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Kinetic friction problem

  1. Mar 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A 425 g block is at rest on a horizontal surface. The coefficient of static friction between the block and the surface is .770, and the coefficient of kinetic friction is .220. A force of magnitude P pushes the block forward and downward. Assume the force is aplied at an angle of 40°

    a. Assuming P is large enought to made the block move, find the acceleration of the block as a function of P (Use P as necessary and round numbers to the third decimal place)

    b. If P=11.13 N and then P=22.26, find the accelerations and the frictional forces exerted on the block

    c. What is the minimum acceleration for the block when P is a minimum and motion is just ready to begin?




    2. Relevant equations



    3. The attempt at a solution

    a. ΣFx=Pcosθ -f=ma
    ΣFxy=N-mg-Psinθ=0

    f=μN

    Pcosθ-μΝ=ma

    Pcosθ-μ(mg+sinθ)=ma

    (cosθ/m)P-mg+(μsinθ/m)P

    -mg+μsinθ/m +(cosθ/m)P=a

    (-425g)(9.8m/s^2)+(.220sin40/425g)+(cos40/424)P=a

    a=4165.001+.001P

    b. P=11.13

    a=4165.001+.001(11.13N)
    4165.012

    f=μΝ
    N=(4165.012)(11.13)+425g(9.8m/s^2)

    c. For part c. I was thinking about using Calculus to minimize the equation, would that work?

    Could someone please tell me if this is correct so far and if it isn't could you please show me what to do?

    Thank you very much
     
  2. jcsd
  3. Mar 15, 2008 #2
    Could someone please tell me if this looks correct?

    a.
    μs=.770
    μk=.220

    ΣFx=max

    ΣFx=Pcos40-f
    f=μkN
    may=0
    0=Σfy=N-mg-Py
    =N-mg-Psin40
    =Pcos40-[.220(mg+Psin40)]
    ΣFx=max
    Pcos40-.220(mg+psin40)=max
    pcos40/.425kg-.220(.425(9.8)+psin40)/.425=ax
    1.47p-2.155=ax

    b.P=11.13N
    1.47(11.13N)-2.155=ax
    ax=14.206m/s^2

    f=μkN

    =.220(9.8m/s^2)
    f=2.156

    c. I would then do the same thing for part c, right?

    Thank you very much
     
    Last edited: Mar 15, 2008
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