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Virtual work principle: equilibrium state of a system

  1. Mar 4, 2016 #1
    1. The problem statement, all variables and given/known data
    We have a system (see attached graphics) of two rods with length ##l## and mass ##m## and some external force ##F##. The coefficients ##k## of the two springs are given. The springs have no deformation when ##\varphi =0##.
    a) Find generalized force of the system.
    b) Determine the equilibrium state if ##F=0##.
    Capture.PNG
    2. Relevant equations


    3. The attempt at a solution
    Firstly we have to determine the origin of our cartesian coordinate system. I went for the center point as shown in the graphics above.

    Than I identified all the forces in the system: $$F_0=(0,F)$$ $$F_1=(-kl(1-\cos \varphi),0)$$ $$F_2=(kl(1-\cos \varphi),0)$$ and lastly $$F_4=F_5=(0,-mg)$$
    Than vectors $$r_0=(0,l\sin \varphi)$$ $$r_1=(-l\cos \varphi ,0)$$ $$r_2=(l\cos \varphi ,0)$$ $$r_3=(-\frac l 2 \cos \varphi, \frac l 2 \sin \varphi)$$ and $$r_4=(\frac l 2 \cos \varphi, \frac l 2 \sin \varphi)$$
    Now in order to use the virtual work principle ##\delta W =\sum _i F_i\delta r_i## one has to firstly calculate all the ##\delta r_i##: $$\delta r_0=(0,l\cos \varphi)d\varphi $$ $$\delta r_1=(l\sin \varphi ,0)d\varphi$$ $$\delta r_2=(-l\sin \varphi ,0)d\varphi$$ $$\delta r_3=(\frac l 2 \sin \varphi, \frac l 2 \cos \varphi)d\varphi$$ and $$\delta r_4=(-\frac l 2 \sin \varphi, \frac l 2 \cos \varphi)d\varphi$$
    Now following the virtual work principle: $$\delta W=l\Big [(F-mg)\cos\varphi +kl(\sin(2\varphi)-2\sin\varphi)\Big ]d\varphi$$
    a) Therefore the generalized force is $$Q=l\Big [(F-mg)\cos\varphi +kl(\sin(2\varphi)-2\sin\varphi)\Big ]$$
    b) In order to find the equilibrium state we take ##\delta W=0## for ##F=0##. That leads me to $$mg \cos \varphi +2kl \sin \varphi -2kl \sin \varphi \cos \varphi =0$$

    The problem is that I have no idea what to do now. Therefore I kinda think I did something wrong. Does anybody know what is wrong here or how to continue?

    ps.: Sorry for the wrong part of this forum - this is probably an "introduction" problem, not advanced.
     
    Last edited: Mar 4, 2016
  2. jcsd
  3. Mar 5, 2016 #2
     
  4. Mar 5, 2016 #3
    Hi!
    First,i divide both side of the equation by 2kl,then define a to be equal to mg/2kl,so the equation would be as follows,
    acos(x)+sin(x)-{cos(x).sin(x)}=0,in which i replaxe the fi notation by "x". Now by taking {cos(x).sin(x)}to the right then squaring both sides
    we have, (a^2)((cos(x))^2) +((sin(x))^2) +2acos(x)sin(x)=(cos(x))^2)((sin(x))^2).Here from trigonometry we know
    cos(x))^2=1-((sin(x))^2),so replace all "cos(x))^2" terms by 1-((sin(x))^2),then divide both sides by sin(x).Now we have onlyone term with
    cosine,then take the other terms to the another side and again squar both sides and replace the cos(x))^2 by1-((sin(x))^2).Finally we have
    an equation which contains onle sin(x).I hope it be easily solvable.
    Best,
     
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