# Virtual work principle: equilibrium state of a system

1. Mar 4, 2016

### skrat

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 $F=0$.

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. Mar 5, 2016

3. Mar 5, 2016

### Evin Baxter

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,