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Homework Help: Calculate the force on a circular surface

  1. Dec 28, 2012 #1
    Hi to everybody

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

    (The attachment has an image with this problem)

    A small probe P is gently forced against the circular surface with a vertical force F as shown. Determine the n- and t-components of this force as functions of the horizontal position s.

    2. Relevant equations

    Newton laws and basic trigonometric relationships.

    3. The attempt at a solution

    Well I show in the attachment my reasoning, the force F is applied at the contact point C, I have represented the direction in which this force acts (vertical line). I need to find the proyections of F to the n and t axis, to calculate Fn and Ft in red.

    Well I think that the angle between the vertical line and the line connecting the center O and the point C is the same angle between F and Fn, in the picture they are called A. I simply write:

    Fn= F cos(A)
    Ft= F sen(A)

    And I can calculate A as:


    But I have seen the answers provided in the book and they are:



    Fn= -(F√(r2-s2))/r

    Does anybody know what is Fs?

    And I have no idea where the equation for Fn comes from.

    Why are my answers so different to the ones in the book?

    Thank you

    Attached Files:

  2. jcsd
  3. Dec 28, 2012 #2


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    "Fs" is F*s, not Fs. Looks like they are avoiding using sin, cos in the answer.
  4. Dec 28, 2012 #3


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    Looks good. (The sine function is generally written sin rather than sen.) From A=arcsin(s/r) you get sin(A) = s/r. Try substituting this expression for sin(A) into your expression Ft= F sin(A) and compare to answer in book.
  5. Dec 28, 2012 #4


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    How would you write cos(A) in terms of s and r?
  6. Dec 28, 2012 #5
    Yes you are right, thank you very much.

    The sinus is simply s/r and I get the Ft.

    For the Fn it´s the same, they avoid to use the cos, to calculate the other leg of the triangle I use the pythagorean theorem r2=s2+y2 where I called y the other leg of the triangle. So the cosinus is exactly the answer in the book.

    Thank you very much to both of you, I have more doubts with problems in this book, hope to see you in the other threads.:biggrin:
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