Question

Need help calculating for a coupler

Answer 1

Here is the question:
What I have so far is F1, F2, and F3 figured out at being the following:
f1 = -20j
f2 = 24i
f3 = -16k

Yet f4 is giving me some trouble as it incorporates two different angles

Answer 2

i thought f4 was going to be .3(150cos 30) for my i direction and .3(150sin30) for my j direction yet when getting the magnitude i am sopost to get 71.9 (which i don't)

Answer 3

great bro

Answer 4

.3(150cos 30) would be the case if that moment was on the x-axis, but look at at the dimensions it gave you.
From the picture alone, pipe looks to follow the -x axis only, but if I were to plot the coordinates of where that couple-moment is at, its (.2, .2, .3).

Answer 5

So this couple will have components in the i, j, and k directions because the path of the pipe isn't orthogonal to any of the axises.

***
It's been so long since i've done this, i've messed up somewhere in my work, but I think you'll be able to figure it out by having this jog you memory.
****

I decided to keep the force in it's direction, and use cos(30)*.3 = .26, so .26 is the separation between the couple forces.

Then I found the resultant length to the location where the moment occurs:
sqrt( (-x)^2 + y^2 + z^2) = 0.3
sqrt( (-.2)^2 + .2^2 + .1^2) = 0.3

**** This portion is where I think I went wrong ***
(-.2/.3)*150*.26 = -26k
(.2/.3)*150*.26 = 26i
(-.1/.3)*150*.26 = -13j
moment created by force 4 becomes
{ 26i -13j -26k }

Then i find the magnitude of all the couple system to be:
sqrt( i^2 + j^2 + k^2)
sqrt( (24+26)^2 + (-20+-13)^2 + (-16+-26)^2) = 73.1 NM (which you said it should be 71.9NM, so i'm off somewhere.)

I hope this helps!