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!

Inclined plane and friction

  1. Aug 18, 2008 #1
    1. The problem statement, all variables and given/known data

    There are two wedges, each with mass m, placed on a flat floor. A cube of mass M is balanced on the wedges. Assume there is no friction between wedge and block, and the coefficient of static friction is less than 1 between the floor and wedges. What is the largest M that can be balanced without moving?

    diagram here:
    http://sites.google.com/site/learningphys/

    2. Relevant equations
    3. The attempt at a solution

    If we split the block in half due to symmetry, the question will get easier.

    force of friction is given by:
    [tex]F=\mu (\frac{M}{2}+m)g[/tex]
    the M/2 comes from taking only half of the block

    i want to equate the force that the block has on one of the wedge.

    Still considering half of the diagram, i want the normal force of (M/2)g to the wedge's surface. The normal force is give by the usual inclined plane equation, mgcos(theta)

    So normal force of (M/2) is:

    [tex]N=\frac{Mg}{2}cos(\theta)[/tex]

    this normal force pushes in the opposite direction of the force of friction... but we have to consider the horizontal component, which tags on a sin(theta) term

    I finally get:

    [tex]\mu (\frac{M}{2}+m)g=\frac{Mg}{2}cos(\theta)sin(\theta)[/tex]

    And then i solve for mu. How close am I to being correct?
     
  2. jcsd
  3. Aug 19, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi learning_phys! :smile:

    No … consider the forces on the block

    there is no friction, so the only forces are Mg/2 and N …

    the horizontal components balance anyway …

    so resolve vertically, and that gives you … ? :smile:
     
  4. Aug 19, 2008 #3
    are you sayin N=Mg/2? but the normal force should be perpendicular to the blocks surface right? (that's why I have the cosine term)
     
  5. Aug 19, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yes (because there is no friction). :smile:
    No … resolve vertically.
     
  6. Aug 19, 2008 #5
  7. Aug 19, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    That's it! :smile:
     
  8. Aug 19, 2008 #7
    awesome

    so I would get:
    [tex]\mu (\frac{M}{2}+m)g=\frac{Mg}{2cos(\theta)}sin(\theta )[/tex]
     
  9. Aug 20, 2008 #8

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    :biggrin: Woohoo! :biggrin:
    (you're easily awed! … :wink:)
    And so the largest M is … ? :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Inclined plane and friction
Loading...