Given Information:
The given expressions are as follows,
F(x)=1.6x−0.045x2θ(x)=0.8+0.125x−0.009x2+0.0002x3
Work done in integral form (Refer Sec. 24.4)
W=∫x0xnF(x)dx
If the direction between the force and displacement changes between initial and final position, then the work done is written as,
W=∫x0xnF(x)cos[θ(x)]dx …… (1)
Here, θ(x) is the angle between force and displacement.
Formula Used:
Multiple Segment Trapezoidal Rule.
In=(b−a)f(x0)+2∑i=1n−1f(xi)+f(xn)2(n)
Calculation:
Calculate the work done.
Substitute the value of F(x) and θ(x) in equation (1)
W=∫030(1.6x−0.045x2)cos(0.8+0.125x−0.009x2+0.0002x3)dx
Apply 4-Segment Trapezoidal rule.
Calculate the value of △x, when n=4,xn=30 and x0=0,
△x=xn−x0n=30−04=7.5
Divide the interval from 0 to 30, and △x=7.5,
So, the value of x after each iteration is x=0,7.5,15,22.5 and 30.
Calculate F(0) for x=0,
F(0)=1.6(0)−0.045(0)2=0
Calculate θ(0) for x=0,
θ(0)=0.8+0.125(0)−0.009(0)2+0.0002(0)3=0.8
Calculate F(0)⋅cos(θ(0)) for x=0,
F(0)⋅cos(θ(0))=0⋅cos(0.8)=0
Calculate F(x) for x=7.5,
F(7.5)=1.6(7.5)−0.045(7.5)2=9.46875
Calculate θ(7.5) for x=7.5,
θ(7.5)=0.8+0.125(7.5)−0.009(7.5)2+0.0002(7.5)3=1.315625
Calculate F(7.5)⋅cos(θ(7.5)) for x=7.5,
F(7.5)⋅cos(θ(7.5))=9.46875×cos(1.315625)=2.39002
Calculate F(x), for x=15,
F(15)=1.6(15)−0.045(15)2=13.875
Calculate θ(15) for x=15,
θ(15)=0.8+0.125(15)−0.009(15)2+0.0002(15)3=1.325
Calculate F(15)⋅cos(θ(15)) for x=15,
F(15)⋅cos(θ(15))=13.875×cos(1.325)=3.37619
Similarly, calculate for x=22.5 and 30 thentabulate the values as shown below.
xF(x)θ(x)F(x)cos(θ(x))000.807.59.468751.3156252.390021513.8751.3253.3761822.513.218751.3343753.09616307.51.85−2.06692
Calculate the solution using Trapezoidal rule,
W=△x2(F(0)⋅cos(θ(0))+2F(7.5)⋅cos(θ(7.5))+2F(15)⋅cos(θ(15))+2F(22.5)⋅cos(θ(22.5))+F(30)⋅cos(θ(30)))
Substitute function values from above table for △x=7.5.
W=7.52(0+2⋅2.39002+2⋅3.37618+2⋅3.09616−2.06692)=58.71675
Hence, the value of W after application of 4-segment Trapezoidal rule is W=58.71675.
Apply 8-Segment Trapezoidal rule.
Calculate the value of △x, when n=8,xn=30 and x0=0,
△x=xn−x0n=30−08=3.75
Divide the interval from 0 to 30, and △x=3.75,
So, the value of x after each iteration is x=0,3.75,7.5,11.25,15,18.75,22.5,26.25 and 30.
Calculate F(0) for x=0,
F(0)=1.6(0)−0.045(0)2=0
Calculate θ(0) for x=0,
θ(0)=0.8+0.125(0)−0.009(0)2+0.0002(0)3=0.8
Calculate F(0)⋅cos(θ(0)) for x=0,
F(0)⋅cos(θ(0))=0⋅cos(0.8)=0
Calculate F(3.75) for x=3.75,
F(3.75)=1.6(3.75)−0.045(3.75)2=5.3671875
Calculate θ(3.75) for x=3.75,
θ(3.75)=0.8+0.125(3.75)−0.009(3.75)2+0.0002(3.75)3=1.152734375
Calculate F(3.75)⋅cos(θ(3.75)) for x=3.75,
F(3.75)⋅cos(θ(3.75))=5.3671875×cos(1.152734375)=2.1790249
Calculate F(x) for x=7.5,
F(7.5)=1.6(7.5)−0.045(7.5)2=9.46875
Calculate θ(7.5) for x=7.5,
θ(7.5)=0.8+0.125(7.5)−0.009(7.5)2+0.0002(7.5)3=1.315625
Calculate F(7.5)⋅cos(θ(7.5)) for x=7.5,
F(7.5)⋅cos(θ(7.5))=9.46875×cos(1.315625)=2.39002
Similarly, calculate x=11.25,15,18.75,22.5,26.25 and 30 then tabulate all solutions as shown below.
xF(x)θ(x)F(x)cos(θ(x))000.803.755.36718751.1527342.179027.59.468751.3156252.3900211.2512.30468751.35195352.671351513.8751.3253.3761818.7514.17968751.29804683.8197322.513.218751.3343753.0961626.2510.99218751.49726560.807535307.51.85−2.06692
Calculate the solution using Trapezoidal rule,
W=△x2(F(0)⋅cos(θ(0))+2F(3.75)⋅cos(θ(3.75))+2F(7.5)⋅cos(θ(7.5))⋯+2F(26.25)⋅cos(θ(26.25))+F(30)⋅cos(θ(30)))
Substitute function values from above table △x=3.75.
W=3.752(0+2⋅2.17902+2⋅2.39002+2⋅2.67135+2⋅3.37618+2⋅3.81973+2⋅3.09616+2⋅0.807535−2.06692)=64.8995062
Hence, the value of W after application of 8-segment Trapezoidal rule is W=64.8995062
Apply 16-Segment Trapezoidal rule.
Calculate the value of △x, when n=8,xn=30 and x0=0,
△x=xn−x0n=30−016=1.875
Divide the interval from 0 to 30, and △x=1.875,
So, the value of x after each iteration is,
x=(0,1.875,3.75,5.625,7.5,9.375,11.25,13.125,15,16.875,18.75,20.625,22.5,24.375,26.25,28.125 and 30).
Calculate F(0) for x=0,
F(0)=1.6(0)−0.045(0)2=0
Calculate θ(0) for x=0,
θ(0)=0.8+0.125(0)−0.009(0)2+0.0002(0)3=0.8
Calculate F(0)⋅cos(θ(0)) for x=0,
F(0)⋅cos(θ(0))=0⋅cos(0.8)=0
Calculate F(1.875) for x=1.875,
F(1.875)=1.6(1.875)−0.045(1.875)2=2.841796875
Calculate θ(1.875) for x=1.875,
θ(1.875)=0.8+0.125(1.875)−0.009(1.875)2+0.0002(1.875)3=1.004052734375
Calculate F(1.875)⋅cos(θ(1.875)) for x=1.875,
F(1.875)⋅cos(θ(1.875))=2.841796875×cos(1.004052734375)=1.525725560
Calculate F(3.75) for x=3.75,
F(3.75)=1.6(3.75)−0.045(3.75)2=5.3671875
Calculate θ(3.75) for x=3.75,
θ(3.75)=0.8+0.125(3.75)−0.009(3.75)2+0.0002(3.75)3=1.152734375
Calculate F(3.75)⋅cos(θ(3.75)) for x=3.75,
F(3.75)⋅cos(θ(3.75))=5.3671875×cos(1.152734375)=2.1790249
Similarly, calculate x=(5.625,7.5,9.375,11.25,13.125,15,16.875,18.75,20.625,22.5,24.375,26.25,28.125 and 30) then tabulate all the solutions as shown below.
xF(x)θ(x)F(x)cos(θ(x))000.801.8752.84179681.00405271.525723.755.36718751.1527342.179025.6257.57617181.2539552.360487.59.468751.3156252.390029.37511.0449211.3456542.4657211.2512.30468751.35195352.6713513.12513.2480461.3424312.999161513.8751.3253.37618 16.87514.1855461.307568 3.69106 18.7514.17968751.29804683.8197320.62513.8574211.30434573.6487822.513.218751.3343753.0961624.37512.2636711.3960442.1322026.2510.99218751.49726560.80753528.1259.4042961.645947−0.706077307.51.85−2.06692
Calculate the solution using Trapezoidal rule,
W=△x2(F(0)⋅cos(θ(0))+2F(1.875)⋅cos(θ(1.875))+2F(3.75)⋅cos(θ(3.75))⋯+2F(28.125)⋅cos(θ(28.125))+F(30)⋅cos(θ(30)))
Substitute function values from above table for △x=1.875.
W=1.8752(0+2⋅1.52572+2⋅2.17902+2⋅2.36048+2⋅2.39002+2⋅2.46572+2⋅2.67135+2⋅2.99916+2⋅3.37618+2⋅3.69106+2⋅3.81973+2⋅3.64878+2⋅3.09616+2⋅2.13220+2⋅0.807535+2⋅(−0.706077)−2.06692)=66.41920875
Hence, the value of W after application of 16-segment Trapezoidal rule is W=66.419208.