Rabu, 16 November 2011

Chapter 12 - Static Equilibrium


               
                In Chapters 10 and 11 we studied the dynamics of rigid objects. Part of this current chapter  addresses  the  conditions under which  a  rigid object  is  in equilibrium. The term  equilibrium implies  either  that  the  object  is  at  rest  or  that  its  center  of mass moves with constant velocity relative to the observer. We deal here only with the former  case,  in which  the object  is  in  static  equilibrium. Static equilibrium  represents  a common  situation  in  engineering  practice,  and  the  principles  it  involves  are  of special interest to civil engineers, architects, and mechanical engineers. If you are an engineering student, you will undoubtedly  take an advanced course  in statics  in  the future.The  last  section of  this  chapter deals with how objects deform under  load  conditions. An elastic object returns  to  its original shape when  the deforming  forces are removed. Several elastic constants are defined, each corresponding to a different type of deformation.

12.1 The Conditions for Equilibrium
                Consider  a  single  force F acting  on  a  rigid  object,  as  shown  in  Figure  12.1. The effect  of  the  force  depends  on  the  location  of  its  point  of  application  P.  If  r is  the position vector of this point relative to O, the torque associated with the force F about O is given by Equation 11.1:    
                Recall from the discussion of the vector product in Section 11.1 that the vector Ʈ is perpendicular to the plane formed by r and F. You can use the right-hand rule to determine  the direction of  Ʈ as  shown  in Figure 11.2. Hence,  in Figure 12.1Ʈ is directed toward you out of the page.
As you can  see  from Figure 12.1,  the  tendency of F to  rotate  the object about an axis  through  O depends  on  the moment  arm  d,  as  well  as  on  the magnitude  of  F. Recall that the magnitude of Ʈ is Fd (see Eq. 10.19). According to Equation 10.21, the net torque on a rigid object will cause it to undergo an angular acceleration.

                In  the  current discussion, we want  to  look  at  those  rotational  situations  in which the angular acceleration of a rigid object is zero. Such an object is in rotational equilibrium. Because ∑Ʈ = I for rotation about a fixed axis,  the necessary condition  for rotational equilibrium  is that the net torque about any axis must be zero. We now have two necessary conditions for equilibrium of an object:  
                                        

In the special case of static equilibrium, which  is the main subject of this chapter, the object  is at rest relative to the observer and so has no linear or angular speed (that is, v CM= 0 and w = 0).
     
 
        
           
12.1 (a). The unbalanced torques due to the forces in figure 12.2 cause an angular acceleration even though the linear acceleration is zero.
   
 




12.2 (b) . Notice that the lines of action of all the forces in figure 12.3 intersect at a common point. Thus, the net torque about this point is zero. This zero value of the net torque is independent of the values of the forces. Because no force has a downward component, there is a net force and the object is not in force equilibrium.

The two vector expressions given by Equations 12.1 and 12.2 are equivalent, in general,  to six scalar equations:  three  from  the first condition  for equilibrium, and  three from the second (corresponding to x, y, and z components). Hence, in a complex system  involving several forces acting  in various directions, you could be faced with solving a  set of equations with many unknowns. Here, we  restrict our discussion  to  situations  in which all  the  forces  lie  in  the xy plane. (Forces whose vector representations are in the same plane are said to be coplanar.) With this restriction, we must deal with only three scalar equations. Two of these come from balancing the forces in the x and y directions. The third comes from the torque equation—namely, that the net torque about a perpendicular axis through any point in the xy plane must be zero. Hence, the two conditions of equilibrium provide the equations
                               

                Where the location of the axis of the torque equation is arbitrary, as we now show. Regardless of  the number of  forces  that  are  acting,  if  an object  is  in  translational equilibrium and if the net torque is zero about one axis, then the net torque must also be zero about any other axis. The axis can pass through a point that is inside or outside the boundaries of the object. Consider an object being acted on by several forces such that the resultant force  ∑F = F1 + F2 + F3 =… = 0. Figure 12.4 describes this situation (for clarity, only four forces are shown). The point of application of F1 relative to O is specified by the position vector r1. Similarly, the points of application of F2, F3, ... are specified by r2, r3, ... (not shown). The net torque about an axis through O is
                           
                Now consider another arbitrary point O’ having a position vector r’ relative  to O. The point of application of F1 relative  to O’ is  identified by  the  vector  r1 - r’. Like wise, the point of application of F2 relative to O’ is r2 - r’, and so forth. Therefore, the torque about an axis through O’ is
         
                            Because  the net  force  is assumed  to be  zero  (given  that  the object  is  in  translational equilibrium), the last term vanishes, and we see that the torque about an axis through O’ is equal to the torque about an axis through O. Hence, if an object is in translational equilibrium and the net torque is zero about one axis, then the net torque must be zero about any other axis.
12.2 More on the Center of Gravity
                We have seen that the point at which a force is applied can be critical in determining how  an  object  responds  to  that  force.  For  example,  two  equal-magnitude  but  oppositely directed forces result in equilibrium if they are applied at the same point on an object. However,  if  the point of application of one of  the  forces  is moved, so  that  the two  forces no  longer act along  the same  line of action,  then  the object undergoes an angular acceleration.
                Whenever we deal with  a  rigid  object,  one  of  the  forces we must  consider  is  the gravitational  force  acting  on  it,  and  we must  know  the  point  of  application  of  this force. As we learned in Section 9.5, associated with every object is a special point called its center of gravity. All the various gravitational forces acting on all the various mass elements of  the object are equivalent  to a single gravitational  force acting  through  this point. Thus, to compute the torque due to the gravitational force on an object of mass M, we need only consider the force Mg acting at the center of gravity of the object. How do we find  this  special point? As we mentioned  in Section 9.5,  if we assume that g is uniform over the object, then the center of gravity of the object coincides with its center of mass. To see that this  is so, consider an object of arbitrary shape  lying  in the  xy plane,  as  illustrated  in Figure  12.5. 


Suppose  the object  is divided  into  a  large number  of  particles  of masses m1, m2, m3,  .  .  .  having  coordinates  (x1,  y1),  (x2,  y2), (x3, y3),  .  .  .  . In Equation 9.28 we defined  the x coordinate of  the center of mass of such an object to be
         
                                                        
                We use a similar equation  to define  the y coordinate of  the center of mass, replacing each x with its y counterpart. Let  us now  examine  the  situation  from  another  point  of  view  by  considering  the gravitational force exerted on each particle, as shown in Figure 12.6. 

Each particle contributes a torque about the origin equal in magnitude to the particle’s weight mg multiplied by  its moment arm. For example,  the magnitude of  the  torque due  to  the  force m1g1 is m1g1x1, where g1 is  the value of  the gravitational acceleration at  the position of the particle of mass m1. We wish to locate the center of gravity, the point at which application of the single gravitational force Mg (where M = m1 + m2 + m3 + … is the total mass of  the object) has  the  same effect on rotation as does  the combined effect of all the individual gravitational forces migi. Equating the torque resulting from Mg acting at the center of gravity to the sum of the torques acting on the individual particles gives
                                
                                                        
This expression accounts  for  the  fact  that  the value of g can  in general vary over  the object.  If  we assume  uniform  g over  the  object  (as  is  usually  the  case),  then  the g terms cancel and we obtain

                  Comparing this result with Equation 9.28, we see that the center of gravity is located at the center of mass as long as g is uniform over the entire object. In several examples presented  in  the next  section, we will deal with homogeneous,  symmetric objects. The center of gravity for any such object coincides with its geometric center.
            

12.3 Examples of Rigid Objects in Static Equilibrium
                The photograph of the one-bottle wine holder in Figure 12.7 shows one example of a balanced mechanical  system  that  seems  to  defy  gravity.  For  the  system  (wine  holder plus bottle)  to be  in equilibrium,  the net external  force must be  zero  (see Eq. 12.1) and the net external torque must be zero (see Eq. 12.2). The second condition can be satisfied only when the center of gravity of the system is directly over the support point. When  working  static  equilibrium  problems,  you must  recognize  all  the  external forces acting on the object. Failure to do so results in an incorrect analysis. When analyzing an object in equilibrium under the action of several external forces, use the following procedure.

EXAMPLE..


A seesaw consisting of a uniform board of mass M and length l supports a father and daughter with masses mf and md, respectively, as shown in Figure 12.8. The support (called the fulcrum) is under the center of gravity of the board, the father is a distance d from the center, and thedaughter is a distance    l/2 from the center. Determine the magnitude of the upward force n exerted!
Answer:
Solution First note that, in addition to n, the external forces acting on the board are the downward forces exerted by each person and the gravitational force acting on the board. We know that the board’s center of gravity is at its geometric center because we are told that the board is uniform. Because the system is in static equilibrium, the net force on the board is zero. Thus, the upward force n must balance all the downward forces. From  Fy = 0    , and defining upward as the positive y direction, we have
 

(The equation  also aFy = 0pplies, but we do not need to consider it because no forces act horizontally on the board).

Tidak ada komentar:

Poskan Komentar