Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Work and fluid forces. Constant force formula for work W equal FD, where W is the work done by the force in the body. F the force of constant magnitude in the direction of motion. D is distance along a straight line. Variable force formula for work W equal the integral of A to B of F of X DX. The work done by a variable force is F of X in the direction of motion along the X axis from X equal A to X equal B. Hook's law F equal KX. The force it takes to stretch or compress the spring X length units from its natural or unstressed length is proportional to XK is called the force constant, the pressure depth equation, and a fluid that is standing still. The pressure P at depth H is the fluid's weight density W times the height, so pressure equals weight density times height. Fluid force on a constant depth surface F equal PA, which equals wha, where that W was the weight density, the integral for fluid force. Against a vertical flat plate, the force equals the integral A to B of West, which is the weight density. Again, the strip depth times LY times DY. The strip depth times the strip's length times its thickness. Let's look at a few examples. Stretching a spring. A spring has a natural length of 10 inches. An 800 LB force stretches the spring to 14 inches. Find the force constant, so F equal KX. The force was 800. From the unstressed to the stressed was a distance of four because 14 -, 10, so 800 equal K * 4 or K is 200. Now we want to know how much work is done and stretching the spring from 10 inches to 12 inches. Well, 10 inches is it's unstressed or zero distance. We're stretching it 2 inches past its natural distance or length. So zero to two. We're doing work from 0 to 2, from 10 inches, the unstressed to the stressed of 12. The F of X we just found a minute ago was 200 * X DX. So when we evaluate that integral, we get 100 X squared from zero to 2 or 400 inch pounds. Now we usually give that unit in foot pounds. So to change inch pounds to foot pounds, we realize that there's one foot with 12 inches. So if we put inches on bottom, we have the ratio 1 foot over 12 inches to cancel with the 400 inch pounds, which would leave us 100 / 3 foot pounds. How far beyond its natural length will a 1600 LB force stretch the spring? So 1600 is going to go in for our force. 200 we had found in part A was our constant. So 1600 equal 200 * X. Our X would be 8 inches. Looking at another example, leaky sandbag, a bag of sand originally weighing 144 lbs was lifted at a constant rate. As it rose, sand also leaked out at a constant rate. The sand was half gone by the time the bag had been lifted to 18 feet. So if the sand was half gone, instead of having 144 lbs, it had lost 72 lbs / 18 feet, or it had been losing 4 lbs per foot. How much work was done lifting the sand thus far? So force is going to equal the 144, the original weight minus how many feet off the ground we were and how much we were losing per foot. So 144 - 4 X we're going from zero feet to 18 feet. So the integral zero to 18 of the quantity 144 -, 4 XDX. When we evaluate that integral, we get 1944 foot pounds. The next example, we're going to empty a cistern. The rectangular cistern, a storage tank for rainwater shown, has its top 10 feet below ground level. The cistern, currently full, is to be emptied for inspection by pumping its contents to ground level. How much work will it take to empty the cistern? So the work is going to equal. The weight of our water is 62.4, so we have 62.4 times. If we think about what the area is of each level as we're going our base times our width would be 12 * 20 or 20 * 12. Our height each time is going to be YDY. Now pay attention that we're saying ground level is Ground Zero up here. So if this is 0, we're coming this direction to get our Y each time. So 62.4 * 20 * 12 YDY. We're going from 10 to 20 because we're trying to figure out how far it'll take to empty the cistern. So 10 feet to 20 feet. When we evaluate this, we'd get 747,488 Y squared, 10 to 20 or 2,246,400 foot pounds. That's 62.4 is a standard unit that we use for the weight of water. So Part B says how long will it take a half horsepower pump rated at 275 feet pound per second to pump the tank dry. So we're going to take what we found with the foot pounds and we're going to divide it by that 275 foot pound per second. We're also going to convert seconds to minutes, so one minute is 60 seconds, so about 136 minutes or two hours and 16 minutes. Part C says how long will it take the pump and Part B to empty the tank halfway? It will be less than half the time required to empty the tank completely. So Part C, we're going to do halfway. So we're going to go from 10 to 15 because we're going to still have water from the 15 to 20, so 10 to 1562.4 base times height times depth times the DY. When we get that, we get 936,000 foot pounds. So we're going to divide that by the 275 and multiply by the 1 / 60. So it's going to take us about 57 minutes to do it halfway to complete at half the weight of water. What are the answers to part A through C and the location where water weighs 62.26 lbs and then 62.59. So the only thing that changes is in Part D instead of having 62.4, we're going to change it to 62.26 and then you're going to do the same thing for 62.59. So the information is really just the same. Thank you and have a wonderful day.