Help For the system show, the input is the applied and the output is the tensile

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For the system show, the input is the applied and the output is the tensile force in the spring K2 The lever is ideal and is horizontal when the system is in static equilibrium with fa(t) = 0 and mass m supported by the spring K1. The displacements x1, x2, x3 and theta are measured with respect to the equilibrium position. The lever angle theta remains small. Select a suitable set of state variables and write the corresponding state-variable model (i,e., x1, v2, and theta) Find the output equation y for the reaction force of the pivot on the lever. A block diagram which describes the behavior of the system shown. Use MATLAB'S Simulink to produce the complete response .x(t) of the system. You make take liberty with the values of the system to complete this part of the problem. The figure shows a pendulum mass mp suspended from another mass mc (cart) by means of a light rod of length L. The cart rests on a frictionless surface, bill is "somewhat" restrained by springs k1 and k2 attached to cart's the left and right sides, respectively. The initial conditions for the system are:

Explanation / Answer

THIS SOLVED EXAMPLE WILL HELP YOU

nitially, 50 lbs of salt is dissolved in a tank containing 300 gallons of water. A salt solution with 2 lb/gal concentration is poured into the tank at 3 gal/min. The mixture, after stirring, flows from the tank at the same rate the brine is entering the tank. In all of the problems below, we measure t in minutes.
A)Find the amount of salt Q (in lbs) in the tank as a function of time t.
B)Determine the concentration C of salt (in lbs/gal) in the tank as a function of time t.
C)Determine the steady-state amount of salt (in lbs) in the tank.
D)Find the steady-state concentration (in lbs/gal) of the salt in the tank.

A)

(accumulation, lb/min) = (rate in, lb/min) - (rate out, lb/min)

dQ/dt = 2*3 - (Q/300)*3 = 6 - Q/100

dQ/(6-Q/100) = dt

-100*ln(6-Q/100) = t + D

ln(6-Q/100) = -t/100 + C

6-Q/100 = B*exp(-t/100)

600-Q=A*exp(-t/100)

Q = 600-A*exp(-t/100)

initial condition ==> t=0, Q=50 ==> A=550

Q = 600-550*exp(-t/100) (lb)

B)

C = Q/300

C = (600-550*exp(-t/100))/300

C = 2-11/6*exp(-t/100) (lb/gal)

C)

steady-state ==> dQ/dt=0

dQ/dt = 6 - Q/100

0 = 6 - Q/100

Q = 600 lb

D)

steady-state ==> Q=600 (from part C)

C = Q/300

C = 600/300

C = 2 lb/gal

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