A box being pulled at an angle

  1. Feb 26, 2011 #1
    1. The problem statement, all variables and given/known data
    A 45.0 kg box is pulled with a force of 205 N
    by a rope held at an angle of 46.5° to the
    horizontal. The velocity of the box increases
    from 1.00 m/s to 1.50 m/s in 2.50 s. Calculate
    a) Net force acting horizontally on the box
    b) Frictional force on the box
    c) Horizontal component of the applied force
    d) Co-efficient of kinetic friction between the box and the floor


    2. Relevant equations
    Kinematics equations
    [mu][Fn] = frictional force


    3. The attempt at a solution
    [​IMG]
    a) I found acceleration by: change in velocity = 0.5m/s, change in time = 2.50s
    v
    ---
    t
    = 0.2m/s^2

    Ok I found the summation of the X and Y forces:
    b) Summation of X forces: [(205)(cos46.5) - frictionalF = (45)(0.2)]
    frictionalF = 132.1N [<---]

    Summation of Y forces: [Fn + (205)(sin46.5) - (45.0)(0.2) = 0]
    normalF = 292.74N

    c) Now, I found the horizontal component of the applied force to be:
    [205][cos46.5] = 141.1N, but the answer at the back says 293N. Have I done something wrong?

    d) I am not sure how I am going to get the coefficient.
     
  2. jcsd
  3. Feb 26, 2011 #2
    Mg I included as (45.0)(0.2)...?
     
  4. Feb 26, 2011 #3

    ehild

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    The horizontal component of the applied force is 205 cos(46.5)= 141.1 N it is correct.

    The coefficient of friction is obtained from
    Ffr=μ Fn. Calculate Fn from the equation for the y components of forces.

    ehild
     
    Last edited: Feb 26, 2011
  5. Feb 26, 2011 #4
    Doesn't the applied force y-component count as one of the forces in the Y-summation?
     
  6. Feb 26, 2011 #5

    ehild

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    It is 205 sin(46.5) You have it already in the equation.


    ehild
     
  7. Feb 26, 2011 #6
    Ok so the back of the book is wrong. >_< Thank you.
     
  8. Feb 26, 2011 #7
    so what did you find as fn?
     
  9. Feb 26, 2011 #8
    I found Fn to be 292.74N, after Subtracting appliedforceY from grav.force.
     
  10. Feb 26, 2011 #9
    isnt the normal force equal to the fappy + the force of gravity though, not fappy-fg?
     
  11. Feb 26, 2011 #10
    Fn + Fay - Fg = 0
    so
    Fn = fg - Fay?
     
  12. Feb 26, 2011 #11

    ehild

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    It is correct. Now find the coefficient of friction.

    ehild
     
  13. Feb 26, 2011 #12
    dumb question, would the normal force be different if the object was pushed rather than pulled?
     
  14. Feb 26, 2011 #13

    ehild

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    Yes. The y component of the applied force would change sign.

    ehild
     
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