Virtual work principle: equilibrium state of a system

In summary, the problem is that the author does not understand the virtual work principle and how to apply it.
  • #1
skrat
748
8

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##.
Capture.PNG

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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  • #2
skrat said:

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##.
View attachment 96810

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.
 
  • #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.
 
  • Like
Likes skrat

1. What is the virtual work principle?

The virtual work principle is a fundamental concept in mechanics that states that for a system to be in equilibrium, the virtual work done by external forces must be equal to zero.

2. How is the virtual work principle applied in real-world systems?

The virtual work principle is applied in real-world systems by considering the forces acting on the system and determining if they are in equilibrium, meaning that the virtual work done by these forces is equal to zero.

3. What is the significance of the virtual work principle in mechanics?

The virtual work principle is significant in mechanics because it allows us to analyze the equilibrium state of a system without having to consider the actual displacements of the system. This greatly simplifies the analysis of complex systems.

4. How is the virtual work principle related to the principle of least action?

The virtual work principle is closely related to the principle of least action, which states that a system will always follow the path of least resistance to reach a state of equilibrium. The virtual work principle is a mathematical representation of this concept.

5. Can the virtual work principle be applied to non-mechanical systems?

Yes, the virtual work principle can be applied to non-mechanical systems such as electrical and thermal systems. In these cases, the forces are replaced by their respective analogues (voltage, temperature difference) and the principle still holds true.

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