The weightless bars AB and CE, together with the 5-lb weight BE, form a parallelogram linkage. The ideal spring attached to D has a free length of 2 in. and a stiffness of 7.5 lb/in. Find the two equilibrium positions that are in the range
The equilibrium positions.
Answer to Problem 10.60P
The system is at unstable equilibrium at 29.6o and at 53.1o the system comes in stable equilibrium.
Explanation of Solution
Given Information:
The weight of the load at BE = 5-lb
The stiffness of spring k = 7.5 lb/in
The following figure is given:
Calculation:
Consider the following figure:
To calculate the angle for the equilibrium position, let us calculate the potential energy of the system. The total potential energy of the system consists of potential energy of the weight (Vg) and the potential energy of the springs (Ve)
The total potential energy =
Putting the value of Yg and s in equation (1),
The potential energy of the system comes out to be
Now, let us take the first derivative of the total potential energy of the system,
Now, the principle of minimum potential energy can be used to find the value of angle
The roots of equation
Now, differentiating equation 3 again to get
Thus, system is at stable equilibrium at
Thus, system is at unstable equilibrium at
Thus, system is at stable equilibrium at
Conclusion:
Therefore, thesystem is at unstable equilibrium at 29.6o and at 53.1o the system comes in stable equilibrium.
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Chapter 10 Solutions
International Edition---engineering Mechanics: Statics, 4th Edition
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- International Edition---engineering Mechanics: St...Mechanical EngineeringISBN:9781305501607Author:Andrew Pytel And Jaan KiusalaasPublisher:CENGAGE L