Solving a Difficult System of Equations in Dynamics

In summary, Jack is trying to solve a system of equations but is stuck. One of the equations is inconsistent and he needs the help of someone else to solve it.
  • #1
74baja
3
0
Hi all,

I'm stuck on a system of equations I am left with at the end of a dynamics problem.

[itex]
a=-b+v
[/itex]
[itex]
x=-b+vcos(30)
[/itex]
[itex]
y=vsin(30)
[/itex]
[itex]
171.5=20b^2+5a^2
[/itex]
[itex]
a^2=x^2+y^2
[/itex]
[itex]
10x-40b=0
[/itex]
All in degrees.
I know I have an extra equation, but I thought I'd include it in case it is easier to solve with a certain five of them. I know it's not as simple as solving for one variable in terms of one other, plugging in and solving. I tried looking at it for ways to recombine the equations- no luck.

Thank you,
Jack
 
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  • #2
74baja said:
Hi all,

I'm stuck on a system of equations I am left with at the end of a dynamics problem.

[itex]
a=-b+v
[/itex]
[itex]
x=-b+vcos(30)
[/itex]
[itex]
y=vsin(30)
[/itex]
[itex]
171.5=20b^2+5a^2
[/itex]
[itex]
a^2=x^2+y^2
[/itex]
[itex]
10x-40b=0
[/itex]
All in degrees.
I know I have an extra equation, but I thought I'd include it in case it is easier to solve with a certain five of them. I know it's not as simple as solving for one variable in terms of one other, plugging in and solving. I tried looking at it for ways to recombine the equations- no luck.

Thank you,
Jack

Could you clarify which quantities are variables, and which are constants? Is v a variable, for instance?
 
  • #3
74baja said:
Hi all,

I'm stuck on a system of equations I am left with at the end of a dynamics problem.

[itex]
a=-b+v
[/itex]
[itex]
x=-b+vcos(30)
[/itex]
[itex]
y=vsin(30)
[/itex]
[itex]
171.5=20b^2+5a^2
[/itex]
[itex]
a^2=x^2+y^2
[/itex]
[itex]
10x-40b=0
[/itex]
All in degrees.
I know I have an extra equation, but I thought I'd include it in case it is easier to solve with a certain five of them. I know it's not as simple as solving for one variable in terms of one other, plugging in and solving. I tried looking at it for ways to recombine the equations- no luck.

Thank you,
Jack

You have 6 equations in 5 unknowns a, b, v, x, y, In this case, there is no solution: the equations are inconsistent.

If you leave out the fifth equation (##a^2 = x^2 + y^2##) you can fairly easily solve the remaining equations, just by expressing a,v,x,y in terms of b (using equations 1,2,3,6) and then finding b from equation 4. To see that the original system is inconsistent, just substitute the resulting solution into the missing equation 5.
 
  • #4
I didn't really go into details to calculate the answer, and with 6 equations to find 5 variables it is more than enough and I assume that all equations are correct (i.e. either equation left out can provides same answer), let's use a coordinate system to solve this problem, plot x-y axis and for equation 5, a^2=x^2+y^2, it is a circle, from there you can easily get your answer.

Let's explain more, equation 2, x-b=vcos30, equation 3 y=vsin30, combine can get (x-b)^2+y^2=v^2, another circle, I think this can gives you a clear picture?
 
Last edited:
  • #5
csleong said:
I didn't really go into details to calculate the answer, and with 6 equations to find 5 variables it is more than enough and I assume that all equations are correct (i.e. either equation left out can provides same answer), let's use a coordinate system to solve this problem, plot x-y axis and for equation 5, a^2=x^2+y^2, it is a circle, from there you can easily get your answer.

Let's explain more, equation 2, x-b=vcos30, equation 3 y=vsin30, combine can get (x-b)^2+y^2=v^2, another circle, I think this can gives you a clear picture?

Perhaps you did not read my reply: the equations are inconsistent, so they cannot all be correct.
 
  • #6
Ray Vickson said:
Perhaps you did not read my reply: the equations are inconsistent, so they cannot all be correct.

I think you should really read what I wrote, equation 2 and 3 actually can combine to gives (x+b)^2+y^2=v^2. So there are 5 equations, 5 variables. I don't know where your inconsistent come from, and why they cannot be all correct.

Okay.. yeah previous one I wrote wrongly, it should be x+b instead of x-b.
 
  • #7
a=−b+v (1)
x=−b+vcos30 (2)
y=vsin30 (3)
171.5=20b^2+5a^2 (4)
a^2=x^2+y^2 (5)
10x−40b=0 (6)

From equation 6 => x=4b (7)
From equation 2 => x+b=vcos30 (8)
square(equation 8) + square(equation 3) => (x+b)^2+y^2=v^2 (9)

Sub equation 7 to equation 9 => 25b^2+y^2=v^2 (10)
Sub equation 7 to equation 5 => 16b^2+y^2=a^2 (11)
equation 10 minus equation 11 => 9b^2=v^2-a^2 (12)

From equation 1 => v=a+b (13)
Sub equation 13 to equation 12 => 9b^2=(a+b)^2-a^2
8b^2-2ab=0
b=0 or a=4b

Okay I think Ray is right, my view of the "combining 2 equations" become a circle is a big mistake because of the angle of 30 degree, they are 2 linear equations and a combination of them can't make a circle.

Sorry for the stupid concept.
 
Last edited:

1. What is a difficult system of equations?

A difficult system of equations is a set of two or more equations that cannot be easily solved by traditional methods, such as substitution or elimination. These equations often involve multiple variables and complex expressions.

2. Why are difficult systems of equations important?

Difficult systems of equations are important because they can model real-world problems and situations more accurately than simpler systems. They are also crucial in advanced mathematical concepts and fields, such as physics, engineering, and economics.

3. What are some strategies for solving difficult systems of equations?

Some strategies for solving difficult systems of equations include using substitution, elimination, graphing, or matrices. Other methods include using advanced algebraic techniques, such as Gaussian elimination or Cramer's rule.

4. Can difficult systems of equations have multiple solutions?

Yes, difficult systems of equations can have multiple solutions. Depending on the number of variables and equations in the system, there can be zero, one, or an infinite number of solutions. For example, a system with two equations and two variables can have a unique solution, while a system with three equations and two variables may have no solutions or an infinite number of solutions.

5. How can I check if my solution to a difficult system of equations is correct?

You can check your solution to a difficult system of equations by substituting the values of the variables into each equation and solving for both sides. If the resulting values are equal, then your solution is correct. Another method is to plot the equations on a graph and see if the intersection point matches your solution.

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