Need help with 3D system with applied forces and moment

[Spottyferret]

Homework Statement

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.

Homework Equations

M = F × r

The Attempt at a Solution

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

directions are based on RHR
ƩM_x = 300 lb * 4 ft - 600 lb * 2 ft + 600 lb-ft = 600 lb-ft
ƩM_y = -450 lb * 2 ft - 300 lb * 3 ft = -1800 lb-ft
ƩM_z = 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 F_r vector and set that equal to M_r then just solve for x and y but my math is all wrong with it.

F_r×r=M_r
(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.

 

[KataKoniK]

Hi there,

First of all, great job on setting up the problem and using the right equations. Your mistake seems to be in your understanding of how the line of action intersects the x-y plane.

When we are looking for the line of action, we are trying to find the point where all the forces can be replaced by a single force, without changing the overall effect on the system. In this case, we don't need to use cross products or moments.

To find the line of action, we can simply take the average of the position vectors of all the forces. In this case, that would be:

(3+0+3)/3 = 2 for the x-coordinate
(4+4+4)/3 = 4 for the y-coordinate

So the line of action intersects the x-y plane at (2,4,0).

To find the magnitude of the resultant force, we can use Pythagorean theorem:

|F| = √(450^2 + 600^2 + 300^2) = 750 lb

For the couple moment, we can use the formula you mentioned:

M = F x r

But we need to be careful with the direction of the moment. In this case, since all the forces are acting on the same plane (x-y), the moment will be perpendicular to that plane, which is in the z-direction.

So we can calculate the moment as:

M = (2 ft) x (600 lb) = 1200 lb-ft

I hope this helps clarify your understanding of the problem. Good luck!