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Homework Help: Statics problem disaster =/

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Determine the moment of the 50lb force about point A by resolving the force into horizontal and vertical components.


    2. Relevant equations

    I am using the book statics and strength of materials 2nd edition by Cheng, and its problem 2-31, i have racked my brain and cant figure this problem out, or atleast get it to match the answer in the back of the book which is 64.3 lb*ft clockwise

    3. The attempt at a solution
  2. jcsd
  3. Sep 13, 2007 #2
    My attempt at a solution, forgot that part, probably the most important part.


    M(A)= Fd
    d= tan40x2ft
    d =1.68ft

    M(A)= -84 lb*ft
    M(A) = 84 lb*ft clockwise
  4. Sep 13, 2007 #3


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    Moment about a point (axis) is force times perpendicular distance from the line of action of the force to the point of rotation. This is uually the best way to calculate moments, but for simplicity, please break up the 50 pound force into its x and y components at B, then take the moment of each force about A and sum them , watching plus and minus signs. The perpendicular disance of those forces from A can be calculated from the basic geometry of a 30- 60-90 triangle.
  5. Sep 13, 2007 #4
    You need to get the 50lb arm, at a 90 degree angle to the line AB....then take it times the moment arm....
  6. Sep 13, 2007 #5
    The Moment of a Force is M= r x F

    That would be |M|=r*F* sin(theta)

    Thetha is the smalles angle between the moment arm and the force, and as you can see on the sketch made up by jvk2002 is 40 degrees. Therefore the magnitud |M|= 2 ft * 50 lb*sin(40).

    Make this calculations and you will find that it is 64.3 lb-ft.
    Last edited: Sep 13, 2007
  7. Sep 13, 2007 #6
    All these calculation are to complicated and doesn't get you to the answer.

    You found the very easily the perpendicular vector of the force that is: 50 * sin(40) there was no reason to find the distance between the line of force and the point A.
  8. Sep 14, 2007 #7


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    Since this is a coplanar static problem you can solve it as above, but using the general procedure is

    [tex] \vec{F} = F_{x} \vec{i} + F_{y} \vec{j} [/tex]

    [tex] \vec{r_{AB}} = r_{x} \vec{i} + r_{y} \vec{j} [/tex]

    [tex] \vec{M} = \vec{r_{AB}} \times \vec{F} [/tex]

    The solution of course will be

    [tex] \vec{M} = M_{z} \vec{k} [/tex]
  9. Sep 14, 2007 #8
    Thanks link, and thank you everyone else who helped. i understand how to get the answer now, but i dont think i necessarily understand the logic behind it, I cant find an example problem in the book exactly like this one so that leaves me kind of lost.
  10. Sep 14, 2007 #9


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    There are a number of ways to calculate the moment, one of which is M=(r)(F) sin theta as noted above. This is what we are taught in Physics. However, the question specifically asks you to solve the moment by first breaking up the applied force into its x component (50cos20 = 47 to the right), and y component (50sin20 = 17.1 straight up), then calculate the moment of each of those component forces about A , which is force times perpendicular distance, i.e, using the properties of a 30-60-90 triangle,
    M = 47(sq. root 3) - 17.1(1) = 64.3 clockwise, which of course is the same result. I assune you are taking an engineeering statics course and not PHY 101, and that is why the problem asks you to calcylate the momemts by first btreaking up the applied force into it s components, because generally speaking , this is going to be the simplest way to solve statics problems in most cases.
  11. Sep 14, 2007 #10
    In the book they give me Varignon's theorem which is listed as M(o)=Fy(Xa)- Fx(Ya). Using this equation I get
    M(a) =17.1(1) - 47(1.73)
    M(a) = -64.2lb*ft
    M(a) = 64.2 lb*ft clockwise
  12. Sep 14, 2007 #11


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    Ok, I never heard of Mr. or Mrs. Varignon, but that theorem is correct; you have assumed negative as clockwise, so that's Ok, it's just a matter of convention whether you assume a clockwise momemt to be negative or positive, just stick with the convention once you do. The theorem essentialy just states that the moment of a force about point O is equal to the algebraic sum of the 'y component of the force times the perpendicular x distance to point 0' plus the 'x component of the force times its perpendicular y distance to point O', where a clockwise moment is considered negative, and a counterclockwise momet is considered positive.
  13. Sep 14, 2007 #12
    thanks alot i appreciate the help, i will be posting a problem later 2nite that i need help on even starting, im dumbfounded, just a tip in the right direction might get me started though =).
  14. Sep 14, 2007 #13
    For some reason I didn't notice that the first two post was made by you, sorry.

    Moment of a force or Torque is the distance times the force perpendicular to the moment arm. Ok, think about this, this force would make the body rotate about that point, example point A on your diagram. The other component of the force would accelerate point A lineraly.

    If you still having problems understanding the concept of torque, I think you should review the vector chapter of your book, maybe there is the problem.
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