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Forces and Motion Midterm Review

  1. Jan 21, 2008 #1
    [SOLVED] Forces and Motion Midterm Review

    1. The problem statement, all variables and given/known data

    A bulldoezer drags a log weighing 500N along a rough surface/ The cable attached to the log makes an angle of 30 degress with the ground. The coefficient of static friction between the log and the ground is 0.500. What minimum tension is required in the cable in order for the log to begin to move?

    2. Relevant equations

    Ffriction = (coefficient of static friction)(Fnormal)

    3. The attempt at a solution

    OK after creating a free body diagram I wrote the equations for the sum of the forces in the X and Y dimensions.

    FY = Fg (500) + Ftension(sin30) = Fnormal
    FX = Ftension(cos30) - Ffriction = ma

    Thus, after a lot of substituting and finding the friction, I came out with an answer of 1866.03 Newtons. I don't know what the correct answer should be; I think I'm relatively close, as I recieved a 15/20 for work, etc. Can someone help me find the answer?
  2. jcsd
  3. Jan 21, 2008 #2
    I believe your Fx and Fy equations are set up incorrectly. We're dealing with static friction, and thus zero acceleration in the x-direction.With respect to the y-coordinate, your weight will be pointing in the negative direction, and the normal and Ty forces will point in the positive y-direction. Does this help?
  4. Jan 21, 2008 #3
    I think so... does that mean:
    FX = Ften(cos30) - Ffric = ma (0)


    FY = Ften(sin30) + Fnormal - Fg = ma
  5. Jan 21, 2008 #4
    Yes, except for the 'ma' in the Fy equation. The object does not accelerate vertically. Let me know if you get stuck.
  6. Jan 21, 2008 #5
    Well at first I thought that I needed to combine the equations; I got something like this:

    FX equation rewritten:
    Ften = Ffric/cos30 -->

    FY with substitution of FX equation in Ften:
    (Ffric/cos30)sin30 + Fnormal - Fg = 0

    However, I have two variables because I only know the friction coefficient.
    Also, wouldn't there be acceleration in the Y? That is, gravity's acceleration acting on the log?
  7. Jan 21, 2008 #6
    It would be best to solve for the normal force in your original Fy equation in terms of the tension T. Knowing that [tex] f = \mu(F_{norm}) [/tex], you can plug in the normal force for the Fx equation, and solve for the tension T.

    And yes, the force due to gravity is acting on the log, and in turn we have the presence of the normal force. As a result we have is zero net acceleration in the y direction (and the x-direction too).
  8. Jan 21, 2008 #7
    I'm not sure I follow you on that; I don't know the Fnormal or the Ffriction, and I need one of them plus the static friction coefficient to find the other, right? Normally, Fnormal would just equal Fg, but since there's the Ftension(sin30) force, that also has to be taken into account, right?
  9. Jan 21, 2008 #8
    You have two equations and two unknowns, which is sufficient. Place the following equation in terms of T, solving for Fnormal:

    FY = Ften(sin30) + Fnormal - Fg = 0

    Now you will have eliminated one unknown. Plug Fnormal into your equation:

    FX = Ften(cos30) - Ffric = 0

    which translates to

    FX = Ften(cos30) - (coeff. of fric.)(Fnormal) = 0

    right? Now you can solve for the tension.
  10. Jan 21, 2008 #9
    Hang on, I had a burst of realization; how about this:

    Fnormal = 500N - 0.5Ftension
    Ffriction = 0.866Ftension

    Ffriction = (coefficient)(Fnormal)
    0.866Ftention = (0.500)(500N - 0.5Ftention)
    Ftention = 224.014N
  11. Jan 21, 2008 #10
    I'm afraid that I'm not following your reasoning. You have Fnormal = 500N - 0.5Ftension, but it should be Fnormal = 500N - Ftension(sin30). Now you can plug that into FX = Ften(cos30) - (coeff. of fric.)(Fnormal) = 0, and solve for the tension. Right?
  12. Jan 21, 2008 #11
    See, I took my original FY equation, FY = Ften(sin30) + Fnormal - Fg = 0 and rearranged it to Fnormal = Fg - Ften(sin), or FN = 500 - 0.5Ften.
    Then I took my original FX equation, FX = Ften(cos30) - Ffric = 0 and rearranged it to Ffriction = Ften(cos30), or Ffric = 0.866Ften.

    Then, since I had Ffriction and Fnormal in terms of Ftension, I plugged them into my Ffriction = (coefficient)(Fnormal) equation and solved for Ftension
  13. Jan 21, 2008 #12
    Where did the 0.866 come from in Ffric = 0.866Ften?
  14. Jan 21, 2008 #13
    0.866 is the cosine of 30 degress. Also, 0.5 is the sine of 30, not to be confused with the coefficient of static friction, 0.500.
  15. Jan 21, 2008 #14
    Ok gotcha. You're answer is correct, just remember to put it in sig figs :)
  16. Jan 21, 2008 #15
    It is?! Thanks so much! I'm not so great at physics, but I'm trying. And thank God my teacher doesn't care about sig figs, one less thing to worry about.
  17. Jan 21, 2008 #16
    Haha. You're welcome, keep up the good work :)
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