Equilibrium and Elastacity
 The work of a curious fellow

If everything comes to nothing, here we must remain...
 In general if an object is not accelerated, either in translation or rotation, the object is said to be in equilibrium. Equilibrium comes in two flavors, static equilibrium and dynamic equilibrium. If an object is not moving in our reference frame it is in static equilibrium. If it is moving with constant velocity, both linear and angular, it is in dynamic equilibrium. In this course we are going to focus on static equilibrium. Static equilibrium principles are of most interest to folks who want to determine that their structures are going to stay where they put them. This is an important point for architects, civil engineers, mechanical engineers and carpenters. If there is a net force, F, on an object of mass m, it will be given an acceleration a = F/m. If there is a net torque, t, on an object of moment of inertia I, about any origin we may choose, it will be given an angular acceleration a = t/I. The null acceleration definition of equilibrium above means that there are two necessary conditions for equilibrium. The vector sum of all forces must be zero and the vector sum of all torques must be zero. In the special case of static equilibrium, the velocity of the center of mass is also zero, as is the angular velocity. This equilibrium notion is just a different slant on Newton's first law. Since both forces and torques are vectors in up to three dimensions each, the conditions that forces and torques add up to zero in general involves six independent scalar equations. If several force and torque vectors are applied to an object the resulting expressions can get too messy to handle. To convey the ideas without getting bogged down in the math we will work with the situation where the forces all lie in the (x,y) plane. Many real equilibrium problems may be expressed in terms of coplanar vectors of this sort. With this restriction there are only three equations to deal with, one in x, one in y and one in q. The business about the zero torque condition being true for any origin refers to the fact that torque is defined always relative to some point. The zero torque condition must hold for any point in the (x,y) plane chosen as the origin for torque determinations.
 If an object may be treated as a particle there is no possibility of it rotating so there is only one condition for equilibrium. It must experience no translational acceleration. This is not a very interesting situation, having a particle just lying there, doing nothing, so we will be working with rigid objects that have some noticeable size. In particular, for now, we will be working with a disk about 30 centimeters in diameter with attachment points around the edge and a mass of 2.25 kg. We will pull on it, try to spin it and generally give it a hard time as we explore the idea of static equilibrium. Consider the effect of a force, F, applied straight up the y-axis at the top of a solid disk, by pulling straight up on a pin attached near the edge of the disk. In the absence of any other forces, the disk will accelerate up the y-axis such that d2y/dt2=F/m, where m is the mass of the disk. Pretty clearly this apparatus is not in static equilibrium. In fact in a short time it would disappear off the top of the display, never to be seen again. Since we are working with a mathematical model rather than reality itself, it is not difficult for us to ride along in the reference frame in which the disk remains at rest. Of course such an accelerated reference frame is not inertial. Being in a non-inertial reference frame, we find that unexplained forces crop up. To us it seems that everything, including the disk itself is being pulled in the negative y direction such that anything of mass m has weight F. This works out nicely to provide the reaction force against the pin required by Newton's third law. According to Newton's third law, the reaction force the disk applies to the pin is equal in magnitude and opposite in direction to the applied force. This reaction force is transmitted by the inter-atomic forces in the disk material to the pin from the center of mass of the disk where inertial forces are manifest. The Flying Disk model, illustrated at the left, deals with this situation.
 Fortunately we can avoid having to chase our disk at ever increasing velocity in order to keep it under our observation. The situation we observed when traveling along with the disk is exactly that we find in a stationary laboratory in the presence of a gravitational field. All we have to do is replace the label on our inertial reaction force with weight. The Hanging Disk model reflects this change. It was exactly the line of reasoning we just used that led Einstein to the general theory of relativity. Since gravity is a major player in static equilibrium problems I thought this little digression might be good for us. Notice that there is no such thing as a single force. Either inertia or gravity kicks in an additional force applied at the center of mass.
 Now that we have our apparatus under control, let's consider the effect of having the pin from which the disk hangs, not directly over the center of mass. First some definitions. The line of action of a force is a line containing the force vector and extending as far a necessary in each direction. If two forces have an equal effect on an object they are called "equivalent". For two forces to be equivalent they must be equal in magnitude and direction, and have the same line of action. This means that the two forces may be applied at different points on an object but both points must lie on the line of action of the forces. Forces that share a common line of action may be added to get a net effect For a particle, if two equal and opposite forces were applied they result in no acceleration. For two equal and opposite forces to have no effect on an object of some size they must have the same line of action. An example of forces that may be added to get a net effect would be the applied force at the pin and the weight of the disk in the Hanging Disk model. Even if forces are equal and opposite, if they do not share the same line of action they will result in a torque, giving the object an angular acceleration unless it is constrained. The Pendulum Disk model illustrates this situation. Notice that we display the distance between the line of action of the applied force and the center of mass where the weight acts. This distance is called the moment arm. Two forces in opposite directions, not sharing the same line of action are called a force "couple". If the disk were free to pivot about the pin it would initially swing counterclockwise in this example with a pendulum like motion.
 Now let's alter the apparatus a bit. Imagine our disk, lying on a heavy table and fastened to the table by a pin through the center of the disk. Suppose we have the means to attach a cord to any point we wish on the disk and apply a force by pulling on the cord. The Tugged Disk model allows you to determine both the radial force and the torque that would have to be provided by the central pin to maintain static equilibrium. The algebraic sign of the torque indicates the direction of the torque vector. Remember that torque is a vector quantity with the vector perpendicular to the two vectors from which it is calculated. In this instance the torque is the cross product of the attachment offset vector and the applied force, t=AOXAF. This makes torque point in the z direction, which is awkward for us to display in a 2D model. It turns out that it is more awkward to manipulate the vector magnitude and direction on a 3D model so we are working in 2D where a negative torque indicates that the torque vector points into the screen and a positive torque points out of the screen.
 We have made the point that an object cannot be subject to a single force and that an object subject to two forces can only be in equilibrium if the forces are equal and opposite and have a common line of action. For an object subject to three forces to be in equilibrium, the sum of the three forces must be zero and their lines of action must intersect in a single point or not at all. Forces that meet this condition for equilibrium are said to be "concurrent". Let's leave our disk lying flat on our frictionless table and remove the restraining central pin used in the previous model. Now we will attach two applied forces to the disk and calculate the magnitude, direction and attachment point for reaction force that would maintain static equilibrium. In the Three Force Equilibrium model we start with a nice symmetrical pair of applied forces and the reaction force to balance them into static equilibrium. You will notice on the three-force equilibrium model that the lines of action of the three vectors are shown to help you locate the intersection. Once set, the attachment points for the applied forces remain fixed. The magnitude and direction of the applied forces may be adjusted. As you change the magnitude and direction of the applied force vectors, you will see the magnitude, direction and attachment point of the reaction force vector change as required for static equilibrium change.