Solving System of Equations (including Trig)

In summary, the author is trying to solve for the tension in the strings and is getting stuck. He has used trigonometric functions to get a step by step solution, but is not sure what the final result is.
  • #1
crastinus
78
9

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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  • #2
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.
 
  • #3
I tired it a different way and got it solved. Really, I just made some simple mistakes and oversights. Nothing interesting at the end.
 

Related to Solving System of Equations (including Trig)

1. How do I solve a system of equations?

To solve a system of equations, you must first identify the variables in each equation and their corresponding coefficients. Then, use algebraic methods such as substitution or elimination to manipulate the equations and find the values of the variables that satisfy both equations.

2. Can trigonometric functions be included in a system of equations?

Yes, trigonometric functions can be included in a system of equations. In fact, solving systems of equations with trigonometric functions is a common application in physics and engineering problems. Trigonometric functions can be solved using the same algebraic methods as other equations.

3. Is there a specific method for solving systems of equations with trigonometric functions?

There is no specific method for solving systems of equations with trigonometric functions. However, it is helpful to use identities and relationships between trigonometric functions to simplify the equations and make them easier to solve. It is also important to keep track of units when working with trigonometric functions.

4. What are some common mistakes to avoid when solving systems of equations with trigonometric functions?

Some common mistakes to avoid when solving systems of equations with trigonometric functions include forgetting to use the correct units, misapplying trigonometric identities, and making calculation errors. It is important to check your work and make sure that your solutions make sense in the context of the problem.

5. Can a system of equations with trigonometric functions have more than one solution?

Yes, a system of equations with trigonometric functions can have more than one solution. This is because trigonometric functions are periodic, meaning they repeat their values after a certain interval. Therefore, there may be multiple values that satisfy the equations. It is important to check all possible solutions when solving systems of equations with trigonometric functions.

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