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Incline plane + friction

  1. Oct 2, 2008 #1
    1. The problem statement, all variables and given/known data
    In the figure [​IMG] (sorry for that quick drawing with paint) m_1 (first block) = 4.0kg, m_2 = 5.0kg and the coefficient of kinetic friction between the incline plane and the first block is \mu_k = 0.24. Incline = 30 degree with horizontal

    Find the magnitude of the acceleration of the masses and the tension of the cord


    2. Relevant equations
    (1) [tex]\sum Fnet = ma[/tex]
    (2) [tex]f_k = \mu_k*F_n[/tex]

    3. The attempt at a solution
    I applied (1) to the first block in the x-direction and then in the y-direction
    Found:
    [tex]F_n - m_1*g*\sin\theta = 0 (a_y = 0) => F_n = m_1*g*\sin\theta[/tex] [a]
    [tex]T - f_k - m_1*g*\cos\theta = m_1*a_x[/tex]

    I applied (1) to block 2
    [tex]m_2*g - T = m_2*a_y [/tex]

    Added [a] to
    [tex]m_2*g - f_k - m_1*g*\cos\theta = m_1*a_x + m_2*a_y[/tex]
    <=>[tex]m_2*g - \mu_k *m_1*g*\sin\theta - m_1*g*\cos\theta = (m_1+m_2)a[/tex]
    <=>[tex]g*\frac{m_2 - \mu_k *m_1*\sin\theta - m_1*\cos\theta}{m_1+m_2}=a[/tex]

    What am I doing wrong?
     
  2. jcsd
  3. Oct 2, 2008 #2
    Here's what we got: Force from block (2) = 9.81 * 5 = 49.05N
    weight of block (1) = mg = 9.81 * 4 = 39.24
    Normal reaction force = 39.24cos(30) = 33.98N
    Frictional force = 33.98 * 0.26 = 8.84N
    force due to gravity parallel to inclined plane = 39.24sin(30) = 19.62N
    Total retarding force from block (1) = 19.62N + 8.84N = 28.46N
    Tension in rope = 49.05 - 28.46 = 20.60N

    There is constant tension in the rope => magnitude of acceleration is constant and equal for masses.
    Mass total = 4 + 5 = 9kg
    a = F/m = 20.60/9.00 = 2.29ms-2

    Cross my fingers I didn't read the question wrong or make any stupid mistakes. It is ridiculously late at the moment..
     
  4. Oct 2, 2008 #3
    Your equation has gone a bit wrong..

    Fn - M1*g*cos30 = 0 will be right

    M2*G - T = M2*a

    T - M1*G*SIN30 - FRICTIONAL FORCE = M1*A.

    i think that will give the answer.
     
  5. Oct 2, 2008 #4
    why do you have normal force in the X direction. the normal force only apply on vertical which is y direction.
    BUT THIS FORMULA IS DEPENDS ON YOUR ANGLE.
    so Normal force - cos(teta)mg = 0 in the y dirction
    tension -mg = am

    In the x direction will be sin(teta)mg = am if there is no friction .
     
  6. Oct 2, 2008 #5
    vishal_garg, thank you so much :D I spent the whole night banging my head again my desk trying to figure out what I was doing wrong...

    Thanks for all replies
     
  7. Oct 4, 2008 #6
    No need to say that nvictor...
    well are you studying in school or college.??
     
  8. Oct 4, 2008 #7
    College :)
     
  9. Oct 5, 2008 #8
    great...
     
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