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Newton's Law and Friction (Ramp problem)

  1. Feb 7, 2014 #1
    1. The problem statement, all variables and given/known data
    A block is placed on a incline plane. A 21 N force is required to push the block up the incline with constant velocity. What is the weight of the block? Answer in units of N


    2. Relevant equations
    The force pushing the block=21 N with an angle of 13 degrees. The ramp has an angle of 21 degrees. Coefficient of friction=0.17. Ramp goes up from left to right.
    fs = uN, F = ma, acceleration is 0.

    Angle of force = phi, Angle of ramp = theta


    3. The attempt at a solution
    My attempt: (after drawing my free-body diagram)

    F(y-component): N-mgcos(theta) = 0 --> N=mgcos(theta) --> Fsin(phi)-mgcos(theta) = 0
    F(x-component): Fcos(phi)-mgsin(theta)-fs = 0 --> Fcos(phi)-mgsin(theta)-umgcos(theta) = 0

    I multiplied the y-component with the negative u (or mu) to cancel out umgcos(theta)

    Then: Fcos(phi)-uFsin(phi)-mgsin(theta) = 0, So I solved for mg and my answer was incorrect. What am I doing wrong? Any help appreciated, thanks.
     
  2. jcsd
  3. Feb 7, 2014 #2

    BvU

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    What did you find when the solving was done?
    I notice phi is different from the 21 degrees. Is phi wrt horizontal? Does its y component contribute to the friction force ?
     
  4. Feb 7, 2014 #3
    Well the problem indicated that the force pushing the block up the ramp had an angle of 13 degrees. So denoting that phi, I used Fsin(phi) and Fcos(phi) respectively.

    My answer in the end was about 54.85 N. Which was incorrect. No idea what I am doing wrong.
     
  5. Feb 7, 2014 #4

    BvU

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    I take it the free body diagram decomposes mg along the slope and perpendicular to it.
    You want to do the same with the 21 N force. What is the angle wrt the plane ?
    The component perpendicular to the plane is to be added to mg ##\cos\theta## to determine the normal force.
    That times ##\mu## is the friction force, along the slope, downwards. To be added to mg ##\sin\theta##, also downwards. Call the sum Fdown
    No acceleration (i.e. constant speed) along the slope means the 21 N component along the slope is equal and opposite to this Fdown
     
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