第二模塊 動(dòng)力學(xué)基本原理

1、第二模塊 動(dòng)力學(xué)基本原理（四）牛頓定律，約束、約束力 1、牛頓定律的基本概念。 2、牛頓定律的使用條件 3、質(zhì)點(diǎn)的運(yùn)動(dòng)微分方程 4、約束和約束反力41 牛頓定律1. Newton's Three Fundamental Laws. (28)Formulated by Sir Isaac Newton in the latter part of the seventeenth century these laws can be stated as follows:(1) Newton's FIRST LAW. If the resultant force acting on a

2、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).(2) Newton's SECOND LAW. If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the ma

3、gnitude of the resultant and in the direction of this resultant force.Newtons second law of motion is best understood by imagining the following experiment: A particle is subjected to a force F1 of constant direction and constant magnitude F1. Under the action of that force, the particle is observed

4、 to move in a straight line and in the direction of the force (Fig.4.1.1a). By determining the position of the particle at various instants, we find that its acceleration has a constant magnitude . If the experiment is repeated with forces F2, F3, of different magnitude or direction (Fig.1.1.1b and

5、c), we find each time that the particle moves in the direction of the force acting on it and that the magnitudes , , ,., of the accelerations are proportional to the magnitudes F1, F2, F3,., of the corresponding forces:= constantFig.4.1.1The constant value obtained for the ratio of the magnitudes of

6、 the forces and accelerations is a characteristic of the particle under consideration; it is called the mass of the particle and is denoted by m. When a particle of mass m is acted upon by a force F, the force F and the acceleration a of the particle must therefore satisfy the relationF=ma (4.1.1)Th

7、is relation provides a complete formulation of Newtons second law, it expresses not only that the magnitudes of F and a are proportional but also (since m is a positive scalar) that the vectors F and a have the same direction (Fig.4.1.2). We should note that Eq. (4.1.1) still holds when F is not con

8、stant but varies with time in magnitude or direction. The magnitudes of F and a remain proportional, and the two vectors have the same direction at any given instant. However, they will not, in general, be tangent to the path of the particle.Fig.4.1.2When a particle is subjected simultaneously to se

9、veral forces, Eq. (1.1.1) should be replaced by=ma (4.1.2)where represents the sum, or resultant, of all the forces acting on the particle,(3) Newton's THIRD LAW. The forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense. For eve

10、ry action, there is an equal and opposite reaction; that is, the forces of interaction between two particles are equal in magnitude and opposite in direction.2、牛頓定律的使用條件（1）在運(yùn)動(dòng)學(xué)中參考系可以任意選取，但在動(dòng)力學(xué)中則不能任意選取參考系。（2）慣性參考系：與絕對(duì)靜止空間固連的參考系以及相對(duì)其勻速直線平動(dòng)的參考系。牛頓定律適用于一切慣性參考系。（3）伽利略相對(duì)性原理：一切力學(xué)方程和定律對(duì)所有慣性參考系都是等價(jià)的。3、質(zhì)點(diǎn)的運(yùn)動(dòng)微

11、分方程向量式（牛頓第二運(yùn)動(dòng)定律） 4、約束和約束反力（五）受力分析、受力圖物體的受力分析 受力圖 The free-body diagram (FBD) of a body is a sketch of the body showing all forces that act on it. The term free implies that all supports have been removed and replaced by the forces (reactions) that they exert on the body. 在求解靜力平衡問題時(shí)，必須首先分析物體的受力情況，即進(jìn)行

12、受力分析。根據(jù)問題的已知條件和待求量，從有關(guān)結(jié)構(gòu)中恰當(dāng)選擇某一物體（或幾個(gè)物體組成的系統(tǒng)）作為研究對(duì)象。這時(shí)，可設(shè)想將所選擇的對(duì)象從與周圍的約束（含物體）的接觸中分離出來，即解除其所受的約束而代之以相應(yīng)的約束反力。這一過程稱為解除約束。解除約束后的物體，稱為分離體，畫有分離體及其所受的全部力（包括主動(dòng)力和約束反力）的簡(jiǎn)圖，稱為受力圖。The importance of mastering the FBD technique cannot be overemphasized. Free-body diagrams are fundamental to all engineering discip

13、lines that are concerned with the effects that forces have on bodies. The construction of an FBD is the key step that translates a physical problem into a form that can be analyzed mathematically.Forces that act on a body can be divided into two general categoriesreactive forces (or simply reactions

14、) and applied forces. Reactions are those forces that are exerted on a body by the supports to which it is attached. Forces acting on a body that are not provided by the supports are called applied forced of course, all forces, both reactive and applied, must be shown on be-body diagrams.The followi

15、ng is the general Procedure for constructing a free-body diagram.1. A sketch of the body is drawn assuming that all supports (surfaces of contact, supporting cables, etc.) have been removed.2. All applied forces are drawn and labeled on the sketch. The weigh of the body is considered to be an applie

16、d force acting at the center of gravity, the center of gravity of a homogeneous body coincides with the centroid of its volume.3. The reactions due to each support are drawn and labeled on the sketch. (If the sense of a reaction is unknown, it should be assumed. The Solution will determine the corre

17、ct sense: A positive result indicates that the assumed sense is correct, whereas a negative result means that the correct sense is opposite to the assumed sense.)4. All relevant angles and dimensions are shown on the sketch.When you have completed this Procedure, you will have a drawing (i.e., a be-

18、body drawing) that contains all of the information necessary for writing the equilibrium equations of the body.The most difficult step to master in the construction of FBDs is the determination of the support reactions. Table 4. l shows the reactions exerted by various coplanar supports; it also lis

19、ts the number of unknowns that are introduced on an FBD by the removal of each support. To be successful at drawing FBDs, you must be completely familiar with the contents of Table 4.l. It is also helpful to understand the physical reasoning that determines the reactions at each support, which are d

20、escribed below.畫受力圖是求解靜力學(xué)問題的重要步驟。You should keep the following points in mind when you are drawing free-body diagrams.1. Be neat. Because the equilibrium equations will be derived directly from the free-body diagram, it is essential that the diagram be readable.2. Clearly label all forces, angles, a

21、nd distances with values (if known) or symbols (if the values are not known).3. The support reactions must be consistent with the information presented in Table 4.l.4. Show only forces that are external to the body (this includes support reactions). internal forces occur in equal and opposite pairs

22、and thus will not appear on free-body diagrams.例5一1 重量為G的梯子AB，擱在光滑的水平地面和鉛直墻上。在D點(diǎn)用水平繩索與墻相連，如圖所示。試畫出梯子的受力圖。例52 下圖所示的結(jié)構(gòu)有桿AC、CD與滑輪B鉸接組成。物體重W，用繩子掛在滑輪上。如桿、滑輪及繩子的自重不計(jì)，并忽略各處的摩擦，試分別畫出滑輪B（包括繩索）、桿AC、CD及整個(gè)系統(tǒng)的受力圖。在上例中，當(dāng)取整個(gè)系統(tǒng)為研究對(duì)象時(shí)，AC桿與CD桿在C處不分開，此時(shí)這兩根桿之間的相互作用力稱為內(nèi)力。一般來說，當(dāng)取某個(gè)系統(tǒng)為研究對(duì)象時(shí)，該系統(tǒng)內(nèi)構(gòu)件之間的內(nèi)力都不必畫出，而只畫作用在系統(tǒng)上的外力。需

23、要指出的是，內(nèi)力和外力的區(qū)分不是絕對(duì)的，在一定的條件下，內(nèi)力和外力是可以相互轉(zhuǎn)化的。例如在上例中，如果要畫由桿AC、滑輪B、重物和繩子所組成系統(tǒng)的受力圖，則桿CD對(duì)桿AC的約束反力Fsc（原來整個(gè)系統(tǒng)中兩桿之間的內(nèi)力）就成了該系統(tǒng)的外力。Sample Problem 53The homogeneous，250kg triangular plate in rig（a）is supported by a pin at A and a roller at CDrawn the FBD of the plate and determine the number of unknownsSolutionT

24、he FBD of the plate is shown in Fig（b）The pin and roller supports have been removed and replaced by the reactive forces. The forces acting on he plate re described below. W: The Weight of the PlateThe weight of the plate is W = mg = (250)(9.81) = 2453 N. It acts at the centroid G. Only the horizonta

25、l location of G is shown in the figure, because it is sufficient to determine the line of action of W.Ax and Ay: The Components of the Pin Reaction at AFrom Table 4. l, we see that a pin reaction can be shown as two components A. and Ay, which are equivalent to an unknown force acting at an unknown

26、angle. We have shown Ax acting to the right and Ay acting upward. These directions are chosen arbitrarily; the solution of the equilibrium equations will determine the correct sense for each force. Therefore, the free-body diagram would be correct even if A. or Ay were chosen to act in directions opposite to those shown in Fig. (b).NC: The Normal Reaction at CFrom Table 4. l, the force exerted by a roller support is normal to the inclined surface. Therefore, on the FBD we show the force NC at C, inclined at 30o to the vertical. The FBD contai