B Can the normal force be in vertical direction?

AI Thread Summary
The discussion centers on whether the normal force can act in a vertical direction. One participant mistakenly believes that gravity acts horizontally, which leads to confusion about the nature of normal force. Another participant clarifies that normal force is independent of gravity's direction. The conversation emphasizes the need for a proper understanding of normal force and its definition. Ultimately, the normal force can indeed act vertically, depending on the context of the forces involved.
ahmed emad
Messages
43
Reaction score
2
HI,
Can the normal force be in vertical direction ?
 
Physics news on Phys.org
ahmed emad said:
HI,
Can the normal force be in vertical direction ?
What do you think and why?
 
I think it's not because gravity act in horizontal direction ,but I posted to confirm my understand
 
ahmed emad said:
I think it's not because gravity act in horizontal direction ,but I posted to confirm my understand
It sounds like you have horizontal and vertical mixed up...

https://gawallbeds.files.wordpress.com/2013/07/h-and-v.jpg
h-and-v.jpg
 
ahmed emad said:
I think it's not because gravity act in horizontal direction ,but I posted to confirm my understand
Normal force does not necessarily have anything to do with gravity. Do you understand what normal force IS ?
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top