中英文第十.pptVIP

* * * * * Nanchang Hangkong University * * Nanchang Hangkong University Chapter 10 General Equations of Particle Dynamics 第十章 质点动力学的基本方程 Section 10.1 Fundamental Laws of Dynamics Newton’s First Law of motion (Law of Inertia) If the resultant force acting on a particle is zero, the particle will remain at rest (if originally at rest) or will move with constant speed in a straight line (if originally in motion ). §10-1 动力学的基本定律 第一定律 (惯性定律): 不受力作用的质点，将保持静止或作匀速直线运动。 Newton’s Second Law of motion (Relationship between Force and Acceleration): If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant and in the direction of this resultant force. 第二定律 重力Gravitation 力的单位:牛[顿], Unit of force Newton (N) Newton’s Third Law of motion (Law of Action Force and Reaction Force) The mutual forces of action and reaction between two particles are equal, opposite, and collinear. 第三定律 (作用与反作用定律): 两个物体间的作用力与反作用力总是大小相等,方向相反,沿着同一直线,且同时分别作用在这两个物体上。 Section 10.2 Differential Equations of Motion of a Particle 1. Projection of Orthogonal Coordinates 或 or §10-2 质点的运动微分方程 1 、在直角坐标轴上的投影 2. Projection of Natural Axes 由Noting 有 We have 2、在自然轴上的投影 First problem The motion is known, determine the force Second problem The force is known, determine the motion Composite problem Combination of the first and second problems 3. Fundamental Problems of particle Dynamics 3 、质点动力学的两类基本问题 第一类问题：已知运动求力. 第二类问题：已知力求运动. 混合问题：第一类与第二类问题的混合. 曲柄连杆机构如图所示.曲柄OA以匀角速度w 转动,OA=r,AB=l,当 ? = r / l 比较小时,以O 为坐标原点,滑块B 的运动方程可近似写为 The mechanism of crank-link is shown as the figure. The crank OA rotates with a constant angular velocity w . Where OA=r, AB=l. The motion of the slide B can be described as follows when l = r / l 1. 例10-1 SAMPLE PROBLEM 10.1 Knowing the mass of the slide B is m, the friction between the mechanism and the mass of link AB are negligible. Determine the reaction force of