# 達朗貝爾原理

（重定向自達朗伯特原理

${\displaystyle \sum _{i}\ (\mathbf {F} _{i}+\mathbf {I} _{i})\cdot \delta \mathbf {r} _{i}=0\,\!}$

## 導論

${\displaystyle \mathbf {F} _{i}^{(T)}=m_{i}\mathbf {a} _{i}\,\!}$

${\displaystyle \mathbf {F} _{i}^{(T)}-m_{i}\mathbf {a} _{i}=\mathbf {0} \,\!}$

${\displaystyle \mathbf {I} _{i}=-m_{i}\mathbf {a} _{i}\,\!}$

${\displaystyle \mathbf {F} _{i}^{(T)}+\mathbf {I} _{i}=\mathbf {0} \,\!}$

${\displaystyle \delta W_{i}=(\mathbf {F} _{i}^{(T)}+\mathbf {I} _{i})\cdot \delta \mathbf {r} _{i}=0\,\!}$

${\displaystyle \delta W=\sum _{i}\delta W_{i}=\sum _{i}(\mathbf {F} _{i}^{(T)}+\mathbf {I} _{i})\cdot \delta \mathbf {r} _{i}=0\,\!}$

${\displaystyle \delta W=\sum _{i}\ \mathbf {F} _{i}\cdot \delta \mathbf {r} _{i}+\sum _{i}\ \mathbf {C} _{i}\cdot \delta \mathbf {r} _{i}+\sum _{i}\mathbf {I} _{i}\cdot \delta \mathbf {r} _{i}=0\,\!}$

${\displaystyle \delta W=\sum _{i}(\mathbf {F} _{i}+\mathbf {I} _{i})\cdot \delta \mathbf {r} _{i}=0\,\!}$（1）

${\displaystyle \mathbf {F} _{i}^{eff}=\mathbf {F} _{i}+\mathbf {I} _{i}\,\!}$

${\displaystyle \delta W=\sum _{i}\mathbf {F} _{i}^{eff}\cdot \delta \mathbf {r} _{i}=0\,\!}$

### 適用案例

• 剛體的約束條件是一種完整約束，以方程式表達，${\displaystyle (\mathbf {r} _{i}-\mathbf {r} _{j})^{2}=L_{ij}^{2}}$；其中，剛體內部的粒子${\displaystyle P_{i}}$${\displaystyle P_{j}}$的位置分別為${\displaystyle \mathbf {r} _{i}}$${\displaystyle \mathbf {r} _{j}}$，它們之間的距離${\displaystyle L_{ij}}$是個常數。所以，兩個粒子的虛位移${\displaystyle \delta \mathbf {r} _{i}}$${\displaystyle \delta \mathbf {r} _{j}}$之間的關係為
${\displaystyle \delta (\mathbf {r} _{i}-\mathbf {r} _{j})^{2}=2(\mathbf {r} _{i}-\mathbf {r} _{j})(\delta \mathbf {r} _{i}-\delta \mathbf {r} _{j})=0}$

1、${\displaystyle \delta \mathbf {r} _{i}=\delta \mathbf {r} _{j}}$

${\displaystyle \mathbf {C} _{ij}\cdot \delta \mathbf {r} _{i}+\mathbf {C} _{ji}\cdot \delta \mathbf {r} _{j}=0}$

2、${\displaystyle (\mathbf {r} _{i}-\mathbf {r} _{j})\perp (\delta \mathbf {r} _{i}-\delta \mathbf {r} _{j})}$

${\displaystyle \mathbf {C} _{ij}\cdot \delta \mathbf {r} _{i}+\mathbf {C} _{ji}\cdot \delta \mathbf {r} _{j}=\mathbf {C} _{ij}\cdot \delta \mathbf {r} _{i}-\mathbf {C} _{ij}\cdot \delta \mathbf {r} _{j}=\mathbf {C} _{ij}\cdot (\delta \mathbf {r} _{i}-\delta \mathbf {r} _{j})=0}$

• 思考木塊移動於平滑地面上。因為木塊的重量，而產生的反作用力，是地面施加於木塊的一種約束力。注意到對於這案例，符合約束條件的虛位移必須與地面平行，所以，地面施加的約束力垂直於虛位移，它所作的虛功等於零。可是，假若木塊移動的地面是粗糙的，則會有摩擦力產生。由於虛位移平行於摩擦力，虛功不等於零。所以，達朗貝爾原理不適用於這狀況。但是，假設是一隻輪子純滚动英语rolling於地面上，因為輪子與地面的瞬時接觸點是不動的，符合約束條件的虛位移等於零，所以虛功等於零，達朗貝爾原理又適用了[3]

## 拉格朗日方程式的導引

${\displaystyle \mathbf {r} _{i}=\mathbf {r} _{i}(q_{1},q_{2},\cdots ,q_{n},\ t)\,\!}$

${\displaystyle \delta \mathbf {r} _{i}=\sum _{j}\ {\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\delta q_{j}\,\!}$（2）

${\displaystyle \mathbf {v} _{i}={\frac {d\mathbf {r} _{i}}{dt}}={\frac {\partial \mathbf {r} _{i}}{\partial t}}+\sum _{j}{\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}{\dot {q}}_{j}\,\!}$

${\displaystyle {\frac {\partial \mathbf {v} _{i}}{\partial {\dot {q}}_{j}}}={\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\,\!}$（3）

${\displaystyle \sum _{i}\ m_{i}\mathbf {a} _{i}\cdot \delta \mathbf {r} _{i}=\sum _{i,j}\ m_{i}\mathbf {a} _{i}\cdot {\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\delta q_{j}\,\!}$

${\displaystyle \sum _{i,j}\ m_{i}\mathbf {a} _{i}\cdot {\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\delta q_{j}=\sum _{i,j}\left({\frac {d}{dt}}\left(m_{i}\mathbf {v} _{i}\cdot {\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\right)-m_{i}\mathbf {v} _{i}\cdot {\frac {d}{dt}}\left({\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\right)\right)\delta q_{j}\,\!}$

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\right)=\left({\frac {\partial }{\partial t}}+\sum _{k}{\dot {q}}_{k}{\frac {\partial }{\partial q_{k}}}\right)\left({\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\right)={\frac {\partial ^{2}\mathbf {r} _{i}}{\partial q_{j}\partial t}}+\sum _{k}{\frac {\partial ^{2}\mathbf {r} _{i}}{\partial q_{j}\partial q_{k}}}{\dot {q}}_{k}\,\!}$
${\displaystyle {\frac {\partial \mathbf {v} _{i}}{\partial q_{j}}}={\frac {\partial }{\partial q_{j}}}\left({\frac {\partial \mathbf {r} _{i}}{\partial t}}+\sum _{k}{\frac {\partial \mathbf {r} _{i}}{\partial q_{k}}}{\dot {q}}_{k}\right)={\frac {\partial ^{2}\mathbf {r} _{i}}{\partial q_{j}\partial t}}+\sum _{k}{\frac {\partial ^{2}\mathbf {r} _{i}}{\partial q_{j}\partial q_{k}}}{\dot {q}}_{k}\,\!}$

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\right)={\frac {\partial \mathbf {v} _{i}}{\partial q_{j}}}\,\!}$（4）

${\displaystyle \sum _{i,j}\ m_{i}\mathbf {a} _{i}\cdot {\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\delta q_{j}=\sum _{i,j}\left({\frac {d}{dt}}\left(m_{i}\mathbf {v} _{i}\cdot {\frac {\partial \mathbf {v} _{i}}{\partial {\dot {q}}_{j}}}\right)-m_{i}\mathbf {v} _{i}\cdot {\frac {\partial \mathbf {v} _{i}}{\partial q_{j}}}\right)\delta q_{j}\,\!}$

${\displaystyle T=\sum _{i}\ {\frac {1}{2}}m_{i}\mathbf {v} _{i}\cdot \mathbf {v} _{i}\,\!}$

${\displaystyle \sum _{i,j}\ m_{i}\mathbf {a} _{i}\cdot {\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\delta q_{j}=\sum _{j}\ \left({\frac {d}{dt}}\left({\frac {\partial T}{\partial {\dot {q}}_{j}}}\right)-{\frac {\partial T}{\partial q_{j}}}\right)\delta q_{j}\,\!}$（5）

${\displaystyle \sum _{i}\ \mathbf {F} _{i}\cdot \delta \mathbf {r} _{i}=\sum _{i,j}\ \mathbf {F} _{i}\cdot {\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\delta q_{j}=\sum _{j}\ {\mathcal {F}}_{j}\delta q_{j}\,\!}$（6）

${\displaystyle {\mathcal {F}}_{j}=\sum _{i}\ \mathbf {F} _{i}\cdot {\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\,\!}$

${\displaystyle \sum _{j}\ \left({\frac {d}{dt}}\left({\frac {\partial T}{\partial {\dot {q}}_{j}}}\right)-{\frac {\partial T}{\partial q_{j}}}-{\mathcal {F}}_{j}\right)\delta q_{j}=0\,\!}$（7）

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial T}{\partial {\dot {q}}_{j}}}\right)-{\frac {\partial T}{\partial q_{j}}}-{\mathcal {F}}_{j}=0\,\!}$（8）

${\displaystyle {\mathcal {F}}_{j}={\frac {d}{dt}}\left({\frac {\partial V}{\partial {\dot {q}}_{j}}}\right)-{\frac {\partial V}{\partial q_{j}}}\,\!}$

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial (T-V)}{\partial {\dot {q}}_{j}}}\right)-{\frac {\partial (T-V)}{\partial q_{j}}}=0\,\!}$

${\displaystyle L\ {\stackrel {def}{=}}\ T-V\,\!}$

${\displaystyle {\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {q}}_{j}}}\right)-{\frac {\partial L}{\partial q_{j}}}=0\,\!}$

${\displaystyle {\mathcal {F}}_{j}=-{\frac {\partial V}{\partial q_{j}}}\,\!}$

## 達朗貝爾慣性力原理

${\displaystyle \sum _{i}\ \mathbf {F} _{i}=m\mathbf {a} \,\!}$
${\displaystyle \sum _{i}\ \mathbf {M} _{i}={\boldsymbol {\mathcal {I}}}{\boldsymbol {\alpha }}\,\!}$

${\displaystyle \mathbf {I} =-m\mathbf {a} \,\!}$
${\displaystyle \mathbf {M} =-{\boldsymbol {\mathcal {I}}}{\boldsymbol {\alpha }}\,\!}$

${\displaystyle \mathbf {I} +\sum _{i}\ \mathbf {F} _{i}=\mathbf {0} \,\!}$
${\displaystyle \mathbf {M} +\sum _{i}\ \mathbf {M} _{i}=\mathbf {0} \,\!}$

### 剛體二維平面運動實例

${\displaystyle \mathbf {I} =-m\mathbf {a} \,\!}$
${\displaystyle \mathbf {M} =-{\boldsymbol {\mathcal {I}}}\alpha \,\!}$

${\displaystyle \sum _{i}F_{xi}=0\,\!}$
${\displaystyle \sum _{i}F_{yi}=0\,\!}$
${\displaystyle \sum _{i}M_{i}=0\,\!}$

## 參考文獻

1. Lanczos, Cornelius, The Variational Principles of Mechanics, Dovers Publications, Inc: pp. 90–106, 1970, ISBN 978-0-486-65067-8
2. ^ Torby, Bruce. Advanced Dynamics for Engineers. HRW Series in Mechanical Engineering. United States of America: CBS College Publishing. 1984: pp. 269. ISBN 0-03-063366-4 （英语）.
3. Goldstein, Herbert. Classical Mechanics 3rd. United States of America: Addison Wesley. 1980: pp. 18–21. ISBN 0201657023 （英语）.
4. ^ Beer, Ferdinand; E. Russell Johnston, Jr., William E. Clausen. Vector Mechanics for Engineers 7th. United States of America: Elizabeth A. Jones. 2004: pp. 1029, 1167. ISBN 0-07-230491-X （英语）.