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Finding coeficient of friction with a slope/blocks/a pulley

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Two blocks are connected by a thin inextensible string over a frictionless massless pulley
    as shown on the picture below
    The acceleration of gravity is 9.8 m/s
    Given that the two blocks move at 5.7 m/s
    under an acceleration of 4 m/s^2
    calculate the coefficient µk of kinetic friction between the
    left block and the incline.

    The picture is a slope of 23 degrees
    2. Relevant equations
    Fnet = ma and friction = Fn(mu)

    3. The attempt at a solution
    I found the normal force to be 284.2/cos(23) = 308.7436 N.
    the problem says that its moving at a constant velocity so the net force should equal 0.
    m x a = 92 x 4 = 368 N though which would make mu = 368/308.7436 aka greater than 1 which is wrong.
    Maybe I misunderstood something in the wording of the problem? :/
     

    Attached Files:

    Last edited: Oct 19, 2012
  2. jcsd
  3. Oct 19, 2012 #2

    PhanthomJay

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    The problem is not clearly worded. It can't be moving at constant velocity if it is accelerating at 4m/s/s. I guess it means that at some point in time its velocity is 5.7 m/s, but that is irelevant here.

    Check your normal force calculation, you divided when you should have multiplied.

    It is essential that you draw free body diagrams of each block to identify the forces acting on each block, then use newton's laws on each to solve for the friction coefficient from the resulting 2 equations. Don't try to do it in one fell swoop.
     
  4. Oct 19, 2012 #3
    are you absolutely sure my normal force calculation should be multiplied instead of divided? I can't see how.
    The 23 degree angle should be the same as the angle from Fg(vertical) to -Fnormal(perpendicular to the incline) of the block on the incline.
    then cos x = adj/hyp
    cos 23 = 284.2/h
    h cos 23 = 284.2
    h = 284.2/cos 23

    the other things you said are very useful though :) thanks.
     
  5. Oct 20, 2012 #4

    PhanthomJay

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    you have mixed up the weight with the normal force , while also mixing up the hypotenuse and adjacent side
    Yes, this is corrrect
    The weight acts straight down. You should then break up the weight into its components perpendicular and parallel to the incline. The components can never be greater than the resultant weight. The weight is the hypotenuse. The component of the weight perpendicular to the incline is the adjacent side and thus equal to h(cos23) = (284.2)cos 23. Then since there is no acceleration perpendicular to the incline, the Normal force must be equal to the perpendicular component of the weight, per Newton's First Law.
     
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