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# 物理代写|理论力学代写Theoretical Mechanics代考|PHYS354 Equilibrium Equations for One Rigid Body System

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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 $n$.

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