How Is Tension Direction Calculated in a Non-Symmetrically Loaded Fence Wire?

[ny_aish]

During a storm, a tree limb breaks off and comes to rest across a barbed wire fence at a point that is not in the middle between two fence posts. The limb exerts a downward force of 164 N on the wire. The left section of the wire makes an angle of 12° relative to the horizontal and sustains a tension of 447 N. Find the magnitude and direction of the tension that the right section of the wire sustains.
I got the tension in the rope which is 440N, I am having trouble finding the Direction.

° (counterclockwise from x-axis) ?

I tried 1/cos of 164/440...its wrong




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[lightgrav]

how did you get 440 N?
you _should_ keep vertical Forces and horizontal Forces separate .

 

[ny_aish]

i will look thru my notes and see how i came to it.

 

[ny_aish]

this is how i got the 440N: we break the tensions into x- & y- components

So x- direction...-447N*cos(12) + T2*cos(θ2) = 0...So T2*cos(θ2) = 437.2 [eqn 1]
and in the y- direction 447*sin(12) + T2*sin(θ2) - 146 = 0..
so T2*sin(θ2) = 146 - 447*sin(12) = 53.1 [eqn 2]

Now divide eqn 2 by eqn 1 giving sin(θ2)/cos(θ2) = 53.1/437.1

or tan(θ2) = 0.1214...so θ2 = arctan(0.1214) = 6.92o

Now plug back into eqn 1 to get T2...T2 = 437.1/cos(6.92) = 440N