So I know that for an object on an incline in 2 dimensions, the expressions for the forces of gravity and friction respectively would look like this:
$F_g = mg$ where g is the acceleration due to gravity on earth, $-9.81 m s^{-2}$
The components of that vector on a new coordinate plane using the surface of the incline as the x-axis would be $mg\sin(\theta)$ and $mg\cos(\theta)$ where $\theta$ is the angle of the incline from the horizontal/ground.
$F_f = \mu mg\cos(\theta)$ where $\mu$ is the coefficient of friction between the incline and the object.
But how do I do that in 3 dimensions?
For a project I’m doing, I need to formulate an expression for the net force on an object that lies on a 3-dimensional surface and I’m not entirely sure how to go about that.
I imagine the process used in 2 dimensions could be extended to work something like spherical polar coordinates in 3D since the 2D force components are of the form $x = r\cos(\theta)$ and $y = r\sin(\theta)$.
The 3-dimensional surface I’m working with will not have a constant slope so I’m guessing I’d have to work out $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at that point and then resolve the force vector for gravity using the definitions that allow one to convert from 3D Cartesian to spherical polar:
$x$-component: $mg\sin(\theta)\cos(\phi)$
$y$-component: $mg\sin(\theta)\sin(\phi)$
$z$-component: $mg\sin(\phi)$
Is that technically correct? Are there more efficient or overall better ways of going about it?
Choose your coordinate system to have a plane(say $xy$) defined by the vertical direction and the normal to the plane at the point where you can find your mass $m$ at $t=0$. In the plane your motion is similar to what you had before. In the direction perpendicular to this plane you have no forces (no gravity and no normal force), so the acceleration in this $z$ direction is $0$, or $v_z=v_z(t=0)$