# Statics in 3D, there must a be faster way to do this.

1. Mar 20, 2016

### goonking

Given this, is there a faster way to compute this or must I do 3 separate 3x3 determinants?

I can probably use cross product of each term, for example , (2rk) X (Axi) = (-2rAx j) (not forgetting the negative sign for j)

next, (2rk) X (Ayj) = (2rAyi)

and so on.... but that feels too slow as well and can get very messy.

Last edited: Mar 20, 2016
2. Mar 20, 2016

### PhanthomJay

Break up each of the forces into their x y and z components and sum them separately equal to 0. Then for the sum of moments equals 0 for each force component about an axis, it's

$M_x = F_yz + F_zy$

$M_y = F_xz + F_zx$

$M_z = F_xy + F_yx$

where x y and z are the perpendicular distances from the line of action of the component force to the axis about which moments are being summed.
I don't know if that's easier for you, but it avoids determinants and vector math equations .

3. Apr 21, 2016

### DanielSauza

Divide your 3D problems into 2D problems. Just remember, the forces and moments in a plane must be in equilibrium.