Continuing our discussion of statics, now, I want to talk on the idea of equilibrium. And, these are the subjects or the topics we'll look at. And, in particular in this segment, we'll look at sigma, systems of forces and resultants, the conditions for equilibrium, drawing free body diagrams, analyzing pulleys, and analyzing problems involving friction. So firstly resultants and equilibrium, and here is the section over on the right hand side from the reference handbook. The resultant is similarly, simply the sum of the forces which is acting, or which are acting on an object. So, in this case, let's suppose we have two forces here, F1 and F2, then the resultant force is simply the vector sum of those forces. Or, more generally, the summation overall of the forces which are acting, which is the equation given in the reference handbook here. And again, we can form this resultant by simply adding up the components of the forces in the x and y directions. So for example the x component of the resultant force is equal to the summation of the x components of the individual forces, and the y component is equal to the summation of the y components of the forces. Now, generally speaking we can replace a system of forces by a single force through some point, and a moment. And let me illustrate that. Let's suppose we have a body here which is acted on by three forces, F1, F2, and F3, and we'll assume that these forces do not all pass through a single point. So, if I want to replace that by a single force, then the single resultant is just the summation of those forces, F1 plus F2 plus F3. However, if I move all those forces through a single point like that, then we can see that this is not going to be an equilibrium. Because, if I look at this point O here, each of these forces have a moment about that force. So, to have those two systems be truly equivalent, I need to add a resultant moment. So, therefore, my equivalent system consists of the resulting force here, r, which is the summation of all of those forces passing through some point O here, plus a resultant moment M0, which is the sum of the moments of all of those individual forces about the point. So, in other words, the resultant force is just the summation of the forces which are acting, in this case, F1 plus F2 plus F3, and the resultant moment is the sum of the moments of the individual forces about that point. In other words, this expression here. And these are the formulas which are given in the reference handbook above. Now we can go further than that and replace this force system by a single force acting at some point. So let's suppose that my single force here is R. And if I want this to be a single force here, then I have to, for this system to be in equilibrium, the moment of my resultant force about the point O, in other words, R times d, must be equal to the summation of the individual moments. In other words, the distance here of my resu, single resultant force is equal to the moment, the result in moment divided by the resultant force. So in this way I can replace a force, a system of forces by a single force with no moment. However, I have to calculate where that single force the line of action of that single force. The conditions for equilibrium are, generally speaking, the resultant force or the summation of the forces acting is 0, and the summation of the individual moments about some point is also equal to 0. Those are our basic conditions for a body to be in equilibrium. There are some special cases of that, as shown in this diagram here. The first one, collinear, means that the forces all have the same line of action. So they all pass through one line as shown here. And, in this case, we have only one equation of equilibrium that the sum of the forces along the line of action of those forces, which I'll call the x axis, is equal to 0. The second special case is the case where we have a system of forces which are concurrent. In other words, they all pass through a single point. In that case, in two dimensions at least, we have only two equilibrium conditions, that the sum of the forces in the x direction and the sum of the forces in the y direction is equal to 0. We don't have a moment equation, because all of the forces pass through this point O, therefore they don't have any moments about that point. The third special case is a parallel force system, where all of the forces are parallel to each other. And here again, we have only two equations of equilibrium. We have that the equations of, of the sum of the forces in the direction parallel to the forces, which I'll call the x direction here, is equal to 0. But also the sum of the moments of all of these forces about that point must also be 0. So we have two equations. The most general case, where we have a system of forces and moments, we have three equations, again in two dimensions. We have the sum of the forces in the x direction equals 0, the sum of the forces in the y direction equals 0, and also the sum of the moments about some point is also equal to 0. So, generally speaking, in two dimensions we have three independent ineq, equilibrium equations, which could, for example, be sum Fx is equal to 0,sum Fy is equal to 0, and sum of the moments about some point is equal to 0. However, they don't need to be those three. Equally, we could have sum Fx is equal to 0, sum of moments around some point A is equal to 0, or sum of some moments about B, some point B is equal to 0. So in either case we have only three equilibrium equations, not more. If we have three unknowns, then we have three equations, therefore we can solve for those three unknowns. And we call that kind of a system statically determinate, which is the kind of system we'll be dealing with in this segment on statics. If there are more than three unknowns then we can't solve it from statics alone, and we call that kind of a system statically indeterminate. And to solve that kind of system we need some other information, for example, relationship between the deformations and the forces. And that topic we'll look at later on in mechanics and materials. In three dimensions, generally speaking, we have six independent equations, three force equations and three moment equations. Now the next very important concept in statics is. The ability to draw free body diagrams. And a free body diagram is essentially ends, isolates the body that you're interested in and represents all external forces acting on it and the essential dimensions and their lines of action, et cetera. So, there are different types of support that we'll encounter in two dimensions. Here are some examples. For example the first one here is a flexible cable. So, a cable can be subjected only to a tension force, not a compression force. So it would look something like that. The second type of contact or surface that we might encounter is a smooth surface. So a smooth surface means that there can't be any friction force. In other words, there cannot be any force which is tangential to the surface here. So therefore the only type of contact force there must be normal to the surface. On the other hand, if we have a rough surface, then it that case we can have a component of force which is tangential to the surface. So for rough surface, generally we will have a normal component of force and a tangential component of force as shown here. We'll also encounter roller supports. And a roller support cannot have any force in this direction, which is parallel to or tangential to the surface. It can only have a normal component. So in other words, a roller support will have only a force in this direction. And if there's no friction, which is what we will assume, it can also have no moment. So a roller support will have only a normal component of force, and that is as similar to the kind of force we will get in a freely sub, sliding guide. So in this case, there can not be any direct force parallel to the direction of the guide. And, again, we only have a normal force. We'll also commonly encounter pin connections, and a pin connection will constrain motion in the horizontal and vertical direction. Therefore, we would generally expect to have a horizontal and a vertical component of force. But again, assuming no friction, there will be no component of the force, no moment about that. If there is friction, then we could expect a moment there. But generally for a pin connection, we will neglect that and assume it's frictionless. The most general type of support is shown here, a built-in or fixed support, like a beam which is welded into a, a wall, or cemented into a wall. And this is the most general case. In this case, we would expect a horizontal component of force, a vertical component of force. And, also, some moment reaction. And, the last one, gravitational attraction is a body force which we represent as the weight of the object acting straight downwards through the center of gravity. And, the weight of the object is equal to the mass of the object multiplied by the acceleration due to gravity. So here are some examples of free body diagrams for simple structures. The first one is a plain truss with the force acting as shown, and this is a pin support at A, and a roller support at B. So the free body diagram, the pin support, we have a horizontal and a vertical component of force. The roller support here, we have only a vertical component of force. So, we note that in this case, this is statically determinate. We have three unknowns, Ax, Ay, and By, and three equations, so we can determine the reactions. The second example a cantilever beam, which is concreted or cemented into the wall here and subject to the three forces as shown, the free body diagram in this case looks like this. Here are our forces, F3, F2, F1 and our reactions here. Most generally, we have a vertical and a horizontal component of force, and also a momentary action. And, in addition, we have the weight of the beam which is acting straight downwards, and by symmetry probably acts in the center of the beam. It's acting straight downwards in the center of the beam and has a magnitude of m times g. Case number 3 is a beam with a pin support at B on a horizontal force. And it's acting, it's resting on a support there or a surface, a corner at a with a smooth contact. And, in addition, we have a moment exerted in a clo, in a clockwise direction here. So here's our free body diagram. The unknown reactions at the pin support are Bx and By. And in addition, we have this applied force, P. We have the moment, M, and the weight, which is acting straight downwards through the center, is equal to m times g. Then we have an, a reaction force at N. And we're told that that is a smooth support there. Therefore, the direction of that force is perpendicular as shown there. And again we note that this is a statically determinate system. We have three unknowns, P, By, and N. Therefore, we can solve it, assuming that the other applied forces and moments are known. And the last example number 4, the rigid sy, system of interconnected bodies. We have a roller support at A and a pin support at B, so the free body diagram for that case is as shown here, again, statically determinate. This is a little exercise to complete these diagrams. So the column on the right are all incomplete free body diagrams. So the first one we have a crank supporting a mass, m, with a pin support at A. Here's the free body diagram. But this clearly is not in equilibrium, because there's no horizontal force here to balance the component of the tension through, transmitted through the cable. So the complete diagram here, at the pin support, we have a horizontal and vertical component of force Ay and Ax, so the proper diagram is as shown. And assuming that this is a frictionless pin here, there is no moment exerted there. The second example number 2, we have a control lever here applying a torque to the shaft here at O. Here's the free body diagram, but again, this is obviously not an equilibrium. Because if I take moments about this point O here, there's no moment to balance the moment of the supplied force, P. So the moment is supplied by, I'll call it MO, is the moment exerted by the, the shaft at O on the, on the on the lever. And conversely the moment applied by the lever on the shaft is equal and opposite to M0. Number 3, we have a boom here with a pin joint here, a cable at A, and a weight of mass, m, suspended by a cable at A. And again, looking at the free body diagram here,. We see that this can't be an equilibrium because firstly, we have no horizontal component of force here. And also of course, we must have a vertical component of force. So, the free body diagram looks like that. Again, no moment here because we'll assume that this is, presumably a frictionless pin. The last one, number four, is a crate of mass m leaning against a smooth wall and supported on a rough horizontal surface. And the incomplete free body diagram is right here. But, this is obviously not at equilibrium because there's no horizontal force acting on the crate to balance the reaction force at the wall. So, the horizontal force must be supplied through friction. Let's call it R here for the friction force. Must be supplied by friction to keep that in equilibrium. So the complete free body diagram looks like that where R is obviously equal to the reaction force at A. Now if this was a smooth floor then this could not be in equilibrium because there's nothing to constrain the the force in the horizontal. The only way that this could be in equilibrium in that case with a smooth floor, is if the crate is rotated so that the weight force is vertically above the point B. And then it could potentially be in equilibrium, even with a smooth floor. Now pulleys are analyzed as if they are or usually assumed to be massless, frictionless, either by the cable or at the pin. And so in that case a simple pulley here, if I take moments about this point. We see that the tension in both sides of the cable must be the same for this to be in equilibrium assuming that there is no friction. And similarly summing the forces in the vertical direction here, in the y direction. We see that the upward force on the pin here must be equal to two times the force or the tension force in each cable. So basically, all that we have here to analyze these is that the summation of the forces on each pulley in the vertical direction is equal to zero. And summing the forces in the horizontal direction doesn't add any information because there aren't any forces in the horizontal. So, all we do is apply particle equilibrium, in other words, sum the forces in the vertical direction systematically through the system. Applying this to one pulley at a time. For example, I have here a weight of 200 pounds which is being suspended by two pulleys. And I'm applying a force here to suspend that force. What force, F, do I need to apply to that pulley to lift that weight of 200 pounds? So, we look at the separate elements of this system one at a time. Firstly, the weight is obviously, trivial. And that says that the upward force here must be equal to W. So now, I look at this pulley here. The free body diagram of that pulley looks like this. The downward force here is equal to W. And for vertical equilibrium of that pulley, the tension in the cable here must be equal to W over 2 and W over 2. Next, I look at this pulley right here. And the free body diagram for that pulley looks like this. The tension is continuous, so the tension in each cable is W over 2. Which means that the upward force on that pulley here must be equal to W. So therefore, finally, we see that this force here is equal to the tension in the cable there, it's W over 2. In other words, we have to apply a force of 100 pounds to lift a force of 100, of 200 pounds. And that's how a pulley gives us a mechanical advantage. We can also note that we look at, if we look at this overall system here, and look at the overall equilibrium of this system. Then the forces which are acting here are W acting downwards here, W over 2, which is acting downwards here. W which is acting upwards here, and W over 2, which is acting upwards there. So if we sum the forces in the vertical direction on the whole system we see that indeed that is equal to zero. Which is a useful check of our final answer. The last basic topic I want to look at in friction, in statics is friction. And this we divide into dry friction or static friction, which is between two stationary surfaces. Or fluid friction, which is between moving surfaces due to, movement of a fluid in between the two. But we will only look at static friction. And the magnitude of the static friction is given by this expression here. The maximum magnitude is equal to mu s times N. Where mu s is the coefficient of static friction and N is the normal force between the two surfaces in contact. So, the static friction can be less than that, but it can't be greater than that. So this completes our discussion of basic concepts of statics. And in the next segment, we'll do some examples to illustrate these concepts.
