This is an example worked out in the textbook Matter and Interactions, 4th Edition (pg. 181). The authors assume that solving for two unknowns is no problem, so they don’t show the steps. I’m trying to work it out and am stuck. I’ve used Alpha to get a step by step solution, but it seemed arcane and I’m sure there’s an easier way to do it.
Here goes. A box of mass 350 kg is hanging (no motion) straight down from string 1 that is itself attached two other strings (in a Y junction), string 2 which is attached at a 70 degree angle to x axis (with positive x component) and string 3 which is attached at a 55 degree angle to the x axis (with negative x component).
So, like this:
What is the tension in the strings?
String 1 is easy: 3430 N in the negative y direction.
The textbook gives these two equations (and I see where they come from—-direction cosines and all that):
for d direction:
0= FT2(cos125) + FT3(cos70)
for y direction:
0=FT2(cos35) + FT3(cos20) + (- 3430 N)
I’m sure some kind of trig identity is useful, but I don’t know what it is!
The Attempt at a Solution
Here is my substitution:
FT2=(FT3 cos20 -3430)/cos35
I then substituted that in to the second equation to get this:
0=(FT3cos20 -3430)/cos35 + FT3cos70
Factoring our FT3 actually gets me FT3 times the factor 2402 N, which the text says is the value of FT3.
Where am I going wrong?