Mathematics
Rigid Bodies in Equilibrium
Rigid bodies in equilibrium refer to objects that are not deforming or moving. In this state, the sum of all forces and torques acting on the body is zero, and the body remains at rest or moves with constant velocity. This concept is important in physics and engineering for analyzing the stability and balance of structures and mechanical systems.
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12 Key excerpts on "Rigid Bodies in Equilibrium"
eBook - PDFStatics and Mechanics of Materials
An Integrated Approach
- William F. Riley, Leroy D. Sturges, Don H. Morris(Authors)
- 2017(Publication Date)
- Wiley(Publisher)
EQUILIBRIUM: RIGID AND DEFORMABLE BODIES 6 6-1 INTRODUCTION The concept of equilibrium was introduced in Chapter 3 and applied to a system of forces acting on a particle. Since any system of forces acting on a particle is a concurrent force system, a particle is in equilibrium if the resultant force R of the force system acting on the particle is zero. For the case of a rigid body, it was shown in Chapter 5 that the most general force system can be expressed in terms of a resultant force R and a resultant couple C. Therefore, for a rigid body to be in equilibrium, both the resultant force R and the resultant couple C must van- ish. These two conditions are expressed by the two vector equations R F x i F y j F z k 0 C M x i M y j M z k 0 (6-1) Equations (6-1) can be expressed in scalar form as F x 0 F y 0 F z 0 M x 0 M y 0 M z 0 (6-2) Equations (6-2) are the necessary conditions for equilibrium of a rigid body. If all of the forces acting on a body can be determined from these equations, then they are also the sufficient conditions for equilibrium. The forces and moments that act on a rigid body are either external or in- ternal. Forces applied to a rigid body by another body or by the earth are external forces. Fluid pressure on the wall of a tank or the force applied by a truck wheel to a road surface are examples of external forces. The weight of a body is another example of an external force. If the body of interest is composed of several parts, the forces holding the parts together are defined as internal forces. In addition, in- ternal forces hold the particles forming the body together. In Chapter 4, the inten- sity of these internal forces was called stress. If the intensity of the internal force is too large, the body may rupture or deform excessively. External forces can be divided into applied forces and reaction forces.
No longer available |Learn more- Ping YI, Jun LIU, Feng JIANG(Authors)
- 2022(Publication Date)
- EDP Sciences(Publisher)
Chapter 4 Equilibrium of Rigid Bodies Objectives Study equations of equilibrium for two- or three-dimensional force system. Examine the constraints and draw a proper free-body diagram for a rigid body or a system of rigid bodies. Solve equilibrium problems of a rigid body or a system of rigid bodies. Understand the concepts of redundant constraints, improper constraints and statically determinacy or indeterminacy. Analyze the forces in members of a simple truss. 4.1 Conditions for Rigid-Body Equilibrium From chapter 3, a general force system acting on a body can be simplified to a resultant force F 0 R plus a resultant couple moment M RO at an arbitrarily chosen simplified point O. When the resultant force F 0 R ¼ 0 and the resultant couple moment M RO ¼ 0, the force system is equivalent to zero and imparts neither translational nor rotational effect to the body. Then the rigid body is in equilibrium. From equation (3.10), the necessary and sufficient conditions for equilibrium of a rigid body are X F i ¼ 0 X M O ¼ 0 ( ð4:1Þ It should be noted that P M O is the vector sum of all the couple moments and the moments of all the forces about point O. Resolving forces and moments into their rectangular components, the two vector equilibrium equations can be replaced by the following six scalar equations: DOI: 10.1051/978-2-7598-2901-9.c004 © Science Press, EDP Sciences, 2022 X F x ¼ 0 X M x ¼ 0 X F y ¼ 0 X M y ¼ 0 X F z ¼ 0 X M z ¼ 0 8 > > < > > : ð4:2Þ These six independent scalar equations can be used to solve for at most six unknowns. If it is a three-dimensional (3-D) concurrent force system shown in figure 4.1, three moment equations in equation (4.2) are always satisfied and become useless.
eBook - PDFMechanics
Lectures on Theoretical Physics
- Arnold Sommerfeld(Author)
- 2013(Publication Date)
- Academic Press(Publisher)
In such a case we speak of a spherical top. In a spherical top (cf. Fig. 40c) any axis is a principal axis. § 23. Statics of Rigid Bodies This subject forms the theoretical basis for the whole field of structural mechanics dealing with such topics as the construction of bridges, trusses, arches, etc., and for this reason it is treated with the greatest detail in the texts of mechanical engineering, both analytically and graphically. Here we shall restrict ourselves to the general features of the subject. (1) The Conditions of Equilibrium These, like all questions of equilibrium, are governed by the principle of virtual work. Since this principle can be regarded as the special case of d'Alembert's principle in which the inertial forces vanish, our present analysis can be directly modeled after that of the principles of linear and angular momentum of § 13. Indeed the virtual displacements (translation and rotation) used there are evidently compatible with the internal connec-tions of the rigid body and correspond to the two component parts of the general motion of a rigid body considered in the preceding section. By deleting the inertial forces in Eqs. (13.3) and (13.9), we obtain the general conditions of equilibrium of a rigid body, (1) 2 F *=°> ]>>*=0. The F k are external forces acting at arbitrary points P k of the rigid body. The first Eq. (1) asks us to lay off the force vectors end to end in arbitrary order and with no regard to their points of application, and to examine the resulting force polygon. According to Eq. (1) for equilibrium the polygon of forces must be closed. The L k are the moments of the F k about a reference point 0 whose choice is arbitrary but which must be the same for all the F k . The second Eq. (1) asks us to replace these L k by their (axial) vector representations 126 The Rigid Body IV.23 (cf. p. 37) and to examine the polygon of torques arising when all these vectors are added vectorially.
eBook - PDFComputational Solid Mechanics
Variational Formulation and High Order Approximation
- Marco L. Bittencourt(Author)
- 2014(Publication Date)
- CRC Press(Publisher)
2 EQUILIBRIUM OF PARTICLES AND RIGID BODIES 2.1 INTRODUCTION In this chapter, we review the equilibrium conditions of particles and rigid bodies subjected to external loads. The main objective is to use the already known equilibrium concepts to introduce some useful notions for later chapters, such as virtual actions and the principle of virtual work. It is assumed that the particles and rigid bodies are static and do not move. To satisfy this as-sumption, a sufficient number of kinematic constraints represented by supports should be used to avoid any rigid motion. Initially, we present diagrammatic conventions for supports and loadings. Subsequently, the clas-sical Newtonian conditions for equilibrium of particles are considered, which are based on the con-cepts of forces and force resultant on the particle. The analytical approach is also introduced, using the concept of virtual motion (or displacement) and the principle of virtual power (or work). These concepts are extended to rigid body equilibrium. At the end of the chapter, several examples for equilibrium of particles and rigid bodies are presented. For notation purposes, all vectorial quantities are represented by lowercase bold letters. 2.2 DIAGRAMMATIC CONVENTIONS As mentioned before, the kinematic constraints are represented by supports. They are needed in a sufficient number to avoid any rigid body motion. The kinematic constraints must be satisfied for any motion of the considered mechanical system constituted by particles and rigid bodies. The following sections present the most used diagrams to represent loads and supports. B Pins Joint A (a) Articulation. A (b) Pin. C A D (c) Roller. D A C (d) Fixed pin. (e) Clamp. Figure 2.1 Supports. 2.2.1 SUPPORTS It is important to define symbols to represent the supports, which are responsible for maintaining a mechanical system at rest when subjected to external loads.
- Barry Goodno, James Gere(Authors)
- 2018(Publication Date)
- Cengage Learning EMEA(Publisher)
If equilibrium is to be achieved, a body must be supported or restrained in such a way that it is stable and at rest, and therefore the resultant of all forces and the resultant of all couples acting on the body each must sum to zero. A restrained body cannot undergo rigid-body motion due to the application of static forces. One exception is a body resting on a rough surface whose motion is opposed by static and kinetic friction. Start by constructing a free-body diagram of the body to display all applied (or active) and reactive forces. To do so, first isolate the body under study and then apply the equations of equilibrium to the free-body diagram to find exter- nal reaction forces and couples or internal forces at points of interest. Begin by writing the basic static equilibrium equations and then applying them to the solution of a variety of structures using both scalar and vector operations. Statics is most often concerned with bodies at rest, so assume that both veloc- ity and acceleration of the body are zero. You can also consider the case of a body in motion—but having zero acceleration—with the motion opposed by kinetic friction forces. 3.2 Free-Body Diagrams A static equilibrium analysis is impossible without a correct free-body diagram. The free-body diagram represents the plan for carrying out an analysis to solve for unknown reactions and possibly for internal forces as well. The free-body dia- gram is drawn during the conceptualization stage of the problem-solving activity prior to writing equilibrium equations or selecting a mathematical approach. This section first examines the equilibrium equations needed for either 2D or 3D solutions in vector or scalar algebra format. Then seven different support or restraint conditions for use in constructing a stable model of the component are presented. Finally, the importance of free-body diagram construction is illus- trated through a number of practical examples.
eBook - PDF- Philip Dyke, Roger Whitworth(Authors)
- 2017(Publication Date)
- Red Globe Press(Publisher)
This ensures no translation, and is also imposed on rigid bodies for the same reason. However, since rigid bodies also have volume, there is a second condition to do with it turning that was totally absent when we discussed particles. Particles have no volume, so who can tell if they twist? The second condition we dwell on here is that for rigid bodies, the total moment of the forces acting on the body about any point must be zero to ensure equilibrium in terms of rotation. In vector form these conditions are: a X F i 0 no translation b X M i X r i F i 0 no rotation Condition (b) applies about any point in the rigid body. Normally there are one or two convenient ones, but each will result in an equivalent condition. In general, these two vector conditions for equilibrium lead to six scalar equations. Here, however, most of the time will be spent applying them to two-dimensional problems in which translation takes place in the plane and rotation takes place about the axis perpendicular to this plane. There are then just three equations, two that arise from no translation and the third that arises from no rotation. Here are a few examples that illustrate the wide variety of problem types to which this theory can be applied. Example 10.5 A uniform ladder AB of weight W rests with the end A against a smooth vertical wall and with B on rough horizontal ground. A window-cleaner of weight 2 W finds that when the ladder is inclined at an angle of 60 he can only climb the ladder to 3 4 of its length without the ladder slipping. Find the value of , the coefficient of friction between the ladder and the horizontal ground. ( continued ) An Introduction to Rigid Body Dynamics 257 Solution Figure 10.8 shows all of the forces acting on the ladder when the window-cleaner is at his highest point P . At the ground, A , the reaction force on the ladder has been divided into two components: F is the friction force and R is the normal reaction force.
- Arthur Haas, T. Verschoyle(Authors)
- 2020(Publication Date)
- De Gruyter(Publisher)
CHAPTER I I I THE MOTION OF RIGID BODIES § 23. The Conception of a Rigid Body. THE conception of a material point, which formed the basis of our previous considerations, provides the simplest form in which we can discuss the motion of bodies which, as they occur in nature, are really solid. It follows that this only suffices for simple cases from the elementary fact of experience that, in a solid body, it is quite possible for some individual parts to be at rest while others are in motion. In our considerations of the motion of solid bodies we may most readily take into account the necessity of imagining a solid body to be composed of separate parts, by substituting the conception of a system of discrete particles for that of the simple particle. At the same time we attribute a special property to this system, which corresponds to the most important property shown, in greater or less approximation, by actual solid bodies. This arises from the fact that the form and internal distribution of mass in solid bodies vary, in general, so little during motion, and under the action of forces, that such variations can be neglected for the purposes of many dynamical investigations. We accordingly define a rigid body as a system of particles whose mutual distances are invariable} Denoting the quite arbitrary number of the particles composing the body by n, it follows that the freedom of motion of the system is limited by equations of condition of the form (1) (z h — x h ) 2 + (y k — y h ) 2 + (z k — z h ) 2 = a hk 2 ; where awe signifies the constant distance between the h' h and the hf h particle whose coordinates, with respect to any given coordinate system, are Xh, yh, Zh and Xk, yk, Zk respec-tively. 1 The fact that we can in this way obtain a picture of the mechanical behaviour of solid bodies, which is sufficient for many purposes, has nothing whatever to do with the question of the molecular structure of matter.
eBook - PDFApplied Mathematics
Made Simple
- Patrick Murphy(Author)
- 2014(Publication Date)
- Butterworth-Heinemann(Publisher)
Rigid Bodies in Equilibrium under the Action of Coplanar Forces 201 CHAPTER ELEVEN RIGID B O D I E S I N E Q U I L I B R I U M U N D E R T H E A C T I O N O F C O P L A N A R F O R C E S Beginners often experience considerable difficulty in applying the principles of moments and resolution to systems of coplanar forces. The difficulty is increased still further when the forces have to be obtained by observing the application of the laws of friction, as well as a general feel for the likely equilibrium position of a rigid body. (1) Point of Concurrency Our first concern is with a system of three coplanar forces acting on a body in equilibrium, and we shall describe either the system of forces or the body as being in equilibrium. Before discussing particular cases let us be quite sure we understand that any three non-parallel forces in equilibrium must be concurrent. For example, suppose we have three coplanar forces F, Q, and R in equilibrium. If we find the resultant of Ρ and Q to be S passing through their point of intersection, then we have reduced the number of forces to two, i.e. S and R, which must be equal and opposite for equilibrium. 2. Two equal masses of 0-5 kg are connected by a light inextensible string. One mass rests on a rough horizontal plane and the other is held at rest with the string taut over a pulley in the manner of Fig. 175. With the coefficient of sliding friction μ = i, find the acceleration of the system when the hanging mass is released. Find also the tension in the string during the motion, (g = 9-81 m s~ 2 ) 3. Given that the body on the horizontal plane in Question 2 is 1 m from the pulley, calculate the time taken for this body to reach the pulley. 4. Two equal masses of 1 kg each are connected by a light inextensible string. The string passes over a smooth pulley fixed to the top of a rough plane inclined at 30° to the horizontal. One mass rests in contact with the plane, the other hangs freely.
eBook - PDF- Richard C. Hill, Kirstie Plantenberg(Authors)
- 2013(Publication Date)
- SDC Publications(Publisher)
The normal force is distributed throughout the contact patch. We are idealizing the normal force and modeling it as a single force acting at a single point. Conceptual Dynamics Kinetics: Chapter 6 – Rigid Body Newtonian Mechanics 6 - 18 An imbalance of forces acting on a body causes it to accelerate. Because a rigid body has size, forces may be applied at some distance from each other. This may result in both an imbalance of forces and an imbalance of moments. These imbalances will cause the body to translate as well as rotate. Pure translation occurs when the moments balance, but the forces do not. This means that the body will translate, but not rotate. As was the case with particles, the translational motion of a rigid body can be described by Newton’s second law given in Equation 6.5-3. Newton’s second law: G m F a (6.5-3) F = external forces acting on the rigid body m = mass of the body a G = acceleration of the center of mass Since the acceleration of each point on a rotating rigid body is not necessarily the same, the acceleration in Newton’s second law refers to the acceleration of the center of mass (G). If, however, the rigid body is undergoing pure translation, then the acceleration of every point on the rigid body is equal. In addition to Equation 6.5-3, an equation is needed to describe how the offset forces prevent the body from rotating. If you remember your course on statics, this would imply that the sum of the moments equals zero, M = 0. In statics, however, the body under consideration is not accelerating like it may be in a dynamic situation. Do you think this fact changes this equilibrium equation? Consider the car shown in Figure 6.5 - 2. Imagine you are driving the car down a road with constant velocity. Under this situation, all of the moments are balanced. But what happens when you suddenly slam on the brakes? The car tips forward*.
eBook - PDFTheoretical Mechanics for Sixth Forms
in Two Volumes
- C. Plumpton, W. A. Tomkys(Authors)
- 2017(Publication Date)
- Pergamon(Publisher)
CHAPTER XX THE EQUILIBRIUM OF RIGID BODIES 20.1. Conditions of Equilibrium In Volume I, § 2.8 we discussed the conditions of equilibrium for a body acted upon by three forces only and in § 6.3 we discussed the equi-librium of a rigid body acted upon by a system of coplanar forces. From the conclusions of Chapter CIC of this volume, together with the statement concerning the equilibrium of a rigid body made in § 19.4, we now classify sets of conditions, each necessary and sufficient to ensure the equilibrium of a rigid body, as follows. 1. For a Rigid Body Acted Upon by Two Forces Only The forces must be equal in magnitude and opposite in direction, and they must act in the same straight line. 2. For a Rigid Body Acted Upon by Three Non parallel Forces Only (a) The forces must act at a point, and (b) The vector sum of the forces must be zero. Condition (b) may be tested by any one of: the triangle of vectors, Lami's theorem, the law of the resolved parts. 3. For a Rigid Body Acted Upon by Any System of Parallel Forces etTHE R (a) the sum of the forces, considered as directed quantities, must be zero, and (b) the sum of the moments of the forces about an arbitrary point in the plane must be zero, OR the sum of the moments of the forces about any two points of the plane, which do not lie in a line parallel to the forces, separately vanish. 596
eBook - PDFIntroduction to Physics
Mechanics, Hydrodynamics Thermodynamics
- P. Frauenfelder, P. Huber(Authors)
- 2013(Publication Date)
- Pergamon(Publisher)
A system is in neutral equilibrium if a small displacement gives rise to no new forces or moments. This type of equilibrium exists when the center of gravity of a rigid body always remains vertically below, above, or at the point of support. For example, a sphere on a horizontal plane and a railroad car on hori-zontal rails are in neutral equilibrium. It is important and interesting to consider whether an equilibrium state is stable or unstable when an object is supported on legs and the center of gravity lies above the supporting surface. We define the tipping lines of the Σ *Ί = 0, and ΣΜ ( = 0. FIG. 2-43. Stable equilibrium. FIG. 2-44. Unstable equilibrium. S T A T I C S O F R I G I D B O D I E S 61 body as the external tangents to the points of contact between the body and the supporting surface. If the body is tipped by a small angle about a possible line of tipping, then, in general, a restoring moment is set up. However, if as a result of the tipping, the center of gravity reaches a point vertically above the tipping line, the equilibrium becomes unstable. Depending on the nature of some accidental disturbance, the object may then return to its original equilibrium state or move to a new state of stable equilibrium. For example, a three-legged stool has three tipping lines (Fig.2-45); atable has four tipping lines (Fig. 2-46); and a pentagonal prism has five tipping lines (Fig. 2-47). The law of moments shows that a rigid body is more stable, V FIG. 2-45 FIG. 2-46 FIG. 2-45. Tipping lines of a three-legged stool. FIG. 2-46. Tipping lines of a table. FIG. 2-47. Tipping lines of a pentagonal prism. (a) the greater the horizontal distance from the center of gravity to the nearest (most dangerous) tipping line, (b) the lower the center of gravity, and (c) the larger the weight. These facts are expressed by the concept of stability.
eBook - PDF- Cho S.(Author)
- 2022(Publication Date)
- Springer(Publisher)
135 C H A P T E R 8 Dynamics of Rigid Bodies 8.1 INTRODUCTION In the last chapter kinematics of rigid bodies has been presented. In the present chapter the dy- namics and kinetics of rigid bodies are considered. Section 8.2 is concerned with the derivation of equations of motion of rigid bodies in 3D space. Equations of motion of rigid bodies in 2D space are presented in Section 8.3. The method of work and energy is introduced in Section 8.4. Impulse, momentum, and angular momentum of rigid bodies are included in Section 8.5. Equa- tions for conservation of momentum and angular momentum are considered in Section 8.6. Section 8.7 has to do with the introduction to impulse motion while eccentric impact of rigid bodies is dealt with in Section 8.8. 8.2 EQUATIONS OF MOTION OF RIGID BODIES IN 3D SPACE The two most fundamental equations for rigid body dynamics are the force and angular momen- tum. In this section the equations for translational motion and rotational motion are presented, respectively, in the following two sections. In Section 8.2.3, equations of constrained motions are introduced. The Euler angles and motion of a gyroscope are considered in Section 8.2.4. The final section is concerned with the analysis of steady precession. 8.2.1 EQUATIONS OF TRANSLATIONAL MOTION These equations can be derived similar to those obtained for a system of particles presented in Chapter 6. Specifically, by making use of Equations (6.14) and (6.24), one has the equation for the forces n X i D1 * F i D m * a D m * a G ; (8.1) where now the term on the lhs is the externally applied forces acting on the rigid body while m is the mass and * a G , the acceleration at the center of mass G of the rigid body. The symbol * a in Equation (6.24) is replaced by * a G to distinguish the accelerations at the mass center of a system of particles there and that in Equation (8.1) of a rigid body, as shown in Figure 8.1 in which OXYZ is the fixed or Newtonian frame of reference (FFR).
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