User:Egm6321.f12.team7.kong/report1

=Problem 3=

Given
The expression of $$c_{0} ( Y^{1},t)$$ is given by, $$\displaystyle c_{0} ( Y^{1},t)=-F^1(1-\bar{R}u^2_{,SS})-F^2u^2_{,S}-\frac{T}{R}+M[(1-\bar{R}u^2_{,SS})(u^1_{,tt}-\bar{R}u^2_{,Stt})+u^2_{,S}u^2_{tt}]$$


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(3.1)
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Find
Analyze the dimension of each term and provide the physical meaning

Dimensional analysis of each term
Left hand side:

$$\displaystyle [c_{0} ( Y^{1},t)]=F $$
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(3.2)
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Right hand side:


 * First term

$$\displaystyle [ F^{1}]=F$$
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(3.3)
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$$\displaystyle [ 1]=1$$
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(3.4)
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$$\displaystyle [ \bar{R}]=L$$
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(3.5)
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$$\displaystyle [ u^2]=L\Rightarrow [u^2_{,SS}]=\frac{L}{L^2}=L^{-1}$$
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(3.6)
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$$\displaystyle [\bar{R}u_{,SS}^{2}]=L\cdot L^{-1}=1$$
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(3.7)
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$$\displaystyle [F^1(1-\bar{R}u^2_{,SS})]=F\cdot 1=F $$
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(3.8)
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 * Second term

$$\displaystyle [F^2]=F$$
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(3.9)
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$$\displaystyle [u^2]=L\Rightarrow [u^2_{,S}]=\frac{L}{L}=1$$
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(3.10)
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$$\displaystyle [F^2u^2_{,S}]=F\cdot{1}=F$$
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(3.11)
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 * Third term

$$\displaystyle [T]=F\cdot{L}$$
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(3.12)
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$$\displaystyle [R]=L$$
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(3.13)
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$$\displaystyle [\frac{T}{R}]=\frac{F\cdot{L}}{L}=F$$
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(3.14)
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 * Fourth term

$$\displaystyle [M]=M$$
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(3.15)
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$$\displaystyle [1-\bar{R}u^2_{,SS}]=1$$
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(3.16)
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$$\displaystyle [u^1_{,tt}]=\frac{L}{T^2}=LT^{-2}$$
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(3.17)
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$$\displaystyle [u^2_{,Stt}]=\frac{L}{L\cdot{T^2}}=T^{-2}\Rightarrow[\bar{R}u^2_{,Stt}]=LT^{-2}$$
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(3.18)
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$$\displaystyle [(1-\bar{R}u^2_{,SS})(u^1_{,tt}-\bar{R}u^2_{,Stt})]=1\cdot{L}\cdot{T^{-2}}=LT^{-2}$$
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(3.19)
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$$\displaystyle [u^2_{,S}]=\frac{L}{L}=1$$
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(3.20)
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$$\displaystyle [u^2_{,tt}]=\frac{L}{T^2}=LT^{-2}$$
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(3.21)
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$$\displaystyle [u^2_{,S}u^2_{,tt}]=1\cdot{L}\cdot{T^{-2}}=LT^{-2}$$
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(3.22)
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$$\displaystyle [M[(1-\bar{R}u^2_{,SS})(u^1_{,tt}-\bar{R}u^2_{,Stt})+u^2_{,S}u^2_{tt}]]=M\cdot{L}\cdot{T^{-2}}=F$$
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(3.23)
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Physical meaning
$$ c_0$$ is the horizontal force which act on wheel/magnet of the train. Other terms are forces from different source.