Isolate for Angle With Variables

  • #1
AvocadosNumber
16
0

Homework Statement



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

Homework Equations



[itex]m_{2}g=m_{1}gsinθ-μm_{1}gcosθ[/itex]

We are given the trigonometric identities:
LhWjADR.png

CwvM9Mh.png


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.

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

Thanks.
 
Last edited:

Answers and Replies

  • #2
nasu
3,957
582
Are you trying to solve for the angle in therm of the other variables?
 
  • #3
AvocadosNumber
16
0
Yes, that is what I am trying to do.
 
  • #4
nasu
3,957
582
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
AvocadosNumber
16
0
Alright. This is what I have tried... I don't know if it's entirely correct but it still seems really clustered.

yv2kIXA.jpg


I don't know what to do next though...
 
  • #6
22,067
5,030

Homework Statement



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

Homework Equations



[itex]m_{2}g=m_{1}gsinθ-μm_{1}gcosθ[/itex]

We are given the trigonometric identities:
LhWjADR.png

CwvM9Mh.png


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.

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

Thanks.
Try this:
[itex](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)[/itex]
where [itex]cot\phi=μ[/itex]
 
  • #7
ehild
Homework Helper
15,543
1,915
Alright. This is what I have tried... I don't know if it's entirely correct but it still seems really clustered.

yv2kIXA.jpg


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

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

ehild
 
  • #8
AvocadosNumber
16
0
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
22,067
5,030
Since no one seemed to have any interest in what I did in response #6, I will continue by completing the solution:

[tex]θ=\arcsin\left({\frac{(m_2/m_1)}{\sqrt{μ^2+1}}}\right)+\arctan(1/μ)[/tex]
 
  • #10
nasu
3,957
582
Since no one seemed to have any interest in what I did in response #6, I will continue by completing the solution:

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

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:
  • #11
ehild
Homework Helper
15,543
1,915
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).

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
ehild
Homework Helper
15,543
1,915
Since no one seemed to have any interest in what I did in response #6, I will continue by completing the solution:

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

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

ehild
 
  • #13
22,067
5,030
It should be arctan(μ).
Yes, it is a clever method if somebody knows it.

ehild

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
ehild
Homework Helper
15,543
1,915
Incidentally, this "clever method" has appeared in many threads on physics forums.

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

ehild
 
  • #15
nasu
3,957
582
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.

I did not say it's new. We used in high school and that was long time ago.:smile:
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
22,067
5,030
The OP did not know it, so some explanation would have been necessary.

ehild

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

Chet
 

Suggested for: Isolate for Angle With Variables

Replies
2
Views
1K
Replies
5
Views
7K
Replies
8
Views
6K
Replies
3
Views
6K
Replies
2
Views
1K
  • Last Post
Replies
4
Views
9K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
7K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
2K
Top