Determine T1 & T2 in Concurrent Forces Diagram

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The discussion focuses on determining the unknown forces T1 and T2 in a bridge truss diagram, specifically addressing the vector sum of the 72 kN and 80 kN forces. Participants suggest calculating the horizontal and vertical components of T1 and T2 by forming a triangle with the vectors and using trigonometric relationships. The equations derived from the equilibrium of forces are crucial: for horizontal forces, the equation is -T1*sin(30) + T2*cos(30) - 80cos(45) = 0. The vertical forces can be similarly expressed, allowing for substitution to solve for T1 and T2.

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Ry122
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How do i determine t1 and t2 in the following diagram?

http://users.on.net/~rohanlal/concforces.jpg
 
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They can be anything because you don't have a question in regards to that picture.
 
Thats basically all it says.
The figure shows a join from a bridge truss. Calculate the unknown forces t1 and t2.
 
calculate the vector sum of the 72 and 80 kN forces. If i and j are unit vectors in the direction to the right and up these are 72j and -80cos(45)i -+80sin(45)j.
now what are the components of the unknown forces t1 and t2? (express them as functions
of t1 and t2). Both the i and the j components of all these forces must add.
 
So after findings the vertical and horizontal forces put the horizontal force and t1 and t2 angles into a triangle and solve for each side, the result will give me the horizontal components for both t1 and t2, then do the same for the vertical force to get the vertical components. After finding the horizontal and vertical components of t1 and t2 put them into a right triangle to determine their resultant forces. Is this correct?
 
I don't think you can get the horizontal components from t1 or t2 using just the horizontal force required. you can form a triangle with the vectors t1, t2 and the required force, and you know all angles and one side of this triangle, so you can compute the other two sides.

What I would do however is just solve the two equations that you get if you use the fact that both the total horizontal and vertical force must be 0.

for horizontal you get -t1*sin(30) + t2*cos(30) - 80cos(45) = 0
for vertical you get ....

you can get the first equation in the form t1 = (expression with only t2)
and substitute that in the second.
 

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