### 数学代写|matlab仿真代写simulation代做|Contour Dynamic Force Analysis with D’Alembert Principle

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• Foundations of Data Science 数据科学基础

## 数学代写|matlab仿真代写simulation代做|Reaction Force F24

The revolute joint at $C$ between link 2 and the link 4 is replaced with the joint reaction force $\mathbf{F}{24}$ as shown in Fig. 5.2. The joint reaction force of the link 2 on the link 4 , $\mathbf{F}{24}$, acts at $C$ and has two unknown components $\mathbf{F}{24}=F{24 x} \mathbf{1}+F_{24 y} \mathbf{J}$.
The sum of the forces on link 4 is equal to zero, Fig. $5.2 \mathrm{a}$, or
$$\sum \mathbf{F}^{(4)}=\mathbf{F}{\text {in } 4}+\mathbf{G}{4}+\mathbf{F}{24}+\mathbf{F}{54}=\mathbf{0}$$
where $\mathbf{F}{54}$ is the reaction force of the slider 5 on the link 4 and is perpendicular to the sliding direction $C E$, i.e., $\mathbf{F}{54} \cdot \mathbf{r}{C E}=0$. Equation (5.5) is scalar multiplied with $\mathbf{r}{C E}$ and the following equation is obtained

Equation (5.6) does not contain the reaction $\mathbf{F}{54}$ because $\mathbf{F}{54} \cdot \mathbf{r}{C E}=0$. Next for the links 4 and 5 the sum of the moments with respect to the revolute joint $E$, Fig. $5.2 \mathrm{~b}$, gives $$\sum \mathbf{M}{E}^{(4 \& 5)}=\mathbf{r}{E C} \times \mathbf{F}{24}+\mathbf{r}{E C{4}} \times\left(\mathbf{F}{\mathrm{in} 4}+\mathbf{G}{4}\right)+\mathbf{M}{\mathrm{in} 4}+\mathbf{M}{\mathrm{in} 5}+\mathbf{M}{e}=\mathbf{0},(5.7)$$ where the external moment on slider 5 is $\mathbf{M}{e}=\mathbf{M}{\text {Sext }}$. Equation (5.7) has only a component on $z$-axis. Equations (5.6) and (5.7) represent a system of two equations with two unknowns $F{24 x}$ and $F_{24 y}$. The MATLAB commands for Eqs. (5.6) and (5.7) are:
$\mathrm{F} 24 \mathrm{x}=\operatorname{sym}$ (‘ $\mathrm{F} 24 \mathrm{x}^{\prime}$, ‘real’);
$\mathrm{F} 24 \mathrm{y}=8 y \mathrm{~m}\left(\mathrm{~F}^{24 y^{\prime} \text {, ‘real’); }}\right.$
F24_=[F24x，F24y, 0]; 뭉 unknown joint force
뭏 E_P: $^{\text {P }}$ (sumF4), CE $=0$
몽 (Find_+G4_+F24_) $\left(r E_{-}-r C_{-}\right)=0$
$\mathrm{SF} 4 \mathrm{CE}=\operatorname{dot}\left(\mathrm{Fin4_{- } + \mathrm { GA }}+\mathrm{F} 2 \mathrm{~A}{-}, \mathrm{rE} \mathrm{E}{-}-\mathrm{rC} \mathrm{C}{-}\right)$; 망 $\mathrm{E}{-} \mathrm{R}$ : sumM45E_ $=0_{-}$

## 数学代写|matlab仿真代写simulation代做|Reaction Force F23

The revolute joint at $D$ between link 2 and the link 3 is replaced with the joint reaction force $\mathbf{F}{23}$ as shown in Fig. 5.3. The joint reaction force of the link 2 on the link 3 , $\mathbf{F}{23}$, acts at $D$ and has two unknown components $\mathbf{F}{23}=F{23 x} \mathbf{1}+F_{23 y} \mathbf{J}$.
The sum of the forces on link 3 is equal to zero, Fig. 5.3a, or

$$\sum \mathbf{F}^{(3)}=\mathbf{F}{\text {in } 3}+\mathbf{G}{3}+\mathbf{F}{23}+\mathbf{F}{03}=\mathbf{0}$$
where $\mathbf{F}{03}$ is the reaction force of the ground on slider 3 and is perpendicular to the sliding direction $x$-axis, i.e., $\mathbf{F}{03} \cdot \mathbf{1}=0$. Equation $(5.8)$ is scalar multiplied with $\mathbf{1}$ and the following relation is obtained
$$\left(\sum \mathbf{F}^{(3)}\right) \cdot \mathbf{1}=\left(\mathbf{F}{\mathrm{in} 3}+\mathbf{G}{3}+\mathbf{F}{23}\right) \cdot \mathbf{1}=0$$ Next for the link 2 , Fig. $5.3 \mathrm{~b}$, the sum of the moments with respect to the revolute joint $B$ gives \begin{aligned} &\sum \mathbf{M}{B}^{(2)}= \ &\mathbf{r}{B D} \times\left(-\mathbf{F}{23}\right)+\mathbf{r}{B C} \times\left(-\mathbf{F}{24}\right)+\mathbf{r}{B C{2}} \times\left(\mathbf{F}{\text {in } 2}+\mathbf{G}{2}\right)+\mathbf{M}{i n 2}=\mathbf{0} \end{aligned} From Eqs. (5.9) and $(5.10)$ the two unknowns $F{23 x}$ and $F_{23 y}$ are calculated. The MATLAB commands for the reaction $\mathbf{F}{23}$ are: $\mathrm{F} 23 \mathrm{x}=$ sym (‘F23x’, ‘real’); $\mathrm{F} 23 \mathrm{y}=\operatorname{sym}\left(\mathrm{F}{2} 3 \mathrm{y}^{\prime}\right.$, ‘real’);
$\mathrm{F23}=[\mathrm{F23x}, \mathrm{F23y}, 0]$; 형 unknown joint force
황 $\mathrm{D}{-} \mathrm{P}: \operatorname{sumF} 3-(1)=0$ 당 $\left{\mathrm{Fin} 3{-}+\mathrm{G} 3_{-}+\mathrm{F} 2_{3}\right)(1)=0$
$\mathrm{SF} 33_{-}=\mathrm{Fin} 3_{-}+\mathrm{G} 3_{-}+\mathrm{F} 2^{3}$ i $\mathrm{SF} 3 \mathrm{x}=\mathrm{SF}{3}$ – (1) ;
망 $\mathrm{B}{-} \mathrm{R}$ : sumM2B $=0_{-}$
망 $\mathrm{BDx}\left(-\mathrm{F} 23_{2}\right)+\mathrm{BCx}\left(-\mathrm{F} 4_{-}\right)+\mathrm{BC} 2 \mathrm{x}\left(\mathrm{Fin} 2_{+}+\mathrm{G} 2_{-}\right)+\mathrm{Min} 2_{-}=0_{-}$
sumM2B_ $=\operatorname{cross}\left(r D_{-}-r B_{-},-F 23_{-}\right)+\operatorname{cross}\left(r C_{-}-r B_{-},-F 24_{-}\right) \ldots$
$+\operatorname{cross}\left(\mathrm{rC} 2_{-}-\mathrm{rB}\right.$, Fin2_+G2_)+Min2_;
sumM2Bz = sumM2 B_(3);
solF23=solve $($ SF $3 x$, sumM2Bz );
F23xs=eval (solF23. F23x);
F23ys=eval $($ solF23. F23y );
$\mathrm{F}{23}=$ [F23xs F23ys 0]; 뭉 $\mathrm{F} 23$ $=[-216.373,148.804,0]$ (N)
$\mathrm{s}$

## 数学代写|matlab仿真代写simulation代做|Reaction Force F12

The revolute joint at $B$ between link 1 and the link 2 is replaced with the joint reaction force $\mathbf{F}{12}$ as shown in Fig.5.4. The joint reaction force of the link 1 on the link 2 , $\mathbf{F}{12}$, acts at $B$ and has two unknown components $\mathbf{F}{12}=F{12 x} \mathbf{1}+F_{12 y} \mathbf{J}$.

For the link 2, Fig. 5.4a, the sum of the moments with respect to the revolute joint $D$ is zero
\begin{aligned} &\sum \mathbf{M}{D}^{(2)}= \ &\mathbf{r}{D B} \times \mathbf{F}{12}+\mathbf{r}{D C} \times\left(-\mathbf{F}{24}\right)+\mathbf{r}{D C_{2}} \times\left(\mathbf{F}{\mathrm{in} 2}+\mathbf{G}{2}\right)+\mathbf{M}{\mathrm{in} 2}=\mathbf{0} \end{aligned} The sum of the forces on the links 2 and 3 is equal to zero, Fig. 5.4b, or $$\sum \mathbf{F}^{(2 \& 3)}=\mathbf{F}{12}-\mathbf{F}{24}+\mathbf{F}{\text {in } 2}+\mathbf{G}{2}+\mathbf{F}{\text {in } 3}+\mathbf{G}{3}+\mathbf{F}{03}=\mathbf{0}$$
where $\mathbf{F}{03}$ is the reaction force of the ground on slider 3 and is perpendicular to the sliding direction $x$-axis, i.e., $\mathbf{F}{03} \cdot \mathbf{1}=0$. Equation ( $5.12$ ) is scalar multiplied with 1 and the following relation is obtained
$$\left(\sum \mathbf{F}^{(2 k 3)}\right) \cdot \mathbf{1}=\left(\mathbf{F}{12}-\mathbf{F}{24}+\mathbf{F}{\mathrm{in} 2}+\mathbf{G}{2}+\mathbf{F}{\mathrm{in} 3}+\mathbf{G}{3}\right) \cdot \mathbf{1}=0 .$$

From Eqs. (5.11) and (5.13) the two unknowns $F_{12 x}$ and $F_{12 y}$ are calculated and the MATLAB commands for the reaction $\mathbf{F}{12}$ are: F12x=sym (‘F12x’, ‘rea1’); F12y=sym (‘F12y’,’rea1′); $\mathrm{F} 12{-}=[\mathrm{F} 12 \mathrm{x}, \mathrm{F} 12 \mathrm{y}, 0]$; 뭏 unknown joint force
뫙 $D_{-} R:$ sumM2D_ $=0_{-}$
왕 $\mathrm{DB}{-} \mathrm{xF12}+\mathrm{DC} \mathrm{C}{-}\left(-\mathrm{F} 24_{-}\right)+\mathrm{DC} 2_{-} \mathrm{x}\left(\mathrm{Fin} 2_{-}+\mathrm{G} 2_{-}\right)+\mathrm{Min} 2_{-}=0_{-}$
$\operatorname{sum} 42 \mathrm{D}{-}=\operatorname{cross}\left(\mathrm{rB}-\mathrm{rD}, \mathrm{F} 12{-}\right)+\operatorname{cross}\left(\mathrm{rC}-\mathrm{rD} \mathrm{D}{-},-\mathrm{F} 2{-} 4_{-}\right) \ldots$
$+\operatorname{cross}\left(r \mathrm{C} 2_{-}-r \mathrm{D}{-}\right.$, Fin2$\left.+\mathrm{G} 2_{-}\right)+$Min2_;
sumM2Dz = sumM2D_(3);
항 D_P $_{-}$: sumF23_(1) $=0$
망 $\left(\mathrm{F} 12_{-}+(-\mathrm{F} 24)+\mathrm{Fin} 2_{-}+\mathrm{G} 2_{-}+\mathrm{Fin} 3_{-}+\mathrm{G} 3_{-}\right)(1)=0$
SF23_ = F12_+(-F24_) $+\mathrm{Fin} 2_{-}+\mathrm{G} 2_{-}+\mathrm{Fin} 3_{-}+\mathrm{G} 3_{-} ;$
SF23x = SF23_(1);
solF12=solve (sumM2Dz, SF2 3x);
F12xs=eval (solF12. F12x);
F12ys=eval (solF12. F12y);
$\mathrm{F} 12_{-}=[\mathrm{F} 12 \mathrm{xs}$ F12ys 0];
$\mathrm{F} 21_{-}=-\mathrm{F} 12_{-}$;
맘 $\mathrm{F}{2} 2{-}=[2024.907,-512.546,0] \quad(\mathrm{N})$

## 数学代写|matlab仿真代写simulation代做|Reaction Force F24

∑F(4)=F在 4+G4+F24+F54=0

F24X=符号 (‘ F24X′， ‘真实的’）;
F24是=8是 米( F24是′， ‘真实的’）;
F24_=[F24x，F24y，0]；뭉 未知联合部队
뭏 E_P:磷 (sumF4), CE=0
몽 (查找_+G4_+F24_)(r和−−rC−)=0

## 数学代写|matlab仿真代写simulation代做|Reaction Force F23

(∑F(3))⋅1=(F一世n3+G3+F23)⋅1=0接下来是链接2，图。5.3 b, 相对于旋转关节的力矩总和乙给∑米乙(2)= r乙D×(−F23)+r乙C×(−F24)+r乙C2×(F在 2+G2)+米一世n2=0从方程式。(5.9) 和(5.10)两个未知数F23X和F23是被计算。用于反应的 MATLAB 命令F23是：F23X=sym (‘F23x’, ‘真实’);F23是=符号⁡(F23是′， ‘真实的’）;
F23=[F23X,F23是,0]; Hyung 未知 联手
HwangD−磷:总和⁡3−(1)=0당 $\left{\mathrm{Fin}3{-}+\mathrm{G}3_{-}+\mathrm{F}2_{3}\right)(1)=0\ mathrm {SF} 33 _ {-} = \ mathrm {Fin} 3 _ {-} + \ mathrm {G} 3 _ {-} + \ mathrm {F} 2 ^ {3一世\mathrm{SF} 3 \mathrm{x}=\mathrm{SF}{3}망–(1);网\ mathrm {B {- \ mathrm {R:s在米米2乙=0_{-}망网\mathrm{BDx}\left(-\mathrm{F} 23_{2}\right)+\mathrm{BCx}\left(-\mathrm{F} 4_{-}\right)+\mathrm{BC} 2 \mathrm{x}\left(\mathrm{Fin} 2_{+}+\mathrm{G} 2_{-}\right)+\mathrm{Min} 2_{-}=0_{-}sumM2B_sumM2B_=\operatorname{cross}\left(r D_{-}-r B_{-},-F 23_{-}\right)+\operatorname{cross}\left(r C_{-}-r B_{-} ,-F 24_{-}\right) \ldots+\operatorname{cross}\left(\mathrm{rC} 2_{-}-\mathrm{rB}\right.,F一世n2+G2)+米一世n2;s在米米2乙和=s在米米2乙(3);s这lF23=s这l在和(小号F3×,s在米米2乙和);F23Xs=和在一种l(s这lF23.F23X);F23是s=和在一种l(s这lF23.F23是);\数学{F} {23} =뭉[F23XsF23是s0];蒙\数学{F} 23=[-216.373,148.804,0](ñ)\数学{s}$

## 数学代写|matlab仿真代写simulation代做|Reaction Force F12

∑米D(2)= rD乙×F12+rDC×(−F24)+rDC2×(F一世n2+G2)+米一世n2=0连杆 2 和 3 上的力之和等于 0，图 5.4b，或∑F(2&3)=F12−F24+F在 2+G2+F在 3+G3+F03=0

(∑F(2ķ3))⋅1=(F12−F24+F一世n2+G2+F一世n3+G3)⋅1=0.

뫙D−R:sumM2D_=0−

+叉⁡(rC2−−rD−, 鳍2+G2−)+Min2_;
sumM2Dz = sumM2D_(3);
항 D_P−: sumF23_(1)=0

SF23_ = F12_+(-F24_)+F一世n2−+G2−+F一世n3−+G3−;
SF23x = SF23_(1);
solF12=求解（sumM2Dz，SF2 3x）；
F12xs=eval (solF12.F12x);
F12ys=eval (solF12.F12y);
F12−=[F12XsF12ys 0];
F21−=−F12−;
맘 $\ mathrm {F} {2} 2 {-} = [2024.907, -512.546,0] \ quad (\ mathrm {N})$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。