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Given: coefficient kinetic and coefficient static frictions

  1. Dec 26, 2007 #1
    1. The problem statement, all variables and given/known data
    A problem in engineering mechanics:
    The blocks shown in the figure are connected by flexible inextensible cords passing over frictionless pulley. At Block A the coefficients of friction are Ms = 0.30 and Mk = 0.20 while at Block B they are Ms = 0.40 and Mk = 0.30. Compute the magnitude and direction of the friction force acting on each block.


    the answer is Friction at A = 48lb
    Friction at B = 36lb

    2. Relevant equations
    R is along the angle of friction w/c has two components:
    Normal force at the vertical
    Friction at the horizontal

    3. The attempt at a solution
    I do not know how to approach the problem since there are two given coefficient of friction and it seems very confusing.....

    however i tried solving at block A using the coefficient of kinetic friction at A = 0.20:

    F = friction
    R = the force along the angle of friction
    the angle of friction here lies along with R which is the coefficient of friction
    tan(theta) = 0.20
    theta = 11.3 degrees

    so i used sine law:
    [tex]\frac{\sin{79}}{300} = \frac{\sin{53}}{R}[/tex]
    R = 244lb

    R has two components: the normal force N and the friction F
    F = (244lb)(sin(11.3))
    F = 48lb <<< at A

    i am not so sure about this..
    can u give some hints its very confusing
    Last edited: Dec 26, 2007
  2. jcsd
  3. Dec 26, 2007 #2

    Doc Al

    User Avatar

    Staff: Mentor

    I don't quite understand what you're doing. Where did the angle of 11.3 come from?

    The first thing I'd do is figure out if this thing moves or not. Does it?
  4. Dec 26, 2007 #3

    so ill try to use static friction
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