# Homework Help: Need help with 3D system with applied forces and moment

1. Jun 3, 2012

### Spottyferret

1. The problem statement, all variables and given/known data
The system is a rectangular box with dimensions of 3 ft x 4 ft x 2 ft. There's an applied couple moment of 600 lb-ft on the x-z plane rotating positively. The moment is 2 ft above the x-y plane in the positive z direction.

Applied force 1 is 450 lbs in the negative x direction with a position vector of (3,4,2). Applied force 2 is 600 lb in the positive y direction with a position vector of (0,4,2). Applied force 3 is 300 lb in the positive z direction with a position vector of (3,4,0).

I'm being asked to find the magnitude of the resultant force and couple moment of a wrench and where the line of action intersects the x-y plane.

2. Relevant equations

M = F × r

3. The attempt at a solution

ƩF = (-450, 600, 300) lb
|F| = (4522+6002+3002)1/2 = 808 lb

directions are based on RHR
ƩMx = 300 lb * 4 ft - 600 lb * 2 ft + 600 lb-ft = 600 lb-ft
ƩMy = -450 lb * 2 ft - 300 lb * 3 ft = -1800 lb-ft
ƩMz = 450lb * 4ft = 1800 lb-ft

I know to get where the line of action intersects the x-y plane I need to cross (x,y,0) with the Fr vector and set that equal to Mr then just solve for x and y but my math is all wrong with it.

Fr×r=Mr
(300y)i - (300x)j + (600x-(-450y))k = 600i - 1800j + 1800k

that gives me 300y = 600 so y = 2, -300x = -1800 so x = 6, but 600 * 6 + 450 * 2 ≠ 1800

I'm guessing I'm screwing up on the moments part but even if I separate the couple moment into two equal and opposite vectors that would cause a rotation there I still get all these numbers. I even tried simplifying all vectors and moments to one point but still I'm getting these numbers.

Not looking for the answer just some help seeing the forest through the trees.