Statics: Moments and Couples

 Let's first define a horizontal plane with the right number of x, y, and z and a curved pipe that lay right in the x-y plane, attached by a rope with a known tension and the attachment was on the xz wall. There were a few numbers to that, with a one quarter circular pipe laying in a square with a tension of 80N. We are trying to figure out the moment about the pivot at the origin. 

We're just trying to figure out the moment about the origin point or point o. Remember r x F again with the force of tension. The cross product will always work for us, and we have 2 vectors we need to put in. 

We get the force relative to where a specific plane is. We can tap into anywhere along the light green line. We could go to the midpoint but we eventually need to pick that. Let's do the vector to the end of the pipe instead. We'll notice that this vector is r0. 

This goes 3 meters in the x direction and 3 meters in the y direction, and 0 meters in the X direction. Now we have to determine the force vector, which is slightly harder. The unit vector from A to C is the same direction of the force vector itself. A is (3, 3, 0) and C is (4, 0, -2) and the difference is <1, -3, -2> in correspondance to i, j, and k, respectively. We get the following equations: 

We then get (0.267i - 0.802j - 0.535k). Then we take the cross product

of <3, 3, 0> x <0.267, -0.802, -0.535> and the magnitude saves a little bit of trouble, knowing you can do that because that's how matrices work multiplied by 80N (the magnitude). 

We get the following result:

This is NOT the magnitude of the moment multiplied by the unit vector of the moment. It does, however, give direction to the unit vector, since its magnitude is greater than 1. We just need to hit the line of action of the force somewhere, and we'll end up getting the same result. 

Now, let's deal with a problem with a pipe flange, with a threaded socked to we can screw it to the wall. Handrails along stairways are done this way a lot of the time. A positive moment among the y axis will tend to unscrew. To figure out what the moment of the y axis only, we can dot the moment with the unit vector of that direction, M0 x j, which will give us the moment in the y direction only. It will unscrew positively and if it's negative it's going to tighten. We just put the vector we want in a particular direction and then do the dot product. By the way, this is what a flange looks like:

Let's say a door is hinged on the x axis and up above the x-y plane. Now there's a hinge. The length of the side of the door is 800mm. We want to hold that up, so we'll attach a chain to a point up on the wall that is up 750 mm and over 850mm. The chan then will attach to A, and there's a cable can withstand 500N. Find out the moment about the X ax
is, or the Line OC. We get that by ficuring out the whoe moment (Roa x F) and dotting it with the oc unit vector (Roa x F) o Noc. O to C is in the negative x direction.

Mo = (Ro x F)

Here's the diagram:

We result in the following equation: 

And let's say the chain points from A to B, so we calculate the distance from A to B as a result. 

B is going to be <-850, 0, 750>  and O to A will be <0, 752, 274> so as  as result (752 is 800 cos 20, 274 is 800 sin 20) which results in (500) <-850, -752, 476> / ((850 ^ 2 + 752 ^ 2 + 476 ^ 2) ^ (1/2)) which the magnitude is 1231 mm.  We then just take <0, 752, 274> x <-345, -305, -193> resulting in

<-61566,-94530,259440> which means the moment is -61566 Nm about the X axis.