# Homework Help: System of equations

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1. Nov 25, 2015

### REVIANNA

1. The problem statement, all variables and given/known data
this is actually one of the physics problems and I have boiled down the numerical to two equations.
But I have trouble manipulating equations

2. Relevant equations

Tsin(theta)=Fcos(theta)-mg
and
Tcos(theta)=(mv^2/Lcos(theta))-Fsin(theta)

F and T are the two unknowns
3. The attempt at a solution

I brought the terms involving m to one side and the trig functions to the other
and tried to add the the equations . But it only gets more complicated as in one eq F has a coefficient of cos(theta)+sin(theta) and in the other cos(theta)-sin(theta).same goes for coefficients of T in both equations.

2. Nov 25, 2015

### tommyxu3

The form is like:
$$T\sin\theta-F\cos\theta=...$$
$$T\cos\theta+F\sin\theta=...$$
Multiplying the both $\sin\theta$ and $\cos\theta$, and the other way around, which may help.

3. Nov 25, 2015

### REVIANNA

I did not understand this

4. Nov 25, 2015

### REVIANNA

F(cos(θ))^2 sin(θ) +Tsin^2(θ) cos(θ)
F sin^2(θ) cos(θ) -T cos^2(θ) sin(θ)

what should I do?

5. Nov 25, 2015

### tommyxu3

$$T\sin\theta-F\cos\theta=...(1)$$
$$T\cos\theta+F\sin\theta=...(2)$$
What I meant are $(1)\cdot\sin\theta+(2)\cdot\cos\theta$ and $(1)\cdot\cos\theta-(2)\cdot\sin\theta.$ Can you get anything from them?

6. Nov 25, 2015

### REVIANNA

It worked !!
How did you think about it?
And THanks

7. Nov 25, 2015

### SammyS

Staff Emeritus
I hope that tommyxu3 did not literally mean what he wrote.

The form he gave was good.
Here's what to do from that point.

Multiply the first equation by $\ \sin(\theta)\$ and the second equation by $\ \cos(\theta) \$, then add the equations to eliminate F . It's essentially the method of elimination. Then solve for T .
...

Last edited: Nov 25, 2015
8. Nov 25, 2015

### REVIANNA

9. Nov 25, 2015

### Ray Vickson

It is easier if you simplify the symbolics: let $s = \sin(\theta), c = \cos(\theta), A = mg, B = \frac{mv^2}{L} \cos(\theta)$. Then your equations read as
$$sT = cF - A\\ cT = -sF + B$$
or
$$\begin{array}{rcl} cF - sT &=& A\\ sF + cT &=& B \end{array}$$
If you know about matrices and matrix inverion you can write down the solution immediately, because in matrix form the system reads as
$$\pmatrix{c & s \\-s & c} \pmatrix{F\\T} = \pmatrix{A\\B}$$
A crucial simplification is that $c^2 + s^2 = 1$, because these constants are the cosine and sine of the same angle.