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Homework Help: Static equilbrium with moments

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the component forces Az, Ay, Bx, By, Cx, Cz:

    http://phga.pearsoncmg.com/onekey/web/hibbelerSD10e/Public_Html/chapter05/H-4ed-5-92a-fbd.gif

    2. Relevant equations

    The downward force at the origin is 250 lb and the moment is 25 ft-lb

    3. The attempt at a solution

    Well first of all I know it's in equilibrium so the sum of all the components must equal zero. What I don't know is how to deal with the distribution of forces for the Az Cz for example and how to deal with the moment. Help would be greatly appreciated....
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Feb 10, 2009 #2

    nvn

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    thunderbird: You have three unlabeled vectors, and you described only two of them. Please describe the other vector. Afterwards, in statics, you have six equilibrium equations, three of which are summation of moment. Start writing out the six equilibrium equations, and solve simultaneously for the six unknowns. For the three moment equilibrium equations, you can choose any point you wish for summing moments. Just start writing, list the relevant equations, and show your work. Don't worry about the distribution of forces; that will all work itself out in the solution.
     
  4. Feb 10, 2009 #3
    Which is the third unlabeled vector?
     
  5. Feb 10, 2009 #4

    nvn

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    See that vector at the tip of the uppermost member? Can you describe it? Or am I misinterpreting something?
     
  6. Feb 10, 2009 #5
    It's just to show the direction of the moment. The counterclockwise arrow and the arrow going through it are the same vector.
     
  7. Feb 10, 2009 #6

    nvn

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    Are you sure? Moment vectors are usually never drawn that way. I will say, the best way to draw moment vectors is using a double arrowhead representation (unless it is pointing normal to the page, in which case the circular arrow is better). Did you draw this diagram? Or was it given that way, with nothing labeled nor mentioned about that third vector?
     
  8. Feb 11, 2009 #7
    I'm sure. I was given it and all the moments in the book are given that way. That's the only information I was given as well...
     
  9. Feb 11, 2009 #8

    nvn

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    OK, thanks for the clarification. So, try the approach in post 2.
     
  10. Feb 12, 2009 #9
    Here is the original picture before the FBD:

    The bent rod is supported at A, B and C by smooth bearings. Compute the x, y, z components of reaction at the bearings if the rod is subjected to a 250-lb vertical force and a 25-lb•ft couple as shown. The journal bearings are in proper alignment and exert only force reactions on the rod.

    http://phga.pearsoncmg.com/onekey/web/hibbelerSD10e/Public_Html/chapter05/9e-5-92a.gif

    So...

    x:
    Bx+Cx+0=0

    y:
    Ay+By+0=0

    z:
    Az+Cz+250=0

    I know there should be three more equations, but I have no idea how to formulate them...
     
    Last edited by a moderator: Apr 24, 2017
  11. Feb 12, 2009 #10

    nvn

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    Your z-direction force equation is incorrect. You currently ignored the fact that the applied force is in the negative direction. I mention the three other equations in post 2. Sum moments about any point you wish, for each coordinate system direction. Moment is force times perpendicular distance to the axis of rotation you are summing moments about; also add to this any applied moment components. If you are not familiar with how to sum moments, read a few pages in your text book, and study a couple of example problems.
     
  12. Feb 12, 2009 #11
    Ok so

    So...

    x:
    Bx+Cx+0=0

    y:
    Ay+By+0=0

    z:
    Az+Cz-250=0

    I'm not really sure how to break the moment that is given up into components...

    Do you have any tips as to which points I should sum my moments about?

    If I had my text this would be a lot easier... Unfortunately it was left in checked luggage which my airline lost and won't be getting to me in Florida... And this problem needs to be done by tomorrow...

    I really need help on this one :S
     
  13. Feb 12, 2009 #12

    nvn

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    Let the applied moment at the upper member tip be a double arrowhead vector where the single arrowhead vector is currently shown. Now break this double arrowhead vector into components using trigonometry. Sum moments, e.g., about each coordinate system axis passing through the origin.
     
  14. Feb 13, 2009 #13

    cjl

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    The easiest way to sum moments is to start by picking a point - you could sum them about the origin, you could sum them about one of the given points, or you could just use some point at random. It doesn't matter mathematically. It can however make it easier if you pick the right points. I like to use one of the supporting points, as that allows some forces to drop out completely (they will have a moment arm of zero). Once you have the point to sum moments about, then you can get your 3 equations by summing the moments about all 3 axes. This will give you the 6 equations you need to solve for your 6 unknowns.
     
  15. Feb 13, 2009 #14
    Ok so I did the problem and got all of the answers wrong...

    Still need help. Here are my equations and answers:

    Force equilibrium:

    Bx+Cx=0
    Ay+By=0
    Az+Cz=250

    Moment Equilibrium:

    x-axis: Cz=By
    y-axis: Bx+3Cx+(12.5)*(3+2*2^-2)=2Az+(12.5)*(3-2*2^-2)
    z-axis: Cx+(12.5)*(3-2*2^-2)=2Ay+(12.5)*(3+2*2^-2)

    And my answers:

    Ay=236.7
    Az=486.7
    Bx=-482.3
    By=-236.7
    Cx=482.3
    Cz=-236.7

    CAN SOMEONE PLEASE LET ME KNOW WHERE I WENT WRONG ASAP. THIS IS DUE IN 6 HOURS!
     
  16. Feb 13, 2009 #15
    As I mentioned earlier I HAVE NO IDEA how to break the moments into components properly... I don't think I'm doing it right.
     
  17. Feb 13, 2009 #16

    cjl

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    OK, just looking at that, there's definitely something wrong there. What point were you trying to sum the moments around?

    If you try to sum them around point B for example (a decent choice - as I said, you want to pick one of the points where force is applied if possible, as then all the forces applied at that point become irrelevant), the equation for the moment around the z-axis is as follows:
    2*Ay-Cx=25*sin(45o)

    This equation comes from the fact that around point B, the moment that the forces from A will exert around the Z axis only depend on the force at A and the perpendicular moment arm. The perpendicular distance that A is from the Z axis is 2 feet, and the component that will exert the moment is the Y component. By the right hand rule, a force in the positive Y direction at A will exert a positive Z moment. Therefore, the component of the Z moment caused by A is equal to 2*Ay. This same process shows that the C component is equal to -1*Cx.

    The reason for the 25*sin(45o) component on the other side of the equation is that the applied external moment is at an angle of 45 degrees, and as theta approaches zero, the z moment from theta would also approach zero. Therefore, the sin function is the correct one to use.

    Repeat this process for the X and Y moments around point B, and you should have the correct equations (remember that for the X and Y moments, point B is 1 foot above the X and Y axes, therefore the correct perpendicular distance to use for the moment arm isn't the one to the axis, but rather the one to a line parallel to the axis that goes through point B).
     
  18. Feb 13, 2009 #17

    nvn

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    thunderbird: You did excellent work and almost got the right answer. Everything you did is exactly correct except for the way you handled the applied moment term. Remember, as mentioned in posts 6 and 12, the best way to draw moment vectors is using a double arrowhead representation. Let the applied moment at the upper member tip be a double arrowhead vector where the single arrowhead vector is currently shown. Now break this double arrowhead vector into components using trigonometry, to obtain My and Mz.

    As mentioned in post 10, add My to your y-axis moment summation equation. Add Mz to your z-axis moment summation equation. Remember, moments My and Mz are already moments; therefore, you do not multiply moment by a moment arm. Simply add My and Mz to the respective y-axis and z-axis moment summation equations.
     
  19. Feb 13, 2009 #18

    nvn

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    cjl: Your equation in post 16 is currently incorrect. Check your algebra.
     
  20. Feb 13, 2009 #19

    cjl

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    Really? Where is it wrong? I'm not immediately seeing the error...
     
  21. Feb 14, 2009 #20
    I was trying to sum the moments around the origin. Thanks for the help, I will try your advice...
     
  22. Feb 14, 2009 #21
    I'm sorry nvn, I'm not sure what a double arrowhead vector is. I treated it as two equal forces in opposite directions each of magnitude 12.5 lb at 0.5ft from the vector...
     
  23. Feb 15, 2009 #22

    cjl

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    That explains the extra terms you had. As for your method of treating it as two equal forces, it should be valid mathematically. The only problem is that two forces each 12.5lb 0.5 feet from the vector will not apply the proper moment. Each one will apply 12.5*0.5, or 6.75 pound feet, for a total applied moment of 12.5 lb-ft. You should change that to be either 2 equal forces of 25 pounds 0.5 feet from the vector, or 2 equal 12.5 pound forces 1 foot from the vector.
     
  24. Feb 15, 2009 #23

    nvn

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    thunderbird: To learn about the double arrowhead representation of a moment vector, which is much better and easier for your problem, see items 3.5 and especially 3.6 at the bottom of http://books.google.com/books?id=idWo2Sw_488C&pg=PA72".

    cjl: Hint: The mistake you made in your post 16 equation is the most common mistake in algebra, and happens to everyone from time to time. The fact that thunderbird did not make any of this particular kind of mistake is impressive, and indicates thunderbird is quite good in algebra.

    thunderbird: If you use other approaches to represent and compute the applied moment at the upper member tip, it is correct only if it produces the same, correct answer. The correct answer begins with Ay = 223.4835 lbf. Please correct me if I am wrong. cjl keeps referring to summing moments about lines passing through a different point than what you used, without realizing yet that the way you already did that part of the problem is the same or similar, cancels out just as many terms, is an excellent approach, and does not produce extra terms. There is no need to change or redo what is already right and in simplest form. Just correct your applied moment in the equilibrium equations you already derived. Therefore, any part of this discussion (from cjl, you, me, whomever) that focuses on how you handle or compute the applied moment at the upper member tip is the only thing you need to focus on and correct.
     
    Last edited by a moderator: Apr 24, 2017
  25. Feb 15, 2009 #24

    cjl

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    You'll note that in post #22, I stated that his method was correct, aside from the incorrect value of his equivalent moment application (which resulted in half the actual moment). If thunderbird uses precisely that method but with the correct values for the equivalent forces, everything should work out properly.

    EDIT: Oh, there's the error. I was looking for a different kind of error than that (it figures that I would drop a sign - I'm usually pretty good about not doing that).

    The corrected equation for reply #16 should actually read as follows: 2*Ay-Cx=-25*sin(45o)

    Everything else in that post should be correct, although the method thunderbird is using is correct too, once the values are fixed.
     
    Last edited by a moderator: Apr 24, 2017
  26. Feb 15, 2009 #25
    The correct answer begins with Ay = 223.4835 lbf.

    Why is it lbf? Shouldn't it be lb?
     
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