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Find the friction force impeding its motion

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data
    A 21.0 kg box is released on a 38.0° incline and accelerates down the incline at 0.267 m/s2. Find the friction force impeding its motion.

    Friction force is correctly calculated as 121N

    2. Relevant equations
    Ffr = [tex]\mu[/tex]FN (Friction force equals mu times normal force, in case the symbols are not in its commonly used form)
    FN=mg (Normal force equals mass times gravity)

    3. The attempt at a solution
    Ffr = [tex]\mu[/tex]FN
    121=[tex]\mu[/tex]FN
    121=[tex]\mu[/tex]mg
    121=[tex]\mu[/tex](21)(9.8)
    5.7619=tex]\mu[/tex](9.8)
    0.588=[tex]\mu[/tex]

    I see this as the correct method, but the answer came out as incorrect, so that means I'm wrong. Is this formula incorrect for finding mu? If so, what is supposed to be used/what am I supposed to do in order to find mu?

    note: numbers are obtained using longer, more accurate numbers of 121.09613162202047395584132290722 / ((21)(9.8)) = 0.5879494655004859086491739552964 but shortened down to look better
     
  2. jcsd
  3. Jan 29, 2009 #2

    AEM

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    Gold Member


    Generally for inclined plane problems one draws a set of coordinate axes with one axis, say the x-axis along the plane and the y-axis perpendicular to it. Then you should draw in your forces: mg for gravity, N for the normal force between the plane and the block, f for the frictional force acting up the plane. You will need to identify the component of mg that accelerates the box down the plane and the component of mg that is equal and opposite to the normal force. You will have some sines and cosines floating around. (I'll leave it to you to figure that out.) The moral of this story: Don't start hunting for equations before you have drawn a picture and identified the forces involved and understand the relationships between them. Understand what's happening before you start using equations!
     
  4. Jan 29, 2009 #3
    121=[tex]\mu[/tex](21)(9.8)(cos38)
    0.747=[tex]\mu[/tex]
    The answer is correct.

    I drew the diagram but forgot the the inclination affects the total force. Thank you.
     
    Last edited: Jan 29, 2009
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