# Isolate for Angle With Variables

1. Oct 16, 2013

1. The problem statement, all variables and given/known data

Isolate for the angle. Do not sub in numbers, isolate the angle, θ . Use of trig identities required.

2. Relevant equations

$m_{2}g=m_{1}gsinθ-μm_{1}gcosθ$

We are given the trigonometric identities:

3. The attempt at a solution

I have attempted everything from squaring both sides, to no avail, as I see no way to use any trig identities so isolate for the angle. It is the μ that makes it quite difficult to isolate for θ.

$m_{2}g=m_{1}gsinθ-μm_{1}gcosθ$
$m_{2}g=m_{1}g(sinθ-μcosθ)$
$\frac{m_{2}g}{m_{1}g}=(sinθ-μcosθ)$
$\frac{m_{2}}{m_{1}}=sinθ-μcosθ$
At this point I am completely stuck... Any help would be appreciated.

Thanks.

Last edited: Oct 16, 2013
2. Oct 16, 2013

### nasu

Are you trying to solve for the angle in therm of the other variables?

3. Oct 16, 2013

Yes, that is what I am trying to do.

4. Oct 16, 2013

### nasu

You could move one of the terms from right side to left and then square both sides.
Then use the last of the formulas in your image to get a quadratic equation in either sin or cos.

5. Oct 16, 2013

Alright. This is what I have tried... I don't know if it's entirely correct but it still seems really clustered.

I don't know what to do next though...

6. Oct 16, 2013

### Staff: Mentor

Try this:
$(sinθ-μcosθ)=\sqrt{μ^2+1}(\frac{1}{\sqrt{μ^2+1}}sinθ-\frac{μ}{\sqrt{μ^2+1}}cosθ)=\sqrt{μ^2+1}(sinθcos\phi-cosθsin\phi)=\sqrt{μ^2+1}\,\,sin(θ-\phi)$
where $cot\phi=μ$

7. Oct 16, 2013

### ehild

You got a quadratic equation for cosθ. Solve with the quadratic formula.

ehild

8. Oct 16, 2013

Hi, just attempted with quadratic formula, but I can't get anywhere (as in I can't simplify it any further it looks so hideous).

9. Oct 16, 2013

### Staff: Mentor

Since no one seemed to have any interest in what I did in response #6, I will continue by completing the solution:

$$θ=\arcsin\left({\frac{(m_2/m_1)}{\sqrt{μ^2+1}}}\right)+\arctan(1/μ)$$

10. Oct 16, 2013

### nasu

This is a clever method. You can even take μ=tan(β) where β is the "friction angle", which has some physical interpretation.

Edit. Actually the ∅ defined as above it is equal to the friction angle:
tan(∅)=μ.

Are you sure that in your result, the last term is not arctan(μ) rather than of 1/μ?

Last edited: Oct 17, 2013
11. Oct 17, 2013

### ehild

You can simplify it a bit, by expanding the parentheses under the square root. And don't worry if it is not simple.

ehild

12. Oct 17, 2013

### ehild

It should be arctan(μ).
Yes, it is a clever method if somebody knows it.

ehild

13. Oct 17, 2013

### Staff: Mentor

Oh. You're right. It should be arctan(μ). Sorry for the error.
Incidentally, this "clever method" has appeared in many threads on physics forums. It is often used to get the phase angle in periodic solutions to problems.

14. Oct 17, 2013

### ehild

The OP did not know it, so some explanation would have been necessary.

ehild

15. Oct 17, 2013

### nasu

I did not say it's new. We used in high school and that was long time ago.
But this does not make it less clever, right? Or at least more elegant that the quadratic equation.

However, as ehild pointed already a couple of times, the poster may not be aware of the trig identities involved. The less elegant methods may feel more comfortable for a beginner.

16. Oct 17, 2013

### Staff: Mentor

Good point. I guess a couple of extra details would have been really helpful. I'll try to remember this for the future.

Chet