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.

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.

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.

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.

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.

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.

Regardless on the number of forces acting on an object, if the object is in translational equilibrium and the net torque is zero relative to one point, then the net torque must be zero relative to any point within the boundaries of the object or outside. The reason this works is as follows.

The first condition requires that the vector sum of the applied forces must be zero so we have:
Equation 1) S_inF_i = 0
Let the point O be that about which the net torque is zero and let r₁, r₂, r₃...r_n point to the locations where F₁, F₂, F₃...F_n are applied to the object, relative to the point O. Now if we pick any other point C and let the vector r_c point to the location of C relative to O, then r₁-r_c is the vector pointing to the application point of F₁ from the point C, and so on for the other applied forces. Since the torque about O is zero, we have:
Equation 2) S_inr_i XF_i = 0
The torque about the point C is:
Equation 3) S_in (r_i-r_c) XF_i = 0
Carrying out the multiplication in Eq3 we get:
Equation 4) S_inr_i XF_i - r_cX S_inF_i = 0
But we know from Eq 1 that:
S_inF_i = 0
so the last term in Equation 4 vanishes, leaving the torque about C equal to the torque about O, zero in both cases.

So if an object is in translational equilibrium and the net torque about any point is zero, it is also zero about any other point. This is a useful fact that allows us to choose the point about which to calculate torques in equilibrium situations so as to minimize our aggravation with the math.

You may experiment with different choices for the torque axis on the Torque Axis Offset model. The model is similar to the three-force equilibrium model but you may select a new location for the torque axis by holding down the Alt key while you click the mouse. Notice that the torque values may change when the torque axis is moved, but the T1, T2 and TR still add up to zero.

In dealing with rigid objects near the surface of the planet, one of the forces we must handle is that of gravity. Gravity acts on an object as though all the object's weight was located at a point called the "center of gravity". As long as an object is small enough that there is no significant variation in the gravitational field over the extent of the object, the center of gravity coincides with the center of mass. For large structures like tall buildings, we may have to be concerned with the difference between the center of gravity and the center of mass. For the kind of problems in introductory physics and engineering courses it is not usually significant. Mostly your problems will deal with small, symmetrical, uniform objects for which the center of gravity can be found by simple geometry.

There is a recipe for solving static equilibrium problems. Within the simplifications we have invoked it goes like this. Make a sketch of the object involved. From the sketch draw a free body diagram showing the approximate direction and magnitude of all the forces acting on the object with the force vectors attached at the point in the object where the forces are applied. For this diagram you can usually represent the object by a simple beam, or stick, to which you attach the vectors.

Resolve all the force vectors into horizontal and vertical components and set the sum of all the horizontal(x) force components to zero and the sum of all the vertical(y) force components to zero. This leaves you with two equations, one in x and one in y. You may get a third equation by setting the sum of all torques, about a convenient torque axis, to zero. If in the problem you have three or fewer unknowns, you may solve for them by the simultaneous solution of the set of three equations. If you have more than three unknowns the best you can hope for is to get some relationships among them. In some problems this is sufficient.

The first example is a beam balance. This model represents a rigid beam, resting on a fulcrum at its center. In the initial setup there are some weights placed on the beam. Load #1 is a weight of 3 units fixed at the left end of the beam. Load #2 is a movable weight of 5 units initially located at the 25% position from the red end toward the green end. We will refer to the ends of the beam by their color spots since we may find the beam in any orientation. The location of this movable weight is one of the system variables to be found when we run the model. The reaction force vector is located at the fulcrum and represents the force of the fulcrum on the beam. Its direction is fixed upward but its magnitude is the other system variable. The Weight vector is attached at the beam center of gravity and points downward. Its magnitude of course is fixed at the weight of the beam.

Clicking the Action button calculates the location of the movable weight and magnitude of the reaction force that will keep this system in static equilibrium. The results are reported on the free-body diagram. This is a problem you could easily solve by inspection. There is no left or right force. The total down force is known and at equilibrium the total up force must be the same. The reaction vector then must be i*0 + j*10. Obviously we must place the movable weight on the other side of the fulcrum such that the product of the weight times the distance to the fulcrum is the same as on the left. That places the movable weight 80% of the beam's length from the red marked end.

The model is capable of handling more interesting examples. To save you time, instead of building the subsequent models from the beam balance we have pre-prepared separate models. The next example uses a ladder leaned against a wall.

As loaded the Ladder model shows a ladder of weight 2 whatevers leaned against a smooth wall at an angle of 70 degrees. A rough floor supports the ladder. The reaction force from the floor balances the weight of the ladder. The friction between the ladder and the floor keeps the ladder from sliding down the wall. The magnitudes of the reaction and friction forces are unknown in this model. Initial values are assigned to define the vector directions and so the vectors show up on the free-body diagram. The actual values of the variables will be calculated when you click the Action button. The instructions on the model page will walk you through some variations.

To find the equilibrium condition, the model first shifts any movable forces to try to balance the system. If the maximum available movement of any movable vectors leaves the system out of equilibrium, then we go to work adjusting the remaining variable forces.

The model will make every effort to find a set of vectors consistent with those you defined, that result in static equilibrium. It is up to you to determine if the resulting forces make physical sense. In the ladder example for instance, you could start with the ladder almost flat and come up with a friction force many times the reaction force causing it. Unless you imagine a coefficient of static friction much greater than 1.0, this would not work. The ladder would fall. If you turn the ladder away from the wall (set a beam angle greater then 90 degrees), the solution will report the magnitude and direction of forces required to hold it in place. If those forces are not available, it will fall. Also you should be aware that the solution might not be unique. The model will report the first solution it comes up with, based on its internal logic. With movable and variable forces, there may be alternative solutions that work.

Open the Drawbridge model and follow the instructions on the model page to answer the questions. This will complete this little discussion of static equilibrium.

One of the assumptions made in our work so far with solid objects was that the objects were rigid, meaning that they retain their shape and dimensions when forces are applied to them. This is only approximately true for any real object. All solids have some degree of elasticity. The elastic modulus of a material is defined as a ratio of the "stress" applied to the material to the resulting "strain". Stress is the force per unit area applied to a material and strain is the fractional change in size or shape.

The origins of all elasticity in solid objects lie in the arrangement of the atoms from which the object is made. Let's limit ourselves in this introduction to objects made of metal atoms or metal compounds. Normally such objects have regions, large on an atomic scale in which the atoms are arranged neatly in rows layers and blocks. This tidy arrangement is called crystalline structure At the left is an actual picture from a scanning transmission electron microscope at the University of Michigan, showing the atomic arrangement in a crystal.

There is nothing mysterious about this well-ordered structure. You can get a similar arrangement by pouring ball bearings into a box. Gravity and the walls of the box try to minimize the distance between the balls and their reluctance to share their space with each other tends to hold them apart. In the case of the atoms in a crystal there are similar forces of attraction and repulsion. The forces in this case are electric in nature and there are subtle but important differences between the force of attraction and that of repulsion.

The repulsive force that holds atoms apart applies between atoms of every sort. Once within a certain small distance, the force of repulsion grows dramatically as the separation decreases. The attractive force is strongest between atoms that live together in the same crystal. It is less strong between atoms that might live in the same crystal but do not. It is weakest to non-existent between atoms that not only do not live in the same crystal but also could not due to physical differences. Such attractive force as does exist first increases with increasing separation, then drops off quite steeply with distance so that it is essentially zero beyond a few nanometers.

Now imagine pushing down on the top layer of atoms and with equal force pushing upwards on the bottom layer. The stress arising from these forces is called compressive stress. The force holding the atoms apart gets stronger as the distance between the layers decreases so the compressive stress results in a new equilibrium separation with the layers slightly closer together than before the application of the force. If the change in layer separation from our compressive stress was 1e-11 meters, for example, then the two layers nearest the middle would each move in by 1e-11 meters. The two next layers then would each move in by 2e-11 meters and so on so the total compression would be the number of shifted layers times 1e-11 meters. If we had a 1-meter long rod with this structure, imagine how many layers of atoms there would be. Even very tiny changes in the separation of the layers would accumulate over the length of the rod to be noticeable on the macroscopic scale. In fact the fractional change in the length of the rod would be equal to the fractional change in the separation of the atomic layers.

As we compress the layers of atoms, there is a means for stress to be relieved somewhat by the possibility for the atoms to spread out slightly within the layers. If we were to press on all six faces of the crystal simultaneously, as for example by placing the crystal in a high-pressure fluid, this relief would not be available and the strain in any one direction resulting from the stress would be less.

In that case, the applied stress acts to increase the separation between the layers of atoms in our sample. Since we are looking at part of a single crystal here, there is a substantial attractive force available. But as the tension increases, it is possible to exceed even the maximum attraction and from then on, the attractive force decreases with distance. If that tensile stress is reached, the sample comes apart at its weakest point. Of course in a pure crystal with no irregularities in the structure, no point is weaker than any other so the tensile stress for failure, called the yield stress, of the sample will be quite high. For tensile stress less than the yield stress value, the layers of atoms find a new equilibrium separation appropriate to the stress applied and once the stress is removed the layers of atoms will return to their preferred separation.

Again in this case, the atoms will arrange themselves so as to minimize the interatomic stress and within a layer they will shift slightly inward. This is visible on the macroscopic scale as a slight decrease in diameter of our test rod with this crystalline structure, as we stretch it.

The microscopic view of elastic behavior is useful for understanding how materials respond to various stresses. It does not, however, allow us to make quantitative predictions. To help in that area, several quantities called collectively elastic moduli have been defined.

The elastic modulus for length changes is called Young's modulus. It is defined as tensile stress over tensile strain.
Y = (Force/Area)/(DL/L)
The tensile stress is the total stretching force divided by the cross section area of the rod. The tensile strain is the increase in length divided by the original length.

The elastic modulus for changes in volume is called the bulk modulus. The bulk modulus is defined as the volume stress over the volume strain.
B = - (Force/Area)/(DV/V)
The minus sign indicates a decrease in volume with an increase in force. The volume stress is the normal force, per unit area of the surface of the object. The volume strain is the change in volume divided by the original volume. The bulk modulus applies to liquids as well as solids.

The elastic modulus for changes in shape is called the shear modulus. The shear modulus is defined as the shear stress over the shear strain.
S = (Force/Area)/(Dx/h)
The shear stress, Force/Area is the tangential force per unit area of the face where the force is applied. The shear strain is the distance moved by the face receiving the force, divided by the height of the block. For metals that are solid at room temperature each of these elastic moduli are in the range of 10¹⁰ to 10¹¹ Newtons/meter².

There are only a limited number of ways to excite the elastic properties of an object. We may squeeze or stretch it along one or more directions, we may shear it as with a pair of shears, we may bend it or we may twist it. That's it. Each to these operations may be described in terms of compressive, tensile or shear stress and strain. Squeezing is a straightforward application of compressive stress. Stretching is a straightforward application of tensile stress. Shearing is a straightforward application of shear stress. Bending applies compressive stress on the inside of the bend radius and tensile stress on the outside. Twisting applies shear stress throughout the region between torque applications. Evidently the forces involved in landing this aircraft induced stresses that in some places exceeded the elastic limit of the material since it did not regain its shape after the stresses were removed.

Are there any questions?