Theoretical mechanics
1 Theoretical Mechanics
理论力学 算十气拿方
2
Dynarnics On the basis of D'Alembert's Principle and of the Theorem of Virtual Displacements in this chapter the general equation of dynamics and the Lagranges equations of the second kind (abbreviated as Lagrange's equations)is deduced. The general equation of dynamics and the Lagrange's equations are effective means to study dynamical problems. They provide very simple, direct and standard ways to solve dynamical problems of unfree systems of particles
3 On the basis of D‘Alembert’s Principle and of the Theorem of Virtual Displacements in this chapter the general equation of dynamics and the Lagrange's equations of the second kind (abbreviated as Lagrange's equations) is deduced. The general equation of dynamics and the Lagrange's equations are effective means to study dynamical problems. They provide very simple, direct and standard ways to solve dynamical problems of unfree systems of particles
学 本章在达朗伯原理和虚位移原理的基础上,进一步导 出动力学普遍方程和拉格朗日第二类方程(简称拉格朗日 方程)。动力学普遍方程和拉格朗日方程是研究动力学问 题的有力手段,在解决非自由质点系的动力学问题时,显 得十分简捷、规范
4 本章在达朗伯原理和虚位移原理的基础上，进一步导 出动力学普遍方程和拉格朗日第二类方程（简称拉格朗日 方程）。动力学普遍方程和拉格朗日方程是研究动力学问 题的有力手段，在解决非自由质点系的动力学问题时，显 得十分简捷、规范
Chapter 17: Lagranges equations 8 17-1 General equation of dynamics 817-2 Lagrange's equations of the second kind D 17-3 Integrals of the Lagrange's equations of the second kind
5 §17–1 General equation of dynamics §17–2 Lagrange's equations of the second kind §17–3 Integrals of the Lagrange's equations of the second kind Chapter 17: Lagrange's equations
第十七章拉格朗日方程 §17-1动力学普遍方程 §17-2拉格朗日第二类方程 §17-3拉格朗日第二类方程的积分
6 §17–1 动力学普遍方程 §17–2 拉格朗日第二类方程 §17–3 拉格朗日第二类方程的积分 第十七章 拉格朗日方程
Dynarnics 8 17-1 General equation of dynamics Suppose that there are n particles in a system, particle i being described by m,E, M,a, 0=-m a. Then we have F2+N1+Q;=0 If the system of particle is under the action of the ideal constraints, the Q can be treated as positive forces and we get ∑(F+)=0 The explicit form is ∑(X1-m1x)1+(H-m2)C+(∠1-m1三)D]=0 This is the general equation of dynamics. 7
7 mi Fi Ni ai Qi = −mi ai , , , ; Suppose that there are n particles in a system, particle i being described by . Then we have Fi +Ni +Qi =0 If the system of particle is under the action of the ideal constraints, the can be treated as positive forces and we get Qi ( + ) i =0 i i F Q r The explicit form is [(Xi −mi x i )xi +(Yi −mi y i )yi +(Zi −mi z i )zi ]=0 §17-1 General equation of dynamics This is the general equation of dynamics
学 §17-1动力学普遍方程 设质点系有n个质点,第个质点M1:m,F,N1,a;Q=-ma F+N+Q1=0 若质点系受有理想约束,将Q作为主动力处理,则: ∑(F+)=0 解析式: ∑(X1-m1x)1+(H-m2)C+(∠1-m1三)D]=0 动力学普遍方程
8 Mi mi Fi Ni ai Qi = −mi ai 设质点系有n个质点，第i个质点 : , , , ; Fi +Ni +Qi =0 若质点系受有理想约束，将 Qi 作为主动力处理，则： ( + ) i =0 i i F Q r 解析式： [(Xi −mi x i )xi +(Yi −mi y i )yi +(Zi −mi z i )zi ]=0 §17-1 动力学普遍方程 动力学普遍方程
Dynarnics Under the action of the ideal constraints the sum of the virtual works of the positive forces and of the inertial forces, acting on every particle of the system, along an arbitrary virtual displacement is zero at any moment. LExample 1 a triangular prism b is slipping along the smooth inclined plane of the triangular prism A. The triangular prism A is placed on the smooth horizontal plane. The weights of A and B are M and m, the angle of inclination is a. Determine the acceleration of A Solution Investigate the system consisting bB e of both triangular prisms. This system is under the action of ideal A 5 constraints. It has two degree of P freedom. 4=Ma 77777 B+0 B 2B n,2B=ma
9 [Example 1] A triangular prism B is slipping along the smooth inclined plane of the triangular prism A. The triangular prism A is placed on the smooth horizontal plane. The weights of A and B are M and m, the angle of inclination is . Determine the acceleration of A. Investigate the system consisting of both triangular prisms. This system is under the action of ideal constraints. It has two degree of freedom. , . r r B e B r B e B B A Q ma Q ma Q Q Q Q Ma = = = + = Under the action of the ideal constraints the sum of the virtual works of the positive forces and of the inertial forces, acting on every particle of the system, along an arbitrary virtual displacement is zero at any moment. Solution
力单 在理想约束的条件下,质点系的各质点在任一瞬时受到的 主动力与惯性力在任意虚位移上所作的虚功之和为零。 例1三棱柱B沿三棱柱A的光滑斜面滑动,三棱柱A置于光 滑水平面上,A和B的质量分别为M和m,斜面倾角为a。试求三 棱柱A的加速度。 解:研究两三棱柱组 LB B 成的系统。该系统受理想 约束,具有两个自由度 6x4 A P O=Ma B Oe+O OB=ma, 2B=ma 10
10 例1 三棱柱B沿三棱柱A的光滑斜面滑动，三棱柱A置于光 滑水平面上，A和B的质量分别为M和m，斜面倾角为 。试求三 棱柱A的加速度。 解：研究两三棱柱组 成的系统。该系统受理想 约束，具有两个自由度。 r r B e B r B e B B A Q ma Q ma Q Q Q Q Ma = = = + = , 在理想约束的条件下，质点系的各质点在任一瞬时受到的 主动力与惯性力在任意虚位移上所作的虚功之和为零
