# Resultant Couple Moments

by VinnyCee
Tags: couple, moments, resultant
 P: 492 Hi everyone. I have this problem and I don't even know how to start it! If you are familiar with my previous posts, you know that I usually have most of a problem completed before posting here with my difficulties, however, I cannot even start this one! I don't think there is enough information to get an answer! The answer is an i-j-k vector, but how am I supposed to resolve the moments given into it's vectors and forces? Please help!
 PF Gold P: 621 Your first step would be to write the moments as vectors. Notice that the direction of the moment is given by the black arrows in the diagram, and the magnitude is given as 60 lb*ft and 80 lb*ft. After you write them in vector notation I think you can figure out how to find their resultant, or sum.
 P: 492 How do I write the moments as vectors? There are no reference points. All I have are the directions of the vectors, but no lengths or anything! Maybe $$M_1 = -60\,\hat{i}$$
 P: 4,777 Resultant Couple Moments I would advise you to go back and read about vectors in chapter 1 or 2. By the time you are working on couples, your expected to know that stuff. Then post any new thoughts\ideas.
PF Gold
P: 621
 Quote by VinnyCee How do I write the moments as vectors? There are no reference points. All I have are the directions of the vectors, but no lengths or anything! Maybe $$M_1 = -60\,\hat{i}$$
You have the magnitudes (or lengths) of the vectors. They are simply the given torques in lb*ft.

Like Cyrus said, this is basically a vector addition problem.
 P: 492 I can do vectors. I have read those chapters, twice, and did all of the problems. Are you hinting that the vector is to be a unit vector? $-\hat{i}$ is the unit vector for $M_1$. So does this mean that $M_1 = \left(60 lb.\right) \left(-\hat{i}\right)$?
 PF Gold P: 621 Almost. M1 points in the positive X direction.
HW Helper
P: 3,015
 Quote by VinnyCee How do I write the moments as vectors? There are no reference points. All I have are the directions of the vectors, but no lengths or anything! Maybe $$M_1 = -60\,\hat{i}$$
It's actually $+60 \, \hat{i}$.

A hint for the other: its z component is $- 80 sin(30)$, right? (the minus is because of the way it is pointing. For the x and y components, the simplest way to visualize is to first find the projection in the xy plane (which involves a factor of cos(30)) and *then* decompose that projection into the x and y components which will give an *additional* factor of sin(45) and cos(45) (which are of course equal).

Patrick
 P: 492 $$M_1 = \left(60 lb.\right) \left(\hat{i}\right)$$ $$M_2 = \left[\left(-80\,cos 30\,sin 45\right) \hat{i} + \left(-80\,cos 30\,cos 45\right) \hat{j} + \left(-80\,sin 30\right) \hat{k}\right] lb. ft.$$ Those are the vectors for the two moments, right? Then I just algebraically add them together? $$M_3 = \left(11 \hat{i} - 49 \hat{j} - 40 \hat{k}\right)$$ Is that correct?
 Quote by VinnyCee $$M_1 = \left(60 lb.\right) \left(\hat{i}\right)$$ $$M_2 = \left[\left(-80\,cos 30\,sin 45\right) \hat{i} + \left(-80\,cos 30\,cos 45\right) \hat{j} + \left(-80\,sin 30\right) \hat{k}\right] lb. ft.$$ Those are the vectors for the two moments, right? Then I just algebraically add them together? $$M_3 = \left(11 \hat{i} - 49 \hat{j} - 40 \hat{k}\right)$$ Is that correct?