Recent content by Tomer

Ok. I think I understand the difference, although I still find it very confusing. But I need to let it sink. Thanks a lot for all the different explanations!

Hmm. Ok. So a plate has two surfaces. The gravitational force cannot be said to apply on one (or both) of the surfaces, since a "surface" has no mass (though the plate is made out of two surfaces which together have a mass!). The key to my mistake is probably the fact that gravitation is not...

Doc Al -- Why can a point mass (0D) have mass and a surface (2D) not? Why should the plate be accelerated by gravity if it has no mass? I don't understand what you mean. We often describe mass distributions of 2D objects in physics. Chestermiller: funny, I must have written such equations for...

Ok, so what you're saying is that the pressure on a surface is the force applied by another surface on it, or something of the sort, which excludes the gravitational force from being able to exert pressure (unless it indirectly causes a normal force). I can accept that, I guess. It's just that...

I think Dmobb's right, and I wish we'd stick to the definition. Exactly like Dmobb said, the free-falling 2D plate has, as far as I know, a force F applied on it, and yet there's no pressure here. ress_watters, you say the force is not applied on the "surface" in this case but on every...

What is it applied to, if the plate is being accelerated by it? :)

:) I understand of course where you're going at and I agree that the pressure would be intuitively zero in these free-falling scenarios. What I still don't understand is, how does this agree with the definition " Pressure is the amount of force acting per unit area" (Wiki. Not that Wiki must be...

The scenario is the same one as described before: a 2D plate is free-falling due to a gravitational force. A force F is thus exerted on the plate. And yet there's no pressure. You then mentioned that pressure is only defined for "contact" forces (something I'm not seeing written in Wiki)...

I get stepped on by elephants, buried under dirt... :) Intuitively, of course I understand. From a definition point-of-view, I have a plate, a force F is exerted on it due to gravitation, and thus the pressure should be F/A, only it's not. Perhaps the reason for the vanishing pressure is the...

Thanks! Ok. Let's remember this statement. So the two sides of an ideal 2D plates are considered to be 2 different surfaces? If this is the case, it definitely helps me understand why the two forces from both ends don't add up/cancel the pressures. Why? Why is the pressure a layer...

Thanks so much for the fast reply! Well, in the first scenario I actually did mean a "floating plate" in this sense. I did not want to write it, but I wanted to imagine the situation of a force acting on only one side of the plate, as opposed to the second scenario. What happens if the...

Hello everybody, this is slightly embarrassing (for a physics student), but I realized in the last couple of days that I am somewhat confused with the concept of "pressure". Pressure is defined in "school-level" as "force per area". So if I have a plate of 2m^2 on which a downwards force...

Thanks a lot, both of you, for the time you give me. Usually the "full" rate of change would be written as: \frac{d}{dx'}(Something), Like gab mentioned in his last equation. The weird thing is that we do the same thing with the partial derivative! You know the feeling that you lose...

Hey again gab, sorry, but apparently there's still something I'm missing. I'm not sure how to get: \nabla' \cdot \vec{J}(\vec{r'},t_{r}) = -\frac{\partial \rho(\vec{r'},t_{r})}{\partial t_{r}} -\frac{1}{c}\dot{\vec{J}}(\vec{r'}, t_{r}) \cdot \nabla'(R) This is what Griffiths suggests we...

Of course! Gotcha! Thanks a lot for the help!