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Homework Help: Rigid Body Problem

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the following diagram
    If the crate produces a force of gravity of 491 N from the center of gravity G, determine the normal force on both of the wheels and the magnitude and direction of the minimum force required at the grip B needed to lift the load.

    2. Relevant equations
    Fx = 0;
    Fy= 0;
    MB = 0;

    3. The attempt at a solution
    Sign Convention: Up and to the right is positive and counter clockwise is positive.
    from Fx = 0
    0= 2Ax - FBcos(theta) (equation 1)
    from Fy = 0
    0= FBsin(theta) - 491N + 2Ay (equation 2)
    from MB = 0
    0 = -2Axcos60(0.1) +2 Aysin60(0.1) - (491cos30)(0.6) (equation 3)

    from 1
    Ax= 0.5FBcos(theta) (1A)
    from 2
    Ay = 0.5(491N-FBsin(theta) (2A)

    Substitute 1A and 2A into 3

    After Simplifying I have

    -0.05FBcos(theta)-0.0866FBsin(theta) =212.61

    I am now stuck as to how to proceed. I am wondering if my initial free body diagram is flawed. Does the wheel at A produce a horizontal force Ax and a Vertical force Ay? Or is Ay the only force produced. The problem with my equations is I seem to have to many unknowns.
  2. jcsd
  3. Nov 1, 2015 #2


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    Consider the moment at which the trolley starts to ascend.
    I see a few errors there. Please explain how you get those terms.
  4. Nov 1, 2015 #3
    the force of Ax projected onto the bar is Axcos60 and Ay projected onto the bar is Axsin60 or axcos30 then multiply them by perpendicular distance. Then multiply the components of Fg by their perpendicular distance. Although I see what you mean it looks like I am missing (491sin30)(0.4).

    Still I think I may have to many unknowns

    The new equation is now
    0 = -2Axcos60(0.1) +2 Aysin60(0.1) - (491cos30)(0.6) + (491sin30)(0.4) (equation 3)
  5. Nov 1, 2015 #4


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    If you consider my first comment in post #2 you will see that Ay is zero.
    Check the sign, and how do you get 0.4?
    Note you have a cos and a sine term for the 491, but only a cos term for the Ax. Does that suggest anything?
  6. Nov 1, 2015 #5
    Note: I just noticed your first statement and now I have some work to do. This eliminates Ay! one of my unknowns!
  7. Nov 1, 2015 #6
    Excellent, thank you so much, haruspex, I was able to get the correct answers. Your suggestions were all correct. I appreciate the guidance!
  8. Nov 2, 2015 #7


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    Well done.
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