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Homework Help: Kinematics - Find static friction coefficient of a coffee cup

  1. Jan 16, 2008 #1
    Hi !

    Can anyone help me I need help ......

    Problem :
    ************************************************** **********************
    1- A coffee cup on the dashboard of a car slides forward on the dash when the driver decelerates from 40 km/h to rest in 3.5 s or less, but not if he decelerates in a longer time. What is the coefficient of static friction between the cup and the dash ?
    ************************************************** ********************
    2- An 18-kg box is released on a 37*(grade) incline and accelerates down the incline at 0.270 m/s(square). Find the friction force impeding its motion. How large is the coefficient of friction
    ************************************************** ********************

    Thanks !!!!
     
    Last edited: Jan 16, 2008
  2. jcsd
  3. Jan 16, 2008 #2
    what have you tried?
     
  4. Jan 16, 2008 #3
    Problem 1:
    I now that :

    F = F
    uN = ma
    umg = ma
    ug = a
    u(9.8) = (40km/h * 1h/3600s * 1000m/km)/3.5s
    u = 0.324

    Is this correct ?

    Can anyone help me for the problem 2 ?

    Thanks
     
    Last edited: Jan 16, 2008
  5. Jan 16, 2008 #4
    Problem 2
    I have solved these problem ::::::


    The weight of the box is (18 kg)(9.80 N/kg) = 176.4 N
    The component of the weight parallel to the plane is
    (176.4 N)sin(37.0o) = 106.16 N

    The component of the weight perpendicular to the plane is
    (176.4 N)cos(37.0o) = 140.88 N
    So now our expression of Newton's second law has the parallel component of gravity of 106.16 N down (-) the plane, and assuming it is moving down the plane as well as accelerating down the plane, the force of friction F is acting up (+) the plane, and the acceleration is .270 m/s/s down (-) the plane, so:
    <F - 106.16 N> = (18.0 kg)(-.270 m/s/s)
    F = 101.3 N = 101 N, and the coefficient of friction is given by
    Ffr = mkFn
    The Normal force in this case is the perpendicular component of gravity 140.88 N, so we have:
    Ffr = mkFn
    101.3 N = mk(140.88 N)
    So mk = .719
     
  6. Jan 16, 2008 #5
    There you go!You're done!
     
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