Solving System of Equations (including Trig)

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The discussion revolves around solving a system of equations related to the tension in strings supporting a hanging mass. The problem involves two strings at specific angles and requires applying trigonometric functions to set up equations for the x and y components of the forces. The user initially struggles with the substitution method and expresses confusion over the calculations leading to incorrect factors. Ultimately, the issue is identified as simple mistakes in the setup, and the user successfully resolves the equations after re-evaluating their approach. The conversation highlights the importance of careful algebraic manipulation in solving systems of linear equations.
crastinus
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Homework Statement



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:

3 2
Y
1

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)

Homework Equations



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 into 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?
 
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You have a system of linear equations. The coefficients (where the trigonometric functions appear) are just numbers. There is no sin(FT2) or something like that involved. You can solve it like every other system of two linear equations. There might be some tricks to simplify the result but I wouldn't worry about that before you solved the equations.

crastinus said:
Factoring our FT3 actually gets me FT3 times the factor 2402 N
Huh? There shouldn't be products of forces involved.
 
I tired it a different way and got it solved. Really, I just made some simple mistakes and oversights. Nothing interesting at the end.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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