How would I find the net force acting on this ring? Click to see...?

Question

How would I find the net force acting on this ring? Click to see?
Here is the figure of the ring and the forces acting on it.
How would you find the net force acting on the ring?
I need help.

I have previously asked this question. Please show me all the steps you asked in coming to the answer.

Picture url:
http://img415.imageshack.us/my.php?image=scanud4.jpg

Answer 1

You’ve got 3 vectors:
U=[-128, 0] – anti parallel to X-axis
V=[+64, 0] - parallel to X-axis
W=[128*(cos(30), 128*sin(30)] – having both X & Y components.
If you have to find a Sum S of 2 vectors A & B, A being =[Ax, Ay], B being [Bx, By], then do as Sx=Ax+Bx & Sy=Ay+By, that is S=[Ax+Bx, Ay+By].
Now F=U+V+W=[-128+64+128*cos(30), 0+0+128*sin(30)], (cos(30)=sqrt(3)/2, sin(30)=1/2)).
So F=[-64+128*sqrt(3)/2, 128/2]=[128*(sqrt(3)-1)/2, 128/2]=
=64*[sqrt(3)-1, 1].
Now how to draw it in your picture?
Abs(F)=sqrt((F*F))=64*sqrt((sqrt(3)-1)^2+1^2)=
=64*sqrt(3-2*sqrt(3)+1+1)=
=64*sqrt(5-2*sqrt(3))=64*1.5359=98.3N – this is absolute value of force.
Now let’s find its components Fx & Fy, or better say angle of F to X-axis.
As F=64*[sqrt(3)-1, 1], then tan(alpha)=sin/cos=1/(sqrt(3)-1)=1.366
alpha=Atan(1.366)=0.939rad=0.939*180/3.1416=53.8deg
So it shall look like a stick of 98.3N long with angle of 53.8deg to X-axis.
(for Swimming Champs only!)

Answer 2

The simplest way to solve this problem is to draw to scale the forces acting on the ring. But before that you already know by looking at the horizontal force of 128N to the left and 64N to the right, that their net force =128-64=64N to the left. Now get a protractor and a scale and graphing paper and draw the remaining forces. There are only 2 now. One force of 128N at 30 degrees from the x axis and the 64N net force. Draw them to scale. Then draw 2 more lines, one parallel to 128N force and one parallel to 64N force. Now draw the diagonal of the resulting parallelogram starting from the intersection of our 2 forces. Measure that to scale. That's the magnitude of the net force. It's direction is upward. That's also called the resultant of the original 3 forces in your diagram.

Of course, if you want to solve this problem mathematically, it can be done also. If you were sitting right beside me, explaining the solution to you would be rather straightforward. But it's different here at yahoo.