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Eli
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your example

enter image description here

The equation of motion are:

$$m_1\,\ddot{x}_1=F_1+Z$$ $$m_2\,\ddot{x}_2=F_2-Z$$

or

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,\underbrace{\begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}}_{\vec{\ddot{y}}}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 1$$ the constraint equation

$$x_2-x_1=0\quad \Rightarrow \dot{x}_2=\dot{x}_1\tag 2$$

so $x_1=q_1$ is the generalized coordinate thus $\vec{\dot{y}}=J\,\dot{q}_1$

where

$J=\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}$

and eq. (1) 

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,J\,\ddot{q}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 3$$

to eliminate the constraint force $Z$ we multiply eq(3) from the left with $J^T$

$$J^T\,C_Z=\begin{bmatrix} 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=0$$

or

$$\delta x_1\,Z-\delta x_2\,Z=0$$

this is the D'Alembert'sVirtual work principle "the constraint forces dose't affect the equations of motion "

we obtain the equations of motion

$$J^T\,M\,J\,\vec{\ddot{q}_1}=J^T\,\vec{F}$$

your example

enter image description here

The equation of motion are:

$$m_1\,\ddot{x}_1=F_1+Z$$ $$m_2\,\ddot{x}_2=F_2-Z$$

or

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,\underbrace{\begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}}_{\vec{\ddot{y}}}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 1$$ the constraint equation

$$x_2-x_1=0\quad \Rightarrow \dot{x}_2=\dot{x}_1\tag 2$$

so $x_1=q_1$ is the generalized coordinate thus $\vec{\dot{y}}=J\,\dot{q}_1$

where

$J=\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}$

and eq. (1) 

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,J\,\ddot{q}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 3$$

to eliminate the constraint force $Z$ we multiply eq(3) from the left with $J^T$

$$J^T\,C_Z=\begin{bmatrix} 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=0$$

or

$$\delta x_1\,Z-\delta x_2\,Z=0$$

this is the D'Alembert's principle "the constraint forces dose't affect the equations of motion "

we obtain the equations of motion

$$J^T\,M\,J\,\vec{\ddot{q}_1}=J^T\,\vec{F}$$

your example

enter image description here

The equation of motion are:

$$m_1\,\ddot{x}_1=F_1+Z$$ $$m_2\,\ddot{x}_2=F_2-Z$$

or

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,\underbrace{\begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}}_{\vec{\ddot{y}}}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 1$$ the constraint equation

$$x_2-x_1=0\quad \Rightarrow \dot{x}_2=\dot{x}_1\tag 2$$

so $x_1=q_1$ is the generalized coordinate thus $\vec{\dot{y}}=J\,\dot{q}_1$

where

$J=\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}$

and eq. (1) 

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,J\,\ddot{q}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 3$$

to eliminate the constraint force $Z$ we multiply eq(3) from the left with $J^T$

$$J^T\,C_Z=\begin{bmatrix} 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=0$$

or

$$\delta x_1\,Z-\delta x_2\,Z=0$$

this is the Virtual work principle

we obtain the equations of motion

$$J^T\,M\,J\,\vec{\ddot{q}_1}=J^T\,\vec{F}$$

Source Link
Eli
  • 12.8k
  • 2
  • 11
  • 31

your example

enter image description here

The equation of motion are:

$$m_1\,\ddot{x}_1=F_1+Z$$ $$m_2\,\ddot{x}_2=F_2-Z$$

or

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,\underbrace{\begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \\ \end{bmatrix}}_{\vec{\ddot{y}}}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 1$$ the constraint equation

$$x_2-x_1=0\quad \Rightarrow \dot{x}_2=\dot{x}_1\tag 2$$

so $x_1=q_1$ is the generalized coordinate thus $\vec{\dot{y}}=J\,\dot{q}_1$

where

$J=\begin{bmatrix} 1 \\ 1 \\ \end{bmatrix}$

and eq. (1) 

$$\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \\ \end{bmatrix}\,J\,\ddot{q}=\begin{bmatrix} F_1 \\ F_2 \\ \end{bmatrix}+\underbrace{\begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}}_{C_Z}\,Z\tag 3$$

to eliminate the constraint force $Z$ we multiply eq(3) from the left with $J^T$

$$J^T\,C_Z=\begin{bmatrix} 1 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix}=0$$

or

$$\delta x_1\,Z-\delta x_2\,Z=0$$

this is the D'Alembert's principle "the constraint forces dose't affect the equations of motion "

we obtain the equations of motion

$$J^T\,M\,J\,\vec{\ddot{q}_1}=J^T\,\vec{F}$$