# What angle should the board be placed?

## Homework Statement

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

f=ma

## 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...

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Sin(Θ) = c + mkcos(Θ)
It looks good up to here.
What is wrong with the statement sin theta = c + mkcostheta ??

and thanks for the link I appreciate it very much!

What is wrong with the statement sin theta = c + mkcostheta ??
I didn't see anything wrong with it. I thought it was correct. and I thought c = 0.6806 was correct also.

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

Can I still approach my problem the same way though?

gneill
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.

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

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

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?
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?

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

gneill
Mentor
Hmm I see. I'm trying to think of a way that a spread sheet program would help, they mean one like excell, right?
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).
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?
Basic numerical methods (like Newton's Method) are covered at the college level and more advanced methods at the university level.

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

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. :)

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.

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

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