- #1
skrat
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Homework Statement
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##.
Homework Equations
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.
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