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## 物理代写|理论力学代写Theoretical Mechanics代考| Free Damped Linear Oscillator

Each real oscillator eventually comes to stop because of the unavoidable frictional forces. We therefore want to now include them into our considerations where, however, we will restrict ourselves to the simplest case of the Stokes’s friction. Then the extended equation of motion reads:
$$m \ddot{x}=-k x-\alpha \dot{x} .$$
One can realize this situation by a ‘tongue’, dipping into a liquid and being fixed to the mass $m$ (Fig. 2.27). While the frictional term in Eq. (2.167) in general represents a certain approximation, there exists an exact non-mechanical realization of the damped harmonic oscillator by the electrical oscillator circuit. The sum of the partial voltages in the circuit sketched in Fig. $2.28$ must be zero. The electrical current therefore obeys the following differential equation:
$$\ddot{L}+R \dot{I}+\frac{1}{C} I=0 .$$
The ohmic resistance $R$ simulates the frictional term.

After division by $m$ we get from (2.167) the following homogeneous differential equation of second order:
$$\ddot{x}+2 \beta \dot{x}+\omega_0^2 x=0 ; \quad \beta=\frac{\alpha}{2 m} .$$
As ansatz an exponential function appears again plausible:
$$x(t)=e^{\lambda t} .$$
It is exactly then a solution if $\lambda$ fulfills the following relation:
$$\lambda^2+2 \beta \lambda+\omega_0^2=0 .$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Damped Linear Oscillator Under the Inﬂﬂuence of an External Force

Because of the unavoidable friction every oscillating process is exponentially damped unless an additional external force acts. We will now include the latter in our considerations. The equation of motion (2.169) is then to be replaced by
$$\ddot{x}+2 \beta \dot{x}+\omega_0^2 x=\frac{1}{m} F(t) .$$
We choose the same denotations as in the last section and restrict ourselves to the important special case of a periodic force:
$$F(t)=f \cos \bar{\omega} t .$$
One can realize the periodic force by, for instance, a wheel spinning with constant angular velocity and being connected via a drive rod to the oscillating body (Fig. 2.31).

Here again we have an exact non-mechanical realization (Fig. 2.32) by the electrical oscillator circuit if one applies to it a periodic alternating voltage $U_0 \sin \bar{\omega} t$ :
$$\ddot{L}+R \dot{I}+\frac{1}{C} I=U_0 \bar{\omega} \cos \bar{\omega} t$$

## 物理代写|理论力学代写Theoretical Mechanics代考| Free Damped Linear Oscillator

$$m \ddot{x}=-k x-\alpha \dot{x} .$$

$$\ddot{L}+R \dot{I}+\frac{1}{C} I=0 .$$

$$\ddot{x}+2 \beta \dot{x}+\omega_0^2 x=0 ; \quad \beta=\frac{\alpha}{2 m} .$$

$$x(t)=e^{\lambda t}$$

$$\lambda^2+2 \beta \lambda+\omega_0^2=0 .$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Damped Linear Oscillator Under the Inflfluence of an External Force

$$\ddot{x}+2 \beta \dot{x}+\omega_0^2 x=\frac{1}{m} F(t) .$$

$$F(t)=f \cos \bar{\omega} t .$$

$$\ddot{L}+R \dot{I}+\frac{1}{C} I=U_0 \bar{\omega} \cos \bar{\omega} t$$

## 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。

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## 物理代写|理论力学代写Theoretical Mechanics代考|Motion in the Homogeneous Gravitational Field

According to the above-given program we have to at first formulate the equation of motion. Using (2.48) together with (2.43) and exploiting the equality of inertial and heavy mass we can write:
$$\ddot{\mathbf{r}}=\mathbf{g} ; \quad \mathbf{g}=(0,0,-g) .$$
The mass is eliminated; in the gravitational field all bodies are therefrom equally accelerated. It results in a
uniformly accelerated motion
as we have discussed it already in Sect. 2.1.2. We can directly take the former results $(2.30)$ and $(2.31)$ :
\begin{aligned} &\mathbf{v}(t)=\mathbf{v}\left(t_0\right)+\mathbf{g} \cdot\left(t-t_0\right) \ &\mathbf{r}(t)=\mathbf{r}\left(t_0\right)+\mathbf{v}\left(t_0\right)\left(t-t_0\right)+\frac{1}{2} \mathbf{g} \cdot\left(t-t_0\right)^2 . \end{aligned}

## 物理代写|理论力学代写Theoretical Mechanics代考|Linear Differential Equations

We refer to
$$x^{(n)}(t)=\frac{d^n}{d t^n} x(t)$$
as the $n$-th derivative of the function $x(t)$. A relation which contains one or more derivatives of a given function, where the $n$-th derivative appears as the highest,
$$f\left(x^{(n)}, x^{(n-1)}, \ldots, \dot{x}, x, t\right)=0,$$

is called a differential equation of $n$-th order. The goal is to derive the solution function $x(t)$ from such a relation. The basic dynamical equation of Classical Mechanics (2.43) written in Cartesian coordinates, e.g., has just this shape:
$$m \ddot{x}_i-F_i\left(\dot{x}_1, \dot{x}_2, \dot{x}_3, x_1, x_2, x_3, t\right)=0, \quad i=1,2,3 .$$
This is a coupled system of three differential equations of second order for the three functions $x_1(t), x_2(t), x_3(t)$.

Let us first focus, however, on a general relation of the type (2.95). The central statement is formulated in the following

Theorem 2.3.1 The general solution of a differential equation of $n$-th order (2.95) is an ensemble of solutions
$$x=x\left(t \mid \gamma_1, \gamma_2, \ldots, \gamma_n\right),$$
which depends on $n$ independent parameters $\gamma_1, \gamma_2, \ldots, \gamma_n$. Every set of $\gamma_i$ ‘s which are fixed in advance then leads to a special (particular) solution.

## 物理代写|理论力学代写Theoretical Mechanics代考|Motion in the Homogeneous Gravitational Field

$$\ddot{\mathbf{r}}=\mathbf{g} ; \quad \mathbf{g}=(0,0,-g) .$$

2.1.2. 我们可以直接取前一个结果 $(2.30)$ 和 $(2.31)$ :
$$\mathbf{v}(t)=\mathbf{v}\left(t_0\right)+\mathbf{g} \cdot\left(t-t_0\right) \quad \mathbf{r}(t)=\mathbf{r}\left(t_0\right)+\mathbf{v}\left(t_0\right)\left(t-t_0\right)+\frac{1}{2} \mathbf{g} \cdot\left(t-t_0\right)^2 .$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Linear Differential Equations

$$x^{(n)}(t)=\frac{d^n}{d t^n} x(t)$$

$$f\left(x^{(n)}, x^{(n-1)}, \ldots, \dot{x}, x, t\right)=0,$$

$$m \ddot{x}_i-F_i\left(\dot{x}_1, \dot{x}_2, \dot{x}_3, x_1, x_2, x_3, t\right)=0, \quad i=1,2,3 .$$

$$x=x\left(t \mid \gamma_1, \gamma_2, \ldots, \gamma_n\right),$$

## 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。

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## 物理代写|理论力学代写Theoretical Mechanics代考|Forces

At the beginning of this chapter we recognized as the elementary task of each physical theory, in particular the Classical Mechanics, to derive conclusions from preformulated postulates (basis definitions, axioms). The axioms and the fundamental definition of mass are now available. The law of motion (2.42) and (2.43), respectively, have become the principal dynamical equation of Classical Mechanics. This equation is to be solved. As a rule, mathematically that means, for a given force $\mathbf{F}$, one has to solve a differential equation of second order.

More precise than the term force in this connection is, strictly speaking, the concept of the
‘Force Field’
$$\mathbf{F}=\mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t) .$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Kinetic Energy

To each space point a force that acts on the mass point is assigned, which in general can even be time dependent and, additionally, may depend on the particle velocity. Dependence on acceleration $\ddot{\mathbf{r}}$, however, will not appear.

All the matter is built by elementary constituents (molecules, atoms, nucleons, electrons, …). Therefore, in the last analysis each force can be traced back to the interactions between these elementary constituents. To do this in all detail, however, is beyond the framework of Classical Mechanics which only asks for the consequences and not for the elementary causes of the forces. Normally one restricts oneself to mathematically as simple as possible and empirically reasoned model representations
Some frequently used examples are listed in the following:
(a) Weight, Gravitational Force
Each body is ‘heavy’. $1 \mathrm{~m}^3$ of iron is ‘heavier’ than $1 \mathrm{~cm}^3$ of iron. By this everyday experience a new material quantity is documented which is denoted as
gravitational (heavy) mass $m_h$.
It manifests itself in the ‘gravitational force’
$$\mathbf{F}_g=m_h \mathbf{g},$$
which acts on a stationary (motionless) mass point in the gravitational field of the earth. $g$ close to the earth’s surface is a nearly constant vector always pointing downwards in direction to the earth’s center. If we define this direction as the negative $x_3$ direction of a Cartesian coordinate system then we write
$$\mathbf{g}=-(0,0, g) ; \quad g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \text { ‘gravity acceleration’. }$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Forces

$$\mathbf{F}=\mathbf{F}(\mathbf{r}, \dot{\mathbf{r}}, t)$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Kinetic Energy

(a) 重量，重力

$$\mathbf{F}_g=m_h \mathbf{g}$$

$$\mathbf{g}=-(0,0, g) ; \quad g=9.81 \mathrm{~m} \mathrm{~s}^{-2} \text { ‘gravity acceleration’. }$$

## 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。

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## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Angular Momentum

The angular momentum of a particle is defined as
$$\boldsymbol{L}{O}=\boldsymbol{r} \times \boldsymbol{P}=\boldsymbol{r} \times(m v) .$$ Similar to the concept of moment, it can also be decomposed into three components in the three axes, i.e., $$\boldsymbol{L}{O}=L_{x} \boldsymbol{i}+L_{y} \boldsymbol{j}+L_{z} \boldsymbol{k} .$$
The derivative of the angular momentum is
\begin{aligned} \dot{\boldsymbol{L}}{O} &=\dot{\boldsymbol{r}} \times m v+\boldsymbol{r} \times m \dot{\boldsymbol{v}} \ &=\boldsymbol{v} \times m \boldsymbol{v}+\boldsymbol{r} \times m a \ &=\boldsymbol{r} \times \boldsymbol{F} \ &=M{O}(\boldsymbol{F}) . \end{aligned}

## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Kinetic Energy

“Work” is defined as the cross product of the force and the corresponding displacement, which is formulated as
\begin{aligned} W &=\boldsymbol{F} \cdot \boldsymbol{r} \ &=F s \cos \varphi, \end{aligned}
where $\boldsymbol{F}$ is a constant force, $s$ is the magnitude of the displacement, and $\varphi$ is the angle between the lines of the applied force and the displacement, as shown in Fig. 13.1. This means that the work represents the actual contribution of the force to the displacement, and it can be understood that the displacement times the projection quantity of the force along the line of displacement. Notably, the work is a scalar, whose unit is $\mathrm{J}$, which is in memory of the great scientist Joule.
For an arbitrary force, the work should be expressed via integration
$$W=\int_{C} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{r}$$

\begin{aligned} &=\int_{C} F \cos \varphi \mathrm{d} s \ &=\int_{C} F v \cos \varphi \mathrm{d} t . \end{aligned}

## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Angular Momentum

$$\boldsymbol{L} O=\boldsymbol{r} \times \boldsymbol{P}=\boldsymbol{r} \times(m v) .$$

$$\boldsymbol{L} O=L_{x} \boldsymbol{i}+L_{y} \boldsymbol{j}+L_{z} \boldsymbol{k} .$$

$$\dot{\boldsymbol{L}} O=\dot{\boldsymbol{r}} \times m v+\boldsymbol{r} \times m \dot{\boldsymbol{v}} \quad=\boldsymbol{v} \times m \boldsymbol{v}+\boldsymbol{r} \times m a=\boldsymbol{r} \times \boldsymbol{F} \quad=M O(\boldsymbol{F}) .$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Theorem of Kinetic Energy

“功”定义为力和相应位移的叉积，公式为
$$W=\boldsymbol{F} \cdot \boldsymbol{r} \quad=F s \cos \varphi,$$

$$\begin{gathered} W=\int_{C} \boldsymbol{F} \cdot \mathrm{d} \boldsymbol{r} \ =\int_{C} F \cos \varphi \mathrm{d} s \quad=\int_{C} F v \cos \varphi \mathrm{d} t . \end{gathered}$$

## 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。

Posted on Categories:Theoretical mechanics, 物理代写, 理论力学

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## 物理代写|理论力学代写Theoretical Mechanics代考|Rotation on a Fixed Axis

As shown in Fig. 8.3, if we impose a moment on a door, the door will be closed with respect to a fixed axis. This type of motion is also a basic motion, termed as rotation with respect to a fixed axis. In this rotation process, there always exists a fixed axis on the rigid body. This type of motion takes some special features. If we look through the positive direction of $z$-axis in Fig. 8.3, we can find the orbit of an arbitrary point on the rigid body in rotation, which is actually a circle.

To depict the motion of the rigid body, we introduce the angular displacement $\varphi$, whose sign also obeys the right-hand screw law. The angular velocity can be defined as
$$\omega=\lim _{\Delta t \rightarrow 0} \frac{\Delta \varphi}{\Delta t}=\dot{\varphi} .$$
The angular velocity can also be related with the frequency via
$$\omega=2 \pi f .$$
Introducing the concept of rotation velocity $n$ with the unit of $1 / \mathrm{min}$, one has

$$\omega=\frac{2 \pi n}{60}=\frac{\pi n}{30} .$$
Similarly, the angular acceleration can be further defined as
$$\varepsilon=\lim _{\Delta t \rightarrow 0} \frac{\Delta \omega}{\Delta t}=\dot{\omega}=\ddot{\varphi} .$$
The above relation can be analogous to the displacement, velocity, and acceleration defined in the last chapter.

If the rigid body is in rotation with respect to an axis, any point on the rigid body is in a circular motion. As shown in Fig. 8.4, the radius $r$ is the vertical distance from the axis to the arbitrary point, $s$ is the arc length, and $\varphi$ is the corresponding angular displacement. We then have the following geometric relation:
$$s=r \varphi .$$
Taking derivatives on both sides of the above equation, one has
\begin{aligned} &v=\dot{s}=r \dot{\varphi}=\omega r, \ &a_{\tau}=\dot{v}=\ddot{s}=r \ddot{\varphi}=\varepsilon r . \end{aligned}

## 物理代写|理论力学代写Theoretical Mechanics代考|Relative Velocity

We consider two points on the planar figure $A$ and $B$, which have the velocity $v_{A}$ and $v_{B}$, respectively, as schematized in Fig. 9.1. We normally name point $A$ as the base point, as it is a reference point. There is a relative velocity $v_{B A}$, with the meaning that $A$ is the base point. As a result, we know that $v_{B A}$ is not equal to $v_{A B}$. According to the velocity superposition, one has
$$v_{B}=v_{A}+v_{B A},$$
where the direction of the relative velocity $\boldsymbol{v}_{B A}$ is perpendicular to the line $A B$, as point $B$ rotates with respect to point $A$ in a circular motion. From the above formula, we can solve the velocity of any point on the planar figure if the base point is given. Therefore, this method of velocity composition is called “Method of base point”.
As a consequence, if we decompose the above velocities in the direction of line $A B$, then one has
\begin{aligned} \left.\boldsymbol{v}{B}\right|{A B} &=\left.\boldsymbol{v}{A}\right|{A B}+\left.\boldsymbol{v}{B A}\right|{A B} \ &=\left.\boldsymbol{v}{A}\right|{A B} . \end{aligned}
This means the projections of the velocities of the two points are equal along their connection line. In fact, this is the second method to study the velocity composition, which is termed as “Method of velocity projection”. As shown in Fig. 9.2, if we have known the angles between the velocities and line $A B$, we then have the relation
$$v_{A} \cos \alpha=v_{B} \cos \beta .$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Rotation on a Fixed Axis

$$\omega=\lim {\Delta t \rightarrow 0} \frac{\Delta \varphi}{\Delta t}=\dot{\varphi}$$ 角速度也可以通过以下方式与频率相关 $$\omega=2 \pi f$$ 介绍旋转速度的概苡以 $1 / \mathrm{min}$,一个有 $$\omega=\frac{2 \pi n}{60}=\frac{\pi n}{30}$$ 类似地，角加速度可以进一步定义为 $$\varepsilon=\lim {\Delta t \rightarrow 0} \frac{\Delta \omega}{\Delta t}=\dot{\omega}=\ddot{\varphi} .$$

$$s=r \varphi .$$

$$v=\dot{s}=r \dot{\varphi}=\omega r, \quad a_{\tau}=\dot{v}=\ddot{s}=r \ddot{\varphi}=\varepsilon r .$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Relative Velocity

$$v_{B}=v_{A}+v_{B A}$$

$$\boldsymbol{v} B|A B=\boldsymbol{v} A| A B+\boldsymbol{v} B A|A B \quad=\boldsymbol{v} A| A B$$

$$v_{A} \cos \alpha=v_{B} \cos \beta .$$

## 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。

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## 物理代写|理论力学代写Theoretical Mechanics代考|Equilibrium Equations for One Rigid Body System

Let us continue to discuss the equilibrium conditions of the general coplanar force group. As mentioned in Chap. 4, the equilibrium conditions are that the principal vector and principal moment are both zeroes:
\begin{aligned} &\boldsymbol{R}=0 \ &M=0 \end{aligned}
In the Cartesian coordinate system, the above formulas are expressed as
\begin{aligned} &\Sigma X=0, \ &\Sigma Y=0, \ &\Sigma M_{A}=0 . \end{aligned}
This group of formulas includes two equations on force, and one equation on moment, so it is normally named as one moment format. In fact, there are also some other formats on the equilibrium conditions. For example, the two-moment format can be written as
\begin{aligned} &\Sigma X=0 \ &\Sigma M_{A}=0 \ &\Sigma M_{B}=0 \end{aligned}
The three-moment format can be expressed as
\begin{aligned} &\Sigma M_{A}=0 \ &\Sigma M_{B}=0 \ &\Sigma M_{C}=0 \end{aligned}

## 物理代写|理论力学代写Theoretical Mechanics代考|Rigid Multi-body System

The previous discussions are only confined to one rigid body. However, in most cases, we will face a system, which is assembled by several rigid bodies via constraints. This system is named as “rigid multi-body system”. There are two types of forces for this kind of system. One is the external force, which comes from the other objects outside the system, and the other is internal force, which is the interaction force among the rigid bodies in the system. For the rigid multi-body system, we have the following rules:
(1) When the whole system is in equilibrium, every rigid body in the system must be in equilibrium.
(2) We can select a portion of the system as an object, or can choose the whole system to investigate.
(3) When we consider the total system, the internal forces among the subsystems do not appear; and when we analyze one subsystem, it endures the action forces from the other subsystems.
(4) The action and reaction forces always appear together, with the opposite directions.
(5) For a system including $n$ rigid bodies, the number of the individual objects we can choose is $n$.

It can be seen that, for a rigid multi-body system including $n$ rigid bodies, we can write down $3 n$ equilibrium equations totally. If the number of the unknowns is equal to that of the equilibrium equations, all of these unknowns can be solved, and the case is called determinate problem. And vice versa, if the number of the unknowns is bigger than that of the equilibrium equations, not all of these unknowns can be solved, and the case is called indeterminate problem. How to solve the indeterminate problem needs considering the deformation of the object, which is out of the range of Statics.

## 物理代写|理论力学代写Theoretical Mechanics代考|Equilibrium Equations for One Rigid Body System

$$\boldsymbol{R}=0 \quad M=0$$

$$\Sigma X=0, \quad \Sigma Y=0, \Sigma M_{A}=0 .$$

$$\Sigma X=0 \quad \Sigma M_{A}=0 \Sigma M_{B}=0$$

$$\Sigma M_{A}=0 \quad \Sigma M_{B}=0 \Sigma M_{C}=0$$

## 物理代写|理论力学代写Theoretical Mechanics代考|Rigid Multi-body System

(1)当整个㒶统处于平衡状态时，系统中的每个刚体都必须处于平衡状态。
(2)我们可以选择系统的一部分作为对象，也可以选择整个系统进行调育。
（3）当我们考慮整个系统时，子系统之间的内力不出现；当我们分析一个子系统时，它会承受来自其他子系统的作用力。
(4) 作用力和反作用力总是同时出现，方向相反。
(5) 对于一个系统，包括 $n$ 刚体，我们可以选择的单个物体的数量是 $n$.

## 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。

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## 物理代考|理论力学代考THEORETICAL MECHANICS代考|PHY4310 Hinged or pinned support

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## 物理代考|理论力学代考THEORETICAL MECHANICS代考|Hinged or pinned support

If a rod is fixed by the ground, and it can also slightly rotate at this point, this kind of constraint is called hinged or pinned support, as shown in Fig. 3.8. In fact, there are two holes on the ground and the bar, respectively, which are matched by a cylindrical pin. However, in the process of analysis, the pin can be thought to be fixed on the ground or on the bar, and we only consider the match of the two objects, i.e., the ground (or the other bar) and the selected bar. As the bar can rotate with respect to the hinged point $O$, it will endure the reactive force from the ground. It is clear that this is a problem of contact surface, where the reactive force is normal to the tangential line of the contact surface. However, the bar can rotate at any angle at the hinged point, and the direction of the reactive force can’t be a determined one. Therefore, we decompose the reactive force in two directions, i.e., $\boldsymbol{X}{O}$ and $\boldsymbol{Y}{O}$, respectively.
The similar case can happen when two bars are hinged together by a cylindrical pin, as schematized in Fig. 3.9. Each bar can rotate with respect to the hinged point, and therefore the direction of the reactive for each bar is not a determined one, then it can be expressed by the two components along the $x$ – and $y$-coordinate axes, respectively.

## 物理代考|理论力学代考THEORETICAL MECHANICS代考|Roller support

Another special constraint is the roller support, which is shown at point $B$ in Fig. 3.10. The constraint is similar to a small car, which can roll in the horizontal direction, but it can’t move in the vertical direction. As a result, the constraint can only provide one component reactive force $Y_{O}$, which is shown in Fig. 3.10. This structure can be also expressed in two other formats, which are shown in Fig. 3.11. The roller support is especially important in engineering. For example, if we design a bridge, which can be modeled as a beam with two hinged ends, then one end should be designed as a roller support. If both ends are fixed as hinged constraints, then the temperature can induce elongation or compression, and the bridge can’t match this deformation. The roller support can make the bridge slightly move in the horizontal direction, and can overcome the influence of temperature.

## 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。

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## 物理代考|理论力学代考THEORETICAL MECHANICS代考|PHYS111 Constraint

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## 物理代考|理论力学代考THEORETICAL MECHANICS代考|Flexible cable

An object is often constrained by a flexible cable, which means that the cable provides a reactive force to confine the object. As shown in Fig. 3.1, a ball is hung by a rope on the ceiling. If there is no force from the rope, the ball will go downward to the ground. This fact manifests that a reactive force from the cable must exist to balance the gravity of the ball. We first relieve the constraint, only investigating the object. This process is called isolation, and the object is called a free body now. The force from the rope deviates from the object, along the contraction direction of the rope. We use the symbol $\boldsymbol{T}$ to represent the force, with the meaning of tension.

Similarly, the belt of a wheel can transport some objects from one position to another, and the belt is normally in tension. We analyze the wheel, and it endures the pulling force from the belt, as shown in Fig. 3.2. The tension also deviates from the wheel, along the contraction direction of the belt.

## 物理代考|理论力学代考THEORETICAL MECHANICS代考|Smooth contact surface

As shown in Fig. 3.3, an object is lying on a desk. If there is no supporting force from the desk, the object will go downward to the ground. This means the desk provides a reactive force to ensure the constraint of the object. This kind of constraint is called smooth contact surface constraint, where the friction is neglected. The direction of the reactive force from the contact surface is normal to this surface, pointing inside to the object. The action line of the reactive force can also be determined considering that it is perpendicular to the tangential line of the contact surface. As shown in Fig. 3.4, an object is put on a curved surface. We first relieve the constraint, leaving only a free body. To determine the direction of the reactive force from the curved surface, we can draw the tangential line between the ball and the curved surface. Then the force is vertical to this tangential line.

Figure $3.5$ shows the contact of two gears, where the teeth are contacting with each other to transmit torques. Select one tooth, and it undergoes the force from the neighborhood tooth. Relieving the constraint of the neighborhood tooth, the reactive force of the first tooth is perpendicular to the tangent line of the contact surface, as demonstrated in Fig. 3.6.

Another example is a slender bar put in a trough. There are three contact points which can produce reactive forces. These reactive forces are from the smooth contact surfaces, which can be dictated as the schematic displayed in Fig. 3.7. At point $A$, a point from the bar is contacting with a planar surface, and the tangential line is of the same direction with the surface. Point $B$ is similar to $A$. At point $C$, a point from the trough is contacting with the bar surface, and the tangential line can also be marked.

## 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。

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## 物理代考|理论力学代考THEORETICAL MECHANICS代考|PHYS310 Fundamentals of Statics

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## 物理代考|理论力学代考THEORETICAL MECHANICS代考|Force

Force is the mechanical integration of different objects, which will cause the variation of motion state and deformation of the objects. The first influence is the so-called external effect and the second is the internal effect. However, the main object in this course is the rigid body, which is assumed to never deform throughout this textbook. In other words, our goal is only to examine the equilibrium and motion laws for rigid bodies.

In engineering, we often name force as load, such as wind load, snow load, rain load, seismic or earthquake load, and electromagnetic load. The normally seen forces can be divided into two groups, i.e., the concentrated force and the distributed force. As shown in Fig. 2.1, a concentrated force is modeled as being acted at only one point, and a distributed force is abstracted as being applied on an area or a length span of the object. In the real world, there are no concentrated forces, and “concentrated force” is only a perfect model. When the action area is small enough, such as a needle penetrating into the skin can be considered as a concentrated force. As a consequence, the unit of the concentrated force is $\mathrm{N}$ or $\mathrm{kN}$, and that of the distributed force is $\mathrm{N} / \mathrm{m}$ or $\mathrm{kN} / \mathrm{m}$ (force per unit length). Notice that $\boldsymbol{P}$ is a vector and $q(x)$ is a scalar in Fig. 2.1, for $q(x)$ denotes the line density of the load.

As is well known, there are three essential elements of the force, namely, magnitude, direction, and point of action. A force $\boldsymbol{F}$ can be expressed as $\boldsymbol{F}=\boldsymbol{F} \boldsymbol{n}$, where $F$ is the magnitude of the force and $\boldsymbol{n}$ is the unit vector. We then have
$$F=|\boldsymbol{F}| .$$

## 物理代考|理论力学代考THEORETICAL MECHANICS代考|Axioms in Statics

Based upon a lot of observations and practices, scientists and engineers have concluded several axioms in Statics. They have already been proved to be true in practice, and there is no need to prove them from the mathematical viewpoint. We only master the conclusions and don’t deal with the details.
Axiom 1 Two forces in equilibrium axiom
If two forces acting on a rigid body are in equilibrium, they must have the same magnitude, opposite directions, and they are along the same action of line. Otherwise, the rigid body cannot be balanced. In this case, the rigid body can be named as a two-force member or two-force bar, as shown in Fig. 2.5. The key point is that there are only two points on the rigid body enduring forces, and the final resultant forces at these two points have the above relations. We always remember that two points determine one line, and the two forces must be collinear.
The relation of the two forces in Fig. $2.5$ is
\begin{aligned} &\boldsymbol{F}=-\boldsymbol{F}^{\prime}, \ &F=F^{\prime} \end{aligned}

## 物理代考|理论力学代考THEORETICAL MECHANICS代考|Force

$$F=|\boldsymbol{F}| .$$

## 物理代考|理论力学代考THEORETICAL MECHANICS代考|Axioms in Statics

$$\boldsymbol{F}=-\boldsymbol{F}^{\prime}, \quad F=F^{\prime}$$

## MATLAB代写

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