Stu

use a d. It's asked in the question. Alternatively express "Let x equal expenses". Good luck. 

I did define collinear however the diagram mentioning it had two vectors from the same point of origin. Hence my question. Is that defining them as collinear or it would have been something else? 

With regards to resultant force, I asked a lecturer today. When working a problem I have been resolving the " hypotenues of the triangle for example to get the resulting force of Fxi and Fyj etc. But not adding them and just giving the answer. Further example is 5sin45 = ... is the Fr (and times D is torque). Is this correct? The lecturer said the answer was correct, but it sounds like i'm only solving for force, not resulting force. Even today working with a parellelogram and multiple forces this was all I did. Or am I missing the definition of two different things here? 

That is right. 

Thank you. Do the signs on c and d need to be changed when finding the determinant? What was the 1/|A| I was explained to do? Is there another way to find I? I was shown something multiply by first row, and add second row or something like that. Is that what I saw on Kahn Acadamy, reduced row echlon or something to solve for I? Thanks for the post.

Correction to previous post.. 

After the sine, cosine, and tangent is Theata, denoted by the symbol seen for angles, circle with little line for it. As well, Soh, Cah, Toa, is a pnumoic for the relationships only. The correct math equation is now below. Sine for example is multiplied by the angle, thus can also be devided by on both sites, as can the denominator be multiplied on the RHS to get it on the LHS.

Then it is 

Sine( Theata )= Opposite/Hypothenus

Cos (Theata )= Adjcsent/Hypothenus

Tan ( Theata )= Opposite/Adjacent

To make x on its own by doing what you do on one side to both sides of the equals sign. In this case to remove the division of 5 you would multiply and cancle it on the left hand side, then take that to the right hand side as multiplication. There for you have x = 5 * (-6)


To correct the last post it wouldn't be x and x = it would be 1 * x, which to make x on it's own you would still have to take the 1 to the other side as a division. Just imagine it is -6/4 instead and c4ross multiply would give you not x=.

Is F_R in the below image "resultant force"? 
Can some one please verbally interprut the image with more classification.

Image source Victoria University, Collaborate, Engineering Fundementals, 24/07/14

How is the bellow question in the image worked out? 

Image source Victoria University, Collaborate, Engineering Fundementals, 24/07/14

What do you mean?

Thanks. Now I shouldnt get lost.