy is guaranteed to be parallel to the displacement, it is completely useless to add a second reference axis (In the problem at hand, that is). You

That's why I stated that it's a matter of convenience. Not necessity.

 

[nasu]

The OP premise is faulty. And so are the later assumptions like that the angle is zero in general.
Even for one dimensional case the Fdr can be positive or negative. We have 1D problems with friction, don't we?
But this is irelevant to the OP. You cannot make sense of the relationship F=dW/dx, for example. No matter how many dimensions you use.
If you want the force at x=x₁, for example (F(x₁)), what work are going to consider in order to find this force? There is no such think as work at point x₁. There is potential energy at a point but not work.
@Vanhess71 already showed the right relationships a couple of times.