I need to have this Static mechanics theory simplified

AI Thread Summary
The discussion revolves around simplifying static mechanics concepts for a civil engineering student struggling with physics. The student seeks clarity on calculating the resultant's size, direction, and position, specifically how to determine the position graphically. A forum member explains that the resultant vector can be placed anywhere along its action line, and the position is found by using moment summations to equate two force systems. The correct position of the vector is identified as (1.849, 0) m, with a direction of 14.52 degrees from the positive x-axis. This exchange highlights the importance of clear explanations and graphical representations in understanding static mechanics.
Riazy
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First of all thanks for a very good forum. Secondly I must admit that physics was never
my strong side and is still not. During my high school days I had very much problem with physics simply because I was not really a science person (Had all good grades in other subjects than science).

But since joining a civil engineering program I have started to gain interest in physics and despite the fact that I possesses very little knowledge about physics, and have a hard time learning I do still believe that anyone can learn, just if they want, and have the right resources,information and help.

Please excuse me for lacking knowledge of english terminology, as my native language is not English and I don't study in English either, but I will try to be as clear as possible with the best of my ability

Homework Statement


This is a very basic example static problem and I understand much of it
but I don't understand the main point which is aquiring the position of the resultant.

Question: Calculate the resultant size, direction and location ( or maybe position)

variables: See my scanned paper for details and whole of the solution

Variables


Homework Equations



See the scanned paper


The Attempt at a Solution



The solution is present but I merely want someone to explain what's happening when you require
the position. if someone could explain it graphically it would be great.




Scanned papers


[PLAIN]http://img838.imageshack.us/img838/3092/mech1001.jpg

[PLAIN]http://img191.imageshack.us/img191/8014/mech2001.jpg


I am sorry for this very big thread, but I had no choice, and I have no one to ask, and I live one hour away from the university by bus , i would just like to have this theory Simplified a bit.
 
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Riazy: Vector R can be located anywhere along the action line you drew on page 2. Therefore, in item 1 on page 2, you arbitrarily chose to locate the tail of vector R at the x axis. Therefore, yo = 0 m. Now you must determine the corresponding x-coordinate, xo, on the x axis, for the tail of vector R. (You already drew an excellent graphical explanation of this in your diagram on page 2.) You correctly solved for xo in item 4, by computing moment summations, about the origin, for the two equivalent force systems; and then you set these two moment summations equal, and solved for xo. Nice work. All of your answers are correct, and the correct answer for the location (position) of vector R is (xo, yo) = (1.849, 0) m. The direction of R is alpha = 14.52 deg, measured from the positive x axis.
 
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