1. Limited time only! Sign up for a free 30min personal 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!

Static friction between a cylinder and a wall

  1. Jan 26, 2015 #1
    1. The problem statement, all variables and given/known data
    The cylinder in the drawing weighs 100[N]. there is a horizontal rope at the top that prevents it from rolling. what is the coefficient of friction if it's on the verge of movement

    2. Relevant equations
    Friction force: f=μN

    3. The attempt at a solution
    The whole weight of the cylinder lies on the contact point. the coefficient of friction is the tangent of the slope, which is here 250, thus μ=tan 250=0.422
    It should be 0.22
     

    Attached Files:

    • 43.jpg
      43.jpg
      File size:
      7.9 KB
      Views:
      79
  2. jcsd
  3. Jan 26, 2015 #2
    Have you drawn a free body diagram? The normal component of the contact force is not pointing in the vertical direction. So the normal force is not equal to mg. You need to write down the force balance equations for equilibrium, and you probably also need to write down a moment balance equation. Call N the normal contact force, and call F the tangential contact force.

    Chet
     
  4. Jan 26, 2015 #3
    yes i did that but was is a hurry so i used the other method of the tangent which gave the same result.
    $$N=100\cdot \sin 65^0,\ F=100\cdot \cos 65^0$$
    $$F=\mu N\rightarrow \mu=\frac{100\cdot \cos 65^0}{100\cdot \sin 65^0}=0.466$$
     
  5. Jan 26, 2015 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Shouldn't the tension in the rope contribute to these two equations?
     
  6. Jan 26, 2015 #5
    You left out the effect of the rope tension T. That has a component in the normal direction and a component in the tangential direction. The reason that this happened is that you did not draw a free body diagram.

    Chet
     
  7. Jan 26, 2015 #6
    I thought about it but decided that the rope doesn't influence. the rope only balances the moment (torque) and it comes out right according to the book:
    $$100\cdot R\cdot\sin 25^0=T\cdot R(1+\cos 25^0)\rightarrow T=22.17$$
    I will draw a free body diagram soon
     
  8. Jan 26, 2015 #7
    T influences the values of both F and N. Did you really feel that you had advanced beyond the point where you need to draw free body diagrams?

    Chet
     
  9. Jan 26, 2015 #8
    You are right, it comes out good
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted