Continuing our discussion of statics and equilibrium. In this segment, I want to do a series of examples and exercises to illustrate the concepts that we've just discussed. So, first question, we have a bent pipe which is acted on by these three forces that are shown. And we want to replace this by a single force. So we're going to have two questions related to this. What is the magnitude of the single force and where does it act? So, firstly, the magnitude of the resultant force is most nearly which of these alternatives? So here is our general expression, the resultant force is just a summation of all of the forces which are acting. And in this case, well, as in most cases, it's easiest to just sum up the components in the x and y directions. So, the component of the resultant force in the x direction is just equal to the only horizontal force we have is 50 pounds here. And it's in the negative direct, x direction so it's minus 50 pounds. And y is equal to the sum of the y components of the forces, which is plus 60 pounds in this direction and minus because it's downwards 40 pounds in that direction is equal to 20 pounds. So, the force components we have are Rx is in the negative x direction. It's 50 pounds. And Ry is in the positive y direction, is 20 pounds and the magnitude of the resultant force is the square root of the sum of the squares of the components is equal to 53.9 pounds. So the answer is 54 pounds force is the resultant. The second part is where is the x coordinate where the line of action of this single resultant force intersects the x-axis? In other words this is a single force here in this direction but it's going to intersect the x-axis at some location here. Which I'll call X. Where is that location? So, to find this we just takes moments about this point O here, for example. So the moment of the original forces here is equal to the sum of the moments of the 40 pound, the 60 pound and the 50 pound force. Which is given by this expression here. But for this to be in equilibrium, this must be equal to the moment of the resultant and force times the distance different times the distance to it and the easiest way to see that is if I just take the component of the y force here, the component of the x compo, the moment of the x component doesn't have a moment about o because the line of action passes through o. So the only moment I have is the moment of the y component, which is Ry times this unknown distance, x. So rearranging, I get x is equal to this expression divided by Ry. Ry, we've already solved for here, is 20 pounds. So substituting in, we find that x is equal to 65 inches. In other words, the single resultant force intersects the y-axis at a distance 65 inches to the right of the origin O. So the answer is B. Next problem, we have a 2x4 board. It weighs 12 pounds. And we want to compute the vertical component of the force on the carpenter's shoulder. In other words at this point here. So as usual, the first step is to draw the free body diagram, which looks like this. So the board is eight feet long, so assuming it's symmetrical, the weight W, at halfway across here, four feet from the left and right hand sides. The reaction at A here is, I'll call it Ax. And the reaction at Bx here is, I'll call it Bx in the vertical directions. So the easiest way to solve this is just to take moments about this point here, about the point B. Because then I won't have to worry about the unknown force or reaction at B. So the moments are the moments of Ax is Ax times 2 minus the moment of the weight force. Here's the weight force times it's moment arm which is three feet, so rearranging we get Ax is equal to 3w over 2, which is equal to 18 pounds and the closest answer is C, is the vertical component of the reaction on his shoulder. Next one, we have an I-beam which weighs, it's 450 kilograms and it's supporting a drum which has a mass of 220 kilograms. The reaction at the support B here is most nearly which of these alternatives? So here is our free body diagram. At A, we have a pin support. Therefore, generally speaking, we have a vertical and horizontal component of force there. We have the weight of the beam, which I'll call Wb, which acts straight downwards through the center of the beam, which I'll, the center of gravity, which I'll assume is in the center. In other words four meters from the left hand side here and at B, we have a roller support, therefore we only have a vertical component of force there, Bx. So we note that this is a statically determinant system, we have three unknowns, Ax, Ay and Bx and we can solve it. So let's go ahead and do that. And in this case the easiest way to solve it is to take moments about the pin support A. Then we don't have to worry about Ax and Ay because they don't have moments about A. So, taking the moments about A and summing them equal to 0. And I'll assume that this direction clockwise is positive. So, the moment of my weight force here, the weight of the beam is Wb times its moment arm, which is 4 meters, plus the weight of the drum, is Wd times it's moment arm which is 5.6 minus, because I'm assuming it's in the upwards direction. So it's a counter-clockwise moment. By times it's moment arm, which is eight. So, rearranging for, for BY which is what we're trying to solve, we get this equation. And, the weight of course, is the mass times the acceleration due to gravity. So, the weight of the beam is 450 times 9.81, etc. And the answer is 3720 Newtons. So, the best answer is C. And if you want to continue with that then you can also show that the force, the vertical reaction force of A is equal to 2850 Newtons. And also you can see, obviously, that Ax, the horizontal reaction at A must be zero because there aren't any other horizontal forces acting. Next example, we have a bar, which weighs 100 pounds, and is resting on a cubicle block, as shown. And the contact point, here, B. Is smooth. So, we're going to answer two questions about this. The first one, the reaction force at the point of contact here at B is most nearly which of these alternatives. So, here is our free body diagram of the, of the bar. So at A here, we have a vertical component of force Ay and a horizontal component of force Ax. We have the weight of the bar which I'll call Wb. Which acts straight downwards and I'll assume it symmetrical so it's in the middle. In other words ten feet from the point A. Then we have our reaction force B here which I'll call RB. And we're told this is a smooth contact point here therefore the force must be perpendicular to the bar, in other words it acts as an angle theta to the vertical where theta is the angle of the rod to the horizontal. And we also note that we must have some horizontal component of force here, Ax to resist the horizontal component of the reaction force RB. So this force at Ax would be provided either by friction or, some kind of support at that point. So again, similar to the last problem, the easiest way to solve this is to take moments about the point A. And then we don't have to worry about the, two unknowns, Ax and Ay. So, summing the moments about A. And again, I'll take counterclockwise as being the positive direction. We have the moment of the resultant force here, which is Wb times its moment arm, which is 10 cosine theta degree, cosine theta minus the moment of the reaction force RB times its moment arm which is 13 feet. So rearranging, we get the reaction force is WB times 10 cosine theta over 13, and we notice that this triangle here. This is five feet tall. This is five feet thirteen. So this is a 5, 12, 13 triangle. So cosine theta is 12 over 13 and the answer is 71 pounds. So the closest answer is C. Continuing that we're now told that the block weighs 200 pounds and we want to calculate the vertical reaction of the block on the plane. Is it most nearly which of these alternatives? So to solve this we have to draw a separate free body diagram of the block itself. And the free body diagram of the block looks like this. So the reaction force exerted by the rod on the block is equal and opposite to the reaction exerted by the block on the rod. In other words, it's ARB but in the opposite direction here. And also in direction theta to the vertical. We also have the, the weight of the block WB, which acts straight downwards through the center of the block. Then we have the reaction component exerted by the horizontal plane on the block. Which I'll call Rᵧ in the vertical and Rₓ in the horizontal. Now again, we note that this is going to be statically determinant. We can solve this problem because we have three unknowns here and three equations. What are the three unknowns? Well the three unknowns are Ry and Rx, but in addition, we don't know where Ry acts. It acts at some unknown distance from the left-hand side. So we have three unknowns. However, we can solve for them all, because we have three equations. So, in this case we're only asked for the vertical reaction force, in other words, Ry, and we can do that most simply by just summing the form, the forces in the vertical direction, sum Fy is equal to 0, which gives us Ry, the vertical force here, minus because it's acting downwards RB minus because it's also acting downwards, the vertical component of RB, which is RB cosine theta is equal to 0. So, rearranging for Ry which is what we want to solve we get this equation. And plugging in we get, 200 plus RB, which we just solved found 71 pounds, times 12 over 13, and the answer is, 266 pounds. So, the closest answer is B. And, we're not asked for it, but we note that we could also solve for the horizontal component of force Rx is just equal to the horizontal component of RB. And that must be supplied either by friction or by some stop holding the block in place. And we can also find the location of the vertical force by taking moments for example, moments about the point B. Next example, we have a sign which has a mass of 200 kilograms supported by a cable attached by pins at A and C, and the question is the tension in the cable is most nearly which of these alternatives. So here's our free body diagram. This is a pin support here at B. So we have two components of reaction force Bx and By in the horizontal and the vertical. We have the tension in the cable T at angle theta to the horizontal. And we have the weight of the sign, which hangs straight downwards by symmetry through the center or the centroid of that area. First a little bit of geometry, the angle theta here is the arctangent of 1.5 over 2, which is 37 degrees. And to solve this we'll take moments about the point B because then we don't have to worry about the reactions, Bx and By. So taking moments we have the moment and I'll assume that the clockwise direction is positive here. The moment of this force is W, the weight times its moment arm, which is 2 meters, minus the moment of this force. Now, to get the moment of the tension force, we could compute this perpendicular distance here d and compute the moment as d times T. However, it's simpler to recompose this into its horizontal and vertical components here. And the vertical component of that force is T sine theta. The horizontal component is T cosine theta. But we won't have to worry about that. Because T cosine theta of the horizontal component passes through the point B so it doesn't have any moment. So the only moment we have is the moment of this component which is T sine theta multiplied by it's moment on which is 2 meters. So rearranging we have T is equal to W divided by sine theta. The weight in turn is the mass times acceleration due to gravity 200 times 9.81 over sine 37 degrees is equal to 3,260 Newtons and the closest answer is C. Next question, we have a sphere resting between two parallel walls as shown. It's a smooth were, sphere that weighs 20 kilograms and the question is to calculate the contact force at B. Is it most nearly which of these alternatives? So here is our free body diagram, and we're given that to, both of the walls are smooth, which means that the reaction forces here, RA and RB, at these tow, two contacts point, points are perpendicular to the local tangent there. In other words, they pass through the center of the sphere at this point. And by symmetry, the weight acts straight downwards as shown. So in this case I can solve this. I have, in this case I'm only going to have two equations. Some of the forces in the horizontal and the vertical is equal to zero. There's no moment equation because there is no moment about the center here. All of the forces pass through the center, so there's no moment equation. So firstly, summing the forces in the horizontal direction sum fx is equal to 0. We have the horizontal component of RA which is RA cosine 15 degrees minus because it's acting to the left, the horizontal component of RB which is RB sine 30 degrees. So rearranging I get RA is equal to RB sine 30 degrees over cosine 50 degrees. Next, I sum the forces in the vertical direction. Sum fy is equal to 0. So here I have the vertical component of RA which is RA sine 15 degrees plus the vertical component of RB which is RB cosine 30 degrees minus, because it's acting downwards, the weight force is equal to zero. So, here you see, I have two equations, and two unknowns, RA and RB, so I can solve them. And I do that by just substituting this equation for RA back into here which gives me this equation or expression for RB and now I can evaluate RB is equal to 196 Newtons. And next I can take this equation, substitute it back in here to find RA. And the answer is RA is equal to 101 Newtons. But in this case, we're only asked for the, the component of the reaction force at RB. So the an, the answer is B. [BLANK AUDIO]. Next example, we have a cantilever beam, which is concreted into the wall at the left here. And it's subjected to the forces shown, a 1.4 kilonewton force, a 3 kilonewton force, and in addition, a moment of 15 kilonewtons in the counterclockwise direction applied at the point B. And we're going to answer two questions on this. First of all, the horizontal component of the reaction force at the support is most nearly which of these alternatives? So here is the free body diagram showing the forces and the moments. And now in addition we have the weight of the beam, which is acting straight downwards here. And the reaction forces of the support I'll call Ox in the horizontal, Oy in the vertical, and the moment reaction here I'll call M0. 'Kay, so again, this is a statically determinate system. We have three unknowns, Ox, Oy, and MO, and three equilibrium equations. So firstly we compute the weight of the beam. It's the mass times the acceleration due to gravity, is 500 times 9.81 Newtons or divide that by 1000, 4.91 kilonewtons. So, to get the horizontal component to force, all I need to do sum the forces in the horizontal direction. Sum fx equals 0. So in this case the only forces we have are Ox, the horizontal component of force, and the horizontal component of 3 kilonewton force is minus because it's in the negative x direction, 3 sine 30 degrees, in other words Ox is equal to 3 sine 30 which is 1.50 kilonewtons. So the best answer is A. The next part of this, we want to calculate the vertical component of the support force. So it's nearly which of these alternatives? So here, we just sum the forces in the vertical direction equals to zero. So, here we have Oy, the unknown vertical reaction at O plus 1.4 pushing upwards minus, because it's downwards, the weight of the beam and minus because this is also pointing downwards 3 times cosine 30 degrees. So rearranging we have Oy is equal to that, and substituting in the values we have that, that is equal to 6.1 kilonewtons and the closest answer is B. For the last part of this problem, we're going to compute the reaction moment at the support. So in the moment of the support reaction is most nearly which of these alternatives? So here again, the same free body diagram applies and in this case the simplest thing is to take moments about the reaction point here because then I don't have to worry about the moments of Ox and Oy. So the unknown moment is M0 which I'll assume is in the clockwise direction and I'll take as my positive direction for computer moments as clockwise. So the moments are the unknown moment M0 here, minus because this is in the counterclockwise direction, 1.4 times its moment arm, which is 1.2 meters plus the moment of the weight force here. Wb times its moment on is 2.4 plus the moment of this force here. And here, it's again, convenient to compose this or decompose this into its horizontal and vertical components. Because the horizontal component of force has no moment because it passes through o. So all I have is the moment of the vertical component of force here which is 3 cosine 30 degrees multiplied by its moment arm which is 4.8 meters. And then finally I have the moment of the applied moment here which has given us 15 kilonewton meters which is shown as being in the counterclockwise direction. Therefore, it's negative. So, rearranging for the unknown support moment, I get this equation. And substituting in the numbers, I get the answer is 7.57 kilonewton meters, so the best answer is C. Now, I'll do a problem involving pulleys. And in this case, we have a weight here of 100 pounds which is being suspended by three pulleys. And we have finally a cable at the top here, with some tension T at an angle of 30 degrees to the horizontal. And the question is, the tension T in that cable is most nearly equal to which of those alternatives? So, here's the free-body diagram, showing the, free bodies of each of the separate components, each pulley, one at a time. So basically we started one pulley and work our way through the system until we get to the top pulley, so we can compute the tension. So first of all, looking at this bottom pulley here, this free body diagram here, we see that for this to be an equilibrium, the tension in two sides there must be equal, so T1 is equal to T2. And, summing the forces in the vertical direction, we see that T1 must be equal to T2, must be equal to 1000 pounds, the weight of the block divided by 2. In other words, the tension in this portion of the cable here is equal to 500 pounds. Next we continue on up to the second pulley, pulley B here, which is this free-body diagram right here. And in that case, again, the tension there is continuous. In other words, T3 is equal to T4. And summing the forces in the vertical on that pulley is equal to 0. We have T3 is equal to T4 is equal to T2 divided by 2. But T2 we've already solved for here, is 500 pounds. In other words, that is equal to 250 pounds. Finally, we get to the top pulley here. This one, right here. Here is our free-body diagram. And from this, we see that the tension through here is continuous. T is equal to T3. And T3 we've already solved for. Therefore ,T for tension is equal to 250 pounds, and the best answer is A. From which we also noticed that this angle here, given as 30 degrees, is irrelevant. The last problem I'll do is a block resting on an incline. So the mass of the block is 100 kilograms. The angle of the incline is 20 degrees and it's held in place by an unknown mass here of mass m0. So the question is, if the coefficient of static friction is 0.3, the minimum mass of the weight to prevent the block from sliding is most nearly which of these alternatives? So you see here that if this mass is too low, then the tension is not enough and the block will slide down the plane in this direction. On the other hand if this mass was too high, the force would be too large and the block would slide up the incline. But, in this case, we're asked for the, the minimum mass of the weight which is needed to prevent the black, the block from sliding down. So, here's the free-body diagram for that situation. We have a normal force, which I'll denote by N. And the tension in the cable here, at equilibrium, is equal, to the mass of the block multiplied by the weight, in acceleration due to gravity. In other words it's the weight of the block and the tension here is just T equals m0 times g. And, in this case, the friction force to prevent it sliding down must be in this direction. And the maximum value of the friction force is the coefficient of static friction multiplied by the normal force. So, here is the, the friction force is equal to mu S times N. And to find the normal force for equilibrium, we'll just take the components of the forces perpendicular to here. So I'll just rotate the coordinate system x and y to be parallel and perpendicular to the show, the slope. So, summing the forces in the y direction, in other words, this direction, we have N is equal to the weight of the block times cosine 20 degrees. And the weight of the block is mg cosine 20. Or 100 times 9.81 times cosine 20. In other words, the normal force, at the contact is 922 Newtons. Next, I sum the forces in the x direction. In other words, this way. And here, the forces we have are the friction force, maximum friction force which I'll call f max, minus the component of the weight force down the slope, which is W sine 20 degrees plus the tension force in the cable, T. So rearranging for T, which is why we want to find the tension force is W sine 20 minus F max. Which is W sine 20 degrees minus mu S times N, which is 981 sine 20 minus mu, we're given, is the coefficient of friction, is 0.3. The normal force we've solved above is 924, which gives us a tension in the cable of 58.9 Newtons. However, that's not what we're asked for. What we're asked is the mass of the weight, and the mass is tension divided by the acceleration due to gravity. 58.9 divided by 9.81 is equal to 6.0 kilograms, and the answer is C. This is the minimum mass, which will prevent the block from sliding down the incline. And this completes the examples on equilibrium.
