Convert a Cartesian vector to a cylindrical vector

Question

This has been bugging me for the last day or so and I'm pretty sure I'm missing something...

I have a vector with the following magnitudes in cartesian coordinates:

$F_x$ = 1000 Newtons, $F_y$ = 90 Newtons, $F_z$ = 2000 Newtons

I'm trying to convert this to a vector with the same magnitude in cylindrical coordinates. for conversion I used:

$F_r = \sqrt{F_x^2+F_y^2}$

theta (the angle not the circumferential load) = $\arctan(Fy/Fx)$

$F_z = F_z$ as above

We can get the radial and axial components of the force this way, however the issue is how to get the circumferential component of the load? The equation for theta above only gives the angle of the load wrt. the r-axis in cylindrical coordinates.

Any help would be truly appreciated :)

Answer 1

Managed to find the answer after posting this question. For more details check out this link on Youtube:

https://www.youtube.com/watch?v=g9EU7Xb90Vg

Hint: the point at which the force is acting is required, otherwise all vectors are referenced from the origin of the coordinate system.

Answer 2

Given the Cartesian representation of the force,

$$\vec{F} = F_x\vec{i}+F_y\vec{j}+F_z\vec{k}$$

its cylindrical equivalent is

$$\vec{F} = F_0\cos(\theta-\theta_0)\vec{u}_r-F_0\sin(\theta-\theta_0)\vec{u}_\theta+F_z\vec{k}$$

where $F_0=\sqrt{F_x^2+F_y^2}$ and $\theta_0=\arctan\frac{F_y}{F_x}$.

Note that there is a distinction between the coordinate angle $\theta$ and the angle $\theta_0$ of the force $F$. Their difference produces the circumferential load.