# What angle should the board be placed?

1. Mar 27, 2017

1. The problem statement, all variables and given/known data
A wooden block with mass of .0507 kg is placed at the midpoint of a 1m long wooden board, where the coefficient of kinetic friction (mk) = .275, at what angle should the board be placed so that the block takes time t = .387 s to slide to the lower end of the board? You may find a spread sheat program helpful in answer this question

2. Relevant equations

f=ma
3. The attempt at a solution
So I know:

$Fg = .0507 kg * 9.81$
$Fn = Fgcos(\Theta)$

mk = .275
That means Fk (force of kinetic friction) $Fk = mkFgcos(\Theta)$

here I use f = ma
$Fgsin(\Theta) - Fgcos(\Theta)mk = m * a$

$Fg(sin(\Theta) - mkcos(\Theta)) = m * a$

since Fg = mass* 9.81 the two masses cancel leaving

$9.81(sin(\Theta) - mkcos(\Theta)) = a$

Now I use:

$\Delta x = V_{0x}t + (1/2) a_x t^2$

$1/ (.378^2 * 9.81) = (sin(\Theta) - mkcos(\Theta))$

1/(.378^2 * 9.81)= .6806268265 which I will just call "c."

So now I end up with

$Sin(\Theta) = c + mkcos(\Theta)$
$Cos(\Theta) = ( c-sin(\Theta) )/( -mk )$
$\Theta = arcsin(c+ mkcos(\Theta))$
$\Theta = arccos( (c-sin(\Theta) )/( -mk ))$

Now this is as far as I've reached.

$\Theta = arcsin(c+ mkcos(\Theta)) = arccos( (c-sin(\Theta) )/ (-mk))$

but I see no possible way of solving this for Theta...

My biggest hint was:

Which said "x^2 + y^2 = 1"

But I don't understand and don't think I can use this because x and y are sides, correct? While I'm dealing with angles...

Last edited: Mar 27, 2017
2. Mar 27, 2017

### TomHart

3. Mar 27, 2017

What is wrong with the statement sin theta = c + mkcostheta ??

and thanks for the link I appreciate it very much!

4. Mar 27, 2017

### TomHart

I didn't see anything wrong with it. I thought it was correct. and I thought c = 0.6806 was correct also.

5. Mar 27, 2017

Oh okay, ty.

Can I still approach my problem the same way though?

6. Mar 27, 2017

### Staff: Mentor

You're facing one of those cases where the function you want to find the solution for is transcendental. You won't find a simple closed form solution using your usual toolbox of functions and algebra. This is why the problem hinted that a spreadsheet program might be useful. Can you think of way to make use of that tool?

EDIT: I may have been too hasty in declaring the problem transcendental. An algebraic approach is tedious but possible. See later posts.

Last edited: Mar 27, 2017
7. Mar 27, 2017

### TomHart

Why are you asking that? Where did arccos and arcsin come from? Am I missing something?
Part IV gives a nice example.

8. Mar 27, 2017

### TomHart

NOTE: I did not actually try to work out the solution. @gneill may be right that there is not a "simple closed form solution".

9. Mar 27, 2017

Hmm I see. I'm trying to think of a way that a spread sheet program would help, they mean one like excell, right?

Also I think I will close this thread in a minute or two as it seems like it would take me forever to find the solution for theta, but one last question...

At what point in my academic career will I be able to solve such an equation for theta? Obviously it seems very tedious but I am pretty interested in learning to solve these types of equations... What field of math is this?

10. Mar 27, 2017

### Staff: Mentor

I may have been too hasty myself . While algebraically tedious, the method shown in your link will get you to a solution after much toil, and you need to beware of false solutions that crop up due to repeated squaring. So I take back what I said about the "usual toolbox" being inadequate.

It's still much easier to solve numerically though...

11. Mar 27, 2017

### Staff: Mentor

Right. You could do it graphically by plotting the time versus θ or plot $sin(θ) - \mu_k cos(θ)$ and see where it equals c. Or you could use a solver to find the solution numerically, or program your own (a search function like binary search, or a solver like Newton's Method).
Basic numerical methods (like Newton's Method) are covered at the college level and more advanced methods at the university level.

12. Mar 27, 2017

### TomHart

You came up with the equation: sinθ = 0.6806 + 0.275cosθ
Or, rearranging: sinθ - 0.275cosθ - 0.6806 = 0, which I think is correct.

What I did was just made an Excel spreadsheet column to represent the angle θ, that ranged from 0° to 90° in 0.1° increments. Of course, for Excel you have to convert the angle to radians (unless there is a way to perform sin and cos functions using degrees that I don't know about). Then you just make another column that performs the math of the equation for each and every angle. Whatever angle causes that function to equal 0 is the right answer. Without giving away the exact answer, I got somewhere in the 50° to 60° range.
DISCLAIMER: My answers are not always correct. :)

13. Mar 27, 2017

I'm glad to know that using technology on such a question is EXPECTED, because it would be reallly tedious to do such a problem on a test.

14. Mar 27, 2017

### Staff: Mentor

Try Newton's Method. Starting with an initial "guess" of 45 degrees, two iterations will get you an answer good to 3 decimals.

15. Mar 27, 2017

### TomHart

This falls under the blind chicken theory: Even a blind chicken gets a kernel of corn every once in a while. :)