Solving Tension on Strings in Diagram

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SUMMARY

The discussion focuses on calculating the tension in cables AB and AE when a force of 600N is applied through chord AC. The angles involved are critical, with angle AEF at 37° and angle BAC as a right angle. The solution approach involves resolving the forces into horizontal and vertical components, applying the equilibrium condition that the sum of forces in both directions equals zero. The key takeaway is the importance of breaking down applied forces and tension forces into their respective components for accurate calculations.

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Homework Statement




Considering the following diagram:

[PLAIN]http://img545.imageshack.us/img545/6301/physx1.jpg

AC is a force being exerted on chords AB and AE of magnitude 600N. Chord AE is attached to to surface FD. Chord B is attached to a second surface of which no information is given.

Angle BAC is a right angle.

Angle AEF is 37°.

AC is parallel to FD.

Find the tension of AC onto the system. (Ie: On each cable.)


Homework Equations





The Attempt at a Solution



If we elongate AF and AC such that:

[PLAIN]http://img232.imageshack.us/img232/2485/physx2.png

We obtain that angle FEA is equal to angle HAG. (37°)

As such, HAG's complementary angle HAE must be 53°.

If we accept HA to be equal to -600N (Opposite but equal to AC) and EA and BA to be its two components then we should be able to resolve it by stating that:

Horizontal Force = Resultant Vector * cos(Angle)
Vertical Force = Reulstant Vector * sin(Angle)

However, I am confused as how to go on from here? Which is the horizontal and which is the vertical vector?
 
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I'm not sure of your angles, but i assume that AE makes a 37 degree angle with the horizontal.
It is best to solve these problems by breaking up all applied forces and tension forces into their vertical and horizontal components, then sum forces in the horiz and vert direction = 0 to solve for the unknown tensions. This is preferred rather than breaking up the applied force into non perpendicular components.
 

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