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Homework Help: What angle should the board be placed?

  1. Mar 27, 2017 #1
    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

    3. The attempt at a solution
    So I know:

    [itex]Fg = .0507 kg * 9.81[/itex]
    [itex]Fn = Fgcos(\Theta)[/itex]

    mk = .275
    That means Fk (force of kinetic friction) [itex] Fk = mkFgcos(\Theta) [/itex]

    here I use f = ma
    [itex]Fgsin(\Theta) - Fgcos(\Theta)mk = m * a [/itex]

    [itex]Fg(sin(\Theta) - mkcos(\Theta)) = m * a [/itex]

    since Fg = mass* 9.81 the two masses cancel leaving

    [itex] 9.81(sin(\Theta) - mkcos(\Theta)) = a [/itex]

    Now I use:

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

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

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

    So now I end up with

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

    Now this is as far as I've reached.

    [itex] \Theta = arcsin(c+ mkcos(\Theta)) = arccos( (c-sin(\Theta) )/ (-mk)) [/itex]

    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. jcsd
  3. Mar 27, 2017 #2
  4. Mar 27, 2017 #3
    What is wrong with the statement sin theta = c + mkcostheta ??

    and thanks for the link I appreciate it very much!
  5. Mar 27, 2017 #4
    I didn't see anything wrong with it. I thought it was correct. and I thought c = 0.6806 was correct also.
  6. Mar 27, 2017 #5
    Oh okay, ty.

    Hey I noticed in your link that it is using sin/cos instead of arccos(x)/arcsin(y).

    Can I still approach my problem the same way though?
  7. Mar 27, 2017 #6


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    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
  8. Mar 27, 2017 #7
    Why are you asking that? Where did arccos and arcsin come from? Am I missing something?
    Part IV gives a nice example.
  9. Mar 27, 2017 #8
    NOTE: I did not actually try to work out the solution. @gneill may be right that there is not a "simple closed form solution".
  10. Mar 27, 2017 #9
    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?
  11. Mar 27, 2017 #10


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    Staff: Mentor

    I may have been too hasty myself o:). 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...
  12. Mar 27, 2017 #11


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    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.
  13. Mar 27, 2017 #12
    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. :)
  14. Mar 27, 2017 #13
    I think your answer is correct this time though :)

    Was your answer about 56 degrees?

    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.
  15. Mar 27, 2017 #14


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    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.
  16. Mar 27, 2017 #15
    This falls under the blind chicken theory: Even a blind chicken gets a kernel of corn every once in a while. :)

    I got about 56.4.
  17. Mar 27, 2017 #16
    Will do. I will have to learn it by myself first though, apparently Newton's method is taught in Calculus I in most schools, but I was never taught Newtons method -_-

    Thank you all for the help though I wish you all the best in your endeavors
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