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VIRTUAL WORK EXAMPLE

figure pendulum.png
Figure 1 Teeter-totter system with two masses

Problem: Use the principle of virtual work to determine the equilibrium equation for the system shown in Fig. 1↑.
Solution: There are two masses in the system. Consider the unit vector to be pointing down in the figure and pointing to the right. The total forces acting on the two masses are:
(1) F1  = (m1g − kx1) F2  = m2g
The positions of the masses m1 and m2 are
(2) σ1  = x1 + y1 σ2  = x2 + y2
The virtual work done by the external forces is
(3) δW  = 2i = 1 Fiδσi  = (m1g − kx1)(δx1 + δy1) + m2g(δx2 + δy2)  = (m1g − kx1)δx1 + m2gδx2
At this point, the coordinates x1 and x2 are written in terms of the angle θ:
(4) x1  = l1sinθ x2  =  − l2sinθ
The virtual displacement δθ is the variation of θ, so the virtual displacements δx1 and δx2 can be written in terms of δθ by taking the variation of Eq. .
(5) δx1  = l1cosθδθ δx2  =  − l2cosθδθ
Substituting these expressions into the virtual work, the result is
(6) δW = (m1g − kl1sinθ)l1cosθδθ − m2gl2cosθδθ
The principle of virtual work states that the virtual work done by all external forces in a system at equilibrium is zero, provided the virtual displacements are consistent with the constraints and that the constraint forces act perpendicular to the virtual displacements. Therefore, from the principle of virtual work, Eq. 6↑ becomes
(7) (m1g − kl1sinθ)l1cosθδθ − m2gl2cosθδθ = 0
or
(8) [(m1g − kl1sinθ)l1 − m2gl2]δθ = 0
However, since the virtual displacement δθ is arbitrary, the coefficient of δθ in Eq. 8↑ must vanish, so
(9) (m1g − kl1sinθ)l1 − m2gl2 = 0
After some algebra, the equilibrium value of θ is determined:
(10) θ = sin − 1(g)/(kl1)m1 − m2(l2)/(l1)