Kirchhoff's Rules and Spring-Mass Systems in Engineering Problems Assignment

Assignment Task

A common problem in electrical engincenng involves determining the currents and voltages at various locations in resistor circuits. These problems are solved using Kirchhoff's current and voltage rules. The current (or point) rule states that the algebraic sum of all currents entering a node must be zero or di =0

Σ i = 0

where all current entering the node is considered positive in sign. The current rule is an application of the punciple of conservation of charge. The voltage (or loop) rule specifies that the algebraic sum of the potential differences (that is, voltage changes) in any loop must equal zero. For a resistor circuit, this is expressed as ye -Yir =0

Σe - Σ iR = 0

where € is the en (electromotive force) of the voltage sources and A is the resistance of any resistors on the loop. Note that the second term derives from Ohm's law, which states that the voltage drop across an ideal resistor ts equal to the product of the current and the resistance. Kirchhoff's voltage rule is an expression of the conservation of energy.

Idealized spring-mass systems play an important role in mechanical and other engineering problems. Figure 12.11 shows such a system. After they are released, the masses are pulled downward by the force of gravity. Notice that the resulting displacement of each spring in Fig. 12.116 is measured along local coordinates referenced to its initial position in Fig. 12.11.

Newton's second law can be employed in conjunction with force balances to develop a mathematical model of the system. For each mass, the second law can be expressed as

To simplify the analysis, we will assume that all the springs are identical and follow Hooke's law. A free- body diagram for the first mass is depicted in Fig. 12.12. The upward force is merely a direct expression of Hooke's law:

The downward component consists of the two spring forces along with the action of gravity on the mass,

Note how the force component of the two springs is proportional to the displacement of the second mass, corrected for the displacement of the first mass, X. Now, the original differential equation becomes

Thus, we have derived a second-order ordinary differential equation to describe the displacement of the first mass with respect to time. However, notice that the solution cannot be obtained because the model includes a second dependent variable. Consequently, free-body diagrams must be developed for the second and the third masses that can be employed to derive.

The three differential equations above form a system of three differential equations with three unknowns We can obtain the displacements that occur when the system eventually comes to rest, that is, to the steady state. To do this, the derivatives are set to zero to give. 

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