Question

Basic Statics Problem. Will upload a picture in next post. I can not determine how to describe, and why, the 150 lb*ft Moment force in the diagram for Problem 100-101.

Thanks in advance!

Answer 1

Why would M4 (The 150 ft*lb force) be 150[cos(45)sin(45)] in the "i" direction and why would it be 150[-cos(45)cos(45)] in the "j" direction? I dont even think I know why the "k" direction is a negitive sin(45)

Answer 2

However I was able to identify the other moment forces, even though I was unsure of why I had to normalize M3 and not M4

Answer 3

Ok, I hope I can describe this well. This is the sort of thing I do better by pointing my fingers in the relevant direction as I speak.

Answer 4

You did a great job last time!! Thanks man!

Answer 5

The 150 ft.lb moment, M4 is directed along a line in relation to the x,y,z axes that is positive in the i direction and negative in the other two. It is angled to the side at 45º in the xy plane and up at a 45º

Answer 6

If I were to break this up into components, all vector components would be pointing in, so I am not sure why it is positive in the x direction at all. Also, If I take cos(45) that will give me the line lying on the Y axis correct? Multiplying this by sin(45) does what exactly?

Answer 7

In order to get the components of M4, it must be projected into the xy plane (cos45), then along the x-axis itself (sin45). It takes some visualizing to see the angles in relation to the axes. I had to stare at it for a while to see what was going on.

In the j-direction, the same cos45 projection into the xy plane is used and then it is the cos45 that directs it along the y-axis.

Similar things are done for the z-axis.

Answer 8

if it helps, try drawing the individual right triangles for each projection to see whether sine or cosine is used.

Answer 9

Well, that makes sence.. I took your advice in determining the "i" direction. Projection onto the plane first. Then the direction in question. Very nice.

Now for the "j" direction, I am still unsure why it is negative.. It is still the same projection, cos(45) then along the y direction, cos(45), where does the negitive come from?

Answer 10

Just in relation to how the positive direction was defined.

Answer 11

A negative sign just means the direction is opposite of whatever you're calling the positive direction.

Answer 12

Okay, so you are saying if I am going in the direction of increasing value (defined by the coordinate system) it will be positive, however, if I am moving in the direction of a decreasing value (for the component vector at least) I will use a negative value.

You picture is how I visualized the components I was just unsure of how to interpret them. I believe I see this now. And the "z" direction ("k" direction) is negative because its component is pointing down.

Thank you very much!! This problem is MUCH clearer now. And to drive the concept home I will do one more problem before getting ready for work.

Life saver one again!

Answer 13

Well, one last question...

Answer 14

Why would we normalize M3 and not M4? because it contains units?

Answer 15

I don't really know what you mean by that.

Answer 16

For the Moment of M3 I had to normalize it, or take the unit vector.

\[120<2,-2,1>/\sqrt{2^2+(-2)^2+1^2} = [80,-80,40]\]

Answer 17

This would be the magnitude fo the force multiplied by the vector position all over the magnitude of the position (which is the square root of the sum of the squares)

Answer 18

Oh, you mean to find its components because angles weren't given?

Answer 19

I guess thats what it is.. I was unsure of the situations where I would need to normalize the vector. I guess this is onlyj when angles are not given or known.

Answer 20

Yeah, you could first use trig to find the angles, then decompose using the angles, but normalizing cuts a step.

Answer 21

Ah, I see. Well I'm glad I was able to bother you. You're always a help.

Answer 22

I try. Good luck with all that.
Gotta go now.