Solving System of Equations (including Trig)

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SUMMARY

The discussion focuses on solving a system of equations involving tension in strings supporting a 350 kg mass, as presented in the textbook "Matter and Interactions, 4th Edition." The equations derived from the problem include the use of direction cosines, leading to two key equations for the x and y directions. The tension in string 1 is calculated as 3430 N, while the tension in string 3 is found to be 2402 N after proper substitution and simplification. The participant emphasizes that the problem can be treated as a standard system of linear equations without the need for complex trigonometric identities.

PREREQUISITES
  • Understanding of linear equations and systems of equations
  • Familiarity with trigonometric functions and direction cosines
  • Basic knowledge of static equilibrium in physics
  • Experience with algebraic manipulation and substitution methods
NEXT STEPS
  • Study the method of solving systems of linear equations using substitution
  • Learn about direction cosines and their applications in physics problems
  • Explore static equilibrium concepts in mechanics
  • Practice solving similar problems involving tension in multiple strings
USEFUL FOR

Students studying physics, particularly those tackling problems involving static equilibrium and systems of equations, as well as educators looking for examples to illustrate these concepts.

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.
 

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