Explanation Given: The weight of the object is 32 pound, stretching length of the spring is 2 feet and expression for impressed force is f ( t ) = 2 t for t is in between 0 to 5. Definition used: “(1) The Laplace transformation of a linear differential equation with constant coefficient is express as follows: a n L { d n y d t n } = a n [ s n Y ( s ) − s ( n − 1 ) y ( 0 ) − ⋯ − y ( n − 1 ) ( 0 ) ] (2) The linearity property of the Laplace transformation is as below. If a , b be any constant and f , g be any function of t , then L { a f ( t ) + b g ( t ) } = a L { f ( t ) } + b L { g ( t ) } . (3) The linear differential equation for the spring force system is as follows: m x ″ + k x = f ( t ) where, m is the mass and k is spring constant.” Calculation: The value of mass m is calculated as follows: m = W g = 32 32 = 1 The value of spring constant k are calculated as follows: W = k x 32 = k ⋅ 2 k = 16 The expression for the spring force f ( t ) is written as follows: f ( t ) = { 20 t , 0 ≤ t < 5 0 , t ≥ 5 The unit step function for the function f ( t ) is written as follows: f ( t ) = 20 t − 20 t U ( t − 5 ) f ( t ) = 20 t − 20 ( t − 5 ) U ( t − 5 ) − 100 U ( t − 5 ) Substitute the value of m , k and f ( t ) in the equation of spring force system m x ″ + k x = f ( t ) . x ″ + 16 x = 20 t − 20 ( t − 5 ) U ( t − 5 ) − 100 U ( t − 5 ) Use the definition (1) to write the Laplace transform of the above linear differential equation. s 2 L { x } − s x ( 0 ) − x ′ ( 0 ) + 16 L { x } = L { 20 t − 20 ( t − 5 ) U ( t − 5 ) − 100 U ( t − 5 ) } ( 1 ) Put the initial values x ( 0 ) = 0 and x ′ ( 0 ) = 0 in equation (1) and solve. s 2 L { x } + 16 L { x } = 20 s 2 − 20 s 2 e − 5 s − 100 s e − 5 s L { x } = 20 s 2 ( s 2 + 16 ) − 20 s 2 ( s 2 + 16 ) e − 5 s − 100 s ( s 2 + 16 ) e − 5 s The obtained equation is as follows: L { x } = 20 s 2 ( s 2 + 16 ) − 20 s 2 ( s 2 + 16 ) e − 5 s − 100 s ( s 2 + 16 ) e − 5 s ( 2 ) From equation (2) the term 20 s 2 ( s 2 + 16 ) is simplify as follows using partial fraction, 20 s 2 ( s 2 + 16 ) = a 0 s + a 1 s 2 + a 3 s + a 2 s 2 + 16 ( 3 ) Multiply with s 2 ( s 2 + 16 ) in both sides of equation (3). 20 s 2 ( s 2 + 16 ) s 2 ( s 2 + 16 ) = a 0 s 2 ( s 2 + 16 ) s + a 1 s 2 ( s 2 + 16 ) s 2 + s 2 ( a 3 s + a 2 ) ( s 2 + 16 ) s 2 + 16 20 = a 0 s ( s 2 + 16 ) + a 1 ( s 2 + 16 ) + s 2 ( a 3 s + a 2 ) The obtained equation is as follows: 20 = a 0 s ( s 2 + 16 ) + a 1 ( s 2 + 16 ) + s 2 ( a 3 s + a 2 ) ( 4 ) Put s = 0 in equation (4). 20 = a 0 ⋅ 0 ⋅ ( 0 2 + 16 ) + a 1 ( 0 2 + 16 ) + ( a 3 ⋅ 0 + a 2 ) ⋅ 0 2 20=16 a 1 a 1 = 5 4 Substitute the value of a 1 in equation (4). 20 = a 0 s ( s 2 + 16 ) + 5 4 ( s 2 + 16 ) + s 2 ( a 3 s + a 2 ) 20 = a 0 s 3 + 16 a 0 s + 5 4 s 2 + 20 + a 3 s 3 + a 2 s 2 20 = s 3 ( a 0 + a 3 ) + s 2 ( a 2 + 5 4 ) + 16 a 0 s + 20 Compare the similar coefficient of s on both sides of the above expression as follows: 16 a 0 = 0 5 4 + a 2 = 0 a 0 + a 3 = 0 Solve the above system of equation

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ISBN: 9781284105902

Author: Dennis G. Zill

Publisher: Jones & Bartlett Learning

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Chapter 4.3, Problem 71E

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The equation of the motion x(t) using Laplace transform and then use graphic utility to graph x(t) for interval [0,10].

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Chapter 4.3, Problem 70E

Chapter 4.3, Problem 72E

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A crate is supported by three cables as shown. Determine the weight of the crate knowing that the tension in cable AB is 750 lb

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Equilibrium Equations: Two-dimensional 2.28 Ra= Rc= 2.29 (change force to 600N) AC= AB= 2.33 CD= AC= DE= BC= Free Body Diagrams 2.34 Ax= Ay=_ Bx= By= 2.36 Ax= Ay= Bx= By= 2.37 (change middle force to 4000 lbs) Ay=_ Dx= Dy=_ 2.38 (change horizontal force to 2 kN) Ax= Ay= Bx= By=_ 2.40 Ay= By= Dx= 32% Bx= Cy= Dy=

Chapter 4 Solutions

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The equation of the motion x ( t ) using Laplace transform and then use graphic utility to graph x ( t ) for interval [ 0 , 10 ] .

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The equation of the motion x(t) using Laplace transform and then use graphic utility to graph x(t) for interval [0,10].

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Chapter 4 Solutions