1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Friction and critical angle question.

  1. Oct 26, 2006 #1
    I need... a lot of help. I spent an hour on this and a whole page of paper to no avail or certainty.


    Suppose there is a diagonal downward force from top right to bottom left on an object with weight w on a surface with coefficient of friction u (static/kinetic friction aren't distinguished in this question.)

    The diagonal downward force is a vector F_h with angle theta to the horizontal.

    I have already figured out that, for at a certain angle theta, the force F required to push the object to overcome its frictional resistance, is

    F_h = (uw) / (costheta - usintheta)

    However, since every positive change in angle will reduce the horizontal component and increase the vertical component, this will increase the effect normal force on the object, which is N = u + F_hsintheta, thus also increasing the frictional force, F_f = uN.

    At a certain angle, called the Critical Angle, F_hcostheta, which is the horizontal force required to move the object, will be equal to F_f, the frictional force opposing F_h. Increasing that angle will leave F_h < F_f and the object will not be able to move. After a certain interval of increasing degree, F_h will be greater than F_f once more, but the object will now move in the opposite direction.

    Knowing that the Critical Angle forms a singularity at F_h = (uw) / (costheta - usintheta) so that F_h goes to infinity, I know I have to solve for (costheta - usintheta) = 0.

    However, I also need to know the tangent of the Critical Angle, and this is not a happy answer, since my answer for the Critical Angle also included arcsin functions.

    Here are my preliminary results:

    Critical Angle = arcsin(- uw / (Fsqrt(u^2 + 1))) - arctan(-1/u)
    tan(Critical Angle) = (Z + 1/u) / (1 - (-1/u)Z)
    Z = -uw / (sqrt[ (F^2)(u^2 + 1) - (uw^2) ])

    Z being a substitution since it would look horrible in the equation.
  2. jcsd
  3. Oct 26, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper

    You did not state the whole problem. I think I managed to copy and paste your image and blow it up enough to read it. I will get back to you soon about your solution, but perhaps you could edit your post to make the problem clear to everyone.

    I think you are making it much too hard. Set that denominator to zero as you did and solve the equality for the ratio sinθ/cosθ = tanθ. It is a very simple expression involving μ
    Last edited: Oct 26, 2006
  4. Oct 26, 2006 #3


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think i have read the problem corectly, and it appears that you have correctly solved part a, and that your only question relates to part b. You again are correct that solving (costheta =usintheta) will give the critical angle. Try dividing both sides of the equation by costheta to see what you get. What is sintheta/costheta equal to in terms of the trig identities?
  5. Oct 28, 2006 #4
    I have already solved this problem, thank you for all your help.

    I simply had to set costheta = usintheta and then found tantheta by taking tanarccotu = 1/u.

    I was trying to do wave superimposition earlier...
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Friction and critical angle question.
  1. Critical Angle (Replies: 2)

  2. Critical angle (Replies: 2)

  3. Critical Angle (Replies: 2)

  4. Critical Angle (Replies: 3)