学 「例1列车在水平轨道上行驶,车厢内悬挂一单摆,当车厢向 右作匀加速运动时,单摆左偏角度α,相对于车厢静止。求车 厢的加速度a。 16
16 [例1] 列车在水平轨道上行驶，车厢内悬挂一单摆，当车厢向 右作匀加速运动时，单摆左偏角度 ，相对于车厢静止。求车 厢的加速度 a
Dynamic Solution Investigate the single pendulum, add the virtual inertial forces Q=-ma(2=ma) According to the dynamic-static method we have >X=0, mg sin a-Ocosa=0 Solving it we get a=g tga The angle a changes with the acceleration a, when a does not the acceleration a of the train can be calculated. This is the ca change, the angle a does not change, too. If we know the angle theory of a pendulum accelerometer 17
17 Investigate the single pendulum, add the virtual inertial forces. Q = −ma ( Q = ma ). X = 0, mgsin −Qcos = 0. a = g tg. The angle changes with the acceleration , when does not change, the angle does not change, too. If we know the angle the acceleration of the train can be calculated . This is the theory of a pendulum accelerometer. a a a Solution According to the dynamic-static method we have Solving it we get
解:选单摆的摆锤为研究对象 虚加惯性力 O=-ma (o=ma 由动静法,有 ∑X=0, mg Sin a-Qcos=0 解得 a=g tga a角随着加速度a的变化而变化,当a不变时,a角也 不变。只要测出a角,就能知道列车的加速度a。摆式加速 计的原理
18 选单摆的摆锤为研究对象 虚加惯性力 Q = −ma (Q = ma ) X =0, mgsin−Qcos=0 a=gtg 角随着加速度 的变化而变化，当 不变时， 角也 不变。只要测出 角，就能知道列车的加速度 。摆式加速 计的原理。 a a a 解： 由动静法, 有 解得
Dynamics 815-2 D'Alembert's principle for a system of particles A system of particles consists of n particles. For every particle we have F+N1+Q1=0(i=1,2,,n) For the whole system, the positive force system, the system of the reaction forces of constraints and the inertial force system form together of an equivalent force system. This is D'Alembert's principle for a system of particles. It can be expressed by as o∑F+∑N+∑g=0 ∑m(F)+∑m(N)+∑m(Q)=0 Please, note that 2F=0, >mo(F)=0. If we classify the forces by internal and external forces then we have ∑F+∑2=0, ∑而(F)+∑而Q=0。 19
19 §15-2 D’Alembert’s principle for a system of particles A system of particles consists of n particles. For every particle we have F N Q 0 ( i 1,2,......,n ) i + i + i = = For the whole system, the positive force system, the system of the reaction forces of constraints and the inertial force system form together of an equivalent force system. This is D’Alembert’s principle for a system of particles. It can be expressed by as follows. ( ) ( ) ( ) 0. 0, + + = + + = O i O i O i i i i m F m N m Q F N Q Please, note that . If we classify the forces by internal and external forces then we have =0, ( )=0 ( ) (i) O i i Fi m F 。 + = + = ( ) ( ) 0 0 , ( ) ( ) O i e O i i e i m F m Q F Q
学 §15-2质点系的达朗伯原理 设有一质点系由n个质点组成,对每一个质点,有 F+N1+Q=0(i=1,2 对整个质点系,主动力系、约束反力系、惯性力系形式上 构成平衡力系。这就是质点系的达朗伯原理。可用方程表示为: 「∑F+∑N+xQ=0 ∑mo0(F1)+∑mo(N1)+∑mo(Q)=0 注意到∑=0,∑m(F)=0,将质点系受力按内力、外力 划分,则 ∑F+∑Q1=0 ∑m(F)+∑m(Q)=0 20
20 §15-2 质点系的达朗伯原理 对整个质点系，主动力系、约束反力系、惯性力系形式上 构成平衡力系。这就是质点系的达朗伯原理。可用方程表示为： ( ) ( ) ( ) 0 0 + + = + + = O i O i O i i i i m F m N m Q F N Q 设有一质点系由n个质点组成，对每一个质点，有 F N Q 0 ( i 1,2,......,n ) i + i + i = = 注意到 , 将质点系受力按内力、外力 划分, 则 =0, ( )=0 ( ) (i) O i i Fi m F + = + = ( ) ( ) 0 0 ( ) ( ) O i e O i i e i m F m Q F Q