 A buoy is being constructed out of wood and will be used to signal passing boats to boundaries for safety or logical purposes. The primary body of the buoy is made out of a salt water resistant wood, let's call it maple, with a specific gravity of 0.6, and is a 2 inch diameter cylinder that is 12 feet in length. The buoy will be used in seawater which has a specific gravity of 1.025 and needs to float such that the top of the buoy protrudes 18 inches from the water's surface. To accomplish this, an amount of steel will be added to the bottom of the buoy body. How many pounds of steel with a specific gravity of 7.85 should be added? Because this is floating, I can start by recognizing that it must be an equilibrium, so the sum of forces in the y direction must equal 0. I'm going to define my y direction as up in this drawing, therefore the force up must equal the force down. I have one up force, it is the buoyant force, buoyant, and that is counteracted by two weights, the weight of the wood and the weight of the steel. So the buoyant force, which I will abbreviate F-B-O-U-Y which looks suspiciously similar to the word buoy, I wonder if there's a relationship there, and that's equal to the force of wood, that's the weight of the wood pulling down plus the force of steel, which is the weight of the steel. The buoyant force is going to be equal to the weight of the water displaced by the buoy as per Archimedes principle, and I would describe that as the volume displaced, that is the volume of water displaced by the buoy multiplied by the density of seawater, multiplied by gravity, and the weight of the wood I can describe as the volume of the wood in the buoy, multiplied by the density of maple, multiplied by gravity, and the weight of steel I could rewrite in terms of the volume of the steel, multiplied by the density of steel, multiplied by gravity, but I don't need to because what I'm looking for is the actual weight of the steel. So all I have to do is some algebra here, the weight of the steel, which I'm calling F-steel, is equal to the buoyant, nope, try that again, the buoyant force, come on iPad, F-B-O-U-Y minus the weight of the maple, and that's volume displaced, multiplied by the density of water, which is seawater here, multiplied by gravity, minus the force of the wood, which I'm going to write as the volume of the buoy, times the density of the maple, times gravity. The next substitutions that I'll be making are for density, I don't know the density of seawater in this problem I was given a specific gravity, so I'm going to say the specific gravity of the seawater is equal to the density of seawater divided by the density of water at standard temperature and pressure, therefore the density of seawater can be written as the specific gravity of seawater multiplied by the density of water, and following that same logic I can write the density of maple is equal to the specific gravity of maple multiplied by the density of water at standard temperature and pressure, and that's due to the fact that our specific gravity is defined as density of the thing divided by density of water at standard temperature and pressure. So neat thing about this, you know how pretty much every problem so far we've been assuming that the water is at standard temperature and pressure? We don't even have to assume that here because that's how our specific gravities are defined, so we can use that same density without that assumption. Isn't that neat? That's pretty neat. We are still assuming standard gravitational acceleration, we could use 9.81 meters per second squared, but since everything else is an imperial, I probably want to use imperial units, it even says pounds of steel, so I'm going to say 32.2 feet per second squared for both gravity terms. Now for the volumes, let's think about the volume displaced first. So I have a wooden buoy that is a cylinder 12 feet long and it is sticking out of the water such that 18 inches is above the water. Therefore the volume of the wood underneath the water is going to be the cross-sectional area of the circle, which is a circle with a diameter of two inches multiplied by the height under the water, which is 12 feet minus 18 inches, plus I have to account for the steel. So I'm saying 12 feet minus 18 inches, that's the height underneath the water multiplied by five or four times two inches squared. So that's the volume of the wood that is underwater and then I'm adding to that the steel at the bottom. So there's a couple of ways that we could approach this. We could assume that the steel is fabricated as a spike and that it is just rammed into the bottom of the wood such that it pushes the wood fibers out of the way. Therefore the wood and the steel take up the same volume as the wood itself, but we could also approach it as modeling a block of steel of an arbitrary shape underneath the buoy. Now we could dig into the details like how our buoy is constructed and how long would a steel spike rammed into the bottom end grain of the buoy actually last in water, but modeling it as an attachment to the bottom of the buoy is easy enough because we know the density of steel which we can describe as the density of water at standard temperature and pressure multiplied by the specific gravity of steel and then we know f steel, the weight of the steel is the mass of steel multiplied by gravity which would be the volume of the steel. Steel multiplied by the density of steel multiplied by gravity. Therefore the volume of the steel would be the weight of the steel divided by the density of steel times gravity or f steel divided by specific gravity of steel multiplied by the density of water at standard temperature and pressure multiplied by gravity and then if I make that substitution up here, so I'm saying plus the volume of steel and there's my volume displaced by the water. I'm still going to end up with one equation and one unknown so it's okay. The volume of the buoy is just going to be the total length of the buoy 12 feet multiplied by the cross-sectional area of that circle. I guess this should really be called the volume of the wood in the buoy that would be better but what's done is done that would be pi over four times diameter of the buoy squared times the total height which is going to be pi over four times two inches quantity squared times 12 feet. So we have enough now to start combining things into one equation. So I will write f steel is equal to the volume displaced which is 12 feet minus 18 inches times pi over four times two inches quantity squared plus f steel divided by specific gravity of the steel times density of water. I'm going to just drop all the at stp and assume all density of water henceforth is at standard temperature and pressure times gravity that's my volume displaced and then we are multiplying by the specific gravity of seawater times gravity and then we are subtracting pi over four times two inches squared times 12 feet times the specific gravity of maple and I'm going to run out of room so I will drop this down times the density of water times gravity and then next I will distribute the specific gravity of seawater times gravity into that parenthetical statement which point I have f steel is equal to this quantity here multiplied by this quantity here this quantity here I subtract this quantity here and then I can bring f steel over so I have f steel and I'm going to factor that out in one step because I'm starting to run out of room and as you all know it's impossible to add a room on a digital workspace like this so I have steel times the quantity one because I'm factoring out minus specific gravity of seawater times gravity divided by the specific gravity of steel times hdu density times gravity and then that's equal to this quantity minus this quantity so I can say f steel is equal to this whole statement divided by the quantity one minus specific gravity of seawater divided by the specific gravity of steel times the density of water this is alarming why is it alarming because there's so many terms john that's a lot of algebra you're right that is alarming you know what else is alarming I need dimensional homogeneity so in order to take one a unit list number and subtract out something that something must be unitless I have specific gravity divided by a specific gravity which is unitless but then I have that pesky row in the bottom that density is not unitless which means this entire term breaks that is a strong indication that I miss something so I'm going to go back and inspect this term here so I was taking the volume displaced and I was multiplying it by the specific gravity of seawater times gravity and that should have also included the density of water which means that I need to go back to this step and I need to add in the density of water here that is also going to mean when I distributed that term inside of the parentheses I need to write it here as well density of water and density of water good thing I got rid of that stp which means that I should have density of water here hey look it cancels that's nice and density of water here so once again we found ourselves encountering one of those classic totally intentional john mistakes that helps us recognize when we may have made errors remember getting good is making slightly fewer errors and knowing how to recognize them and fix them so this density here disappears which means that I have dimensional homogeneity which as you know is all the rage and also there should be a density of water here which makes sense because in these sorts of problems often you can factor out the density of water and gravity cool you want to do that now let's do that now let's factor out the density of water and gravity and since I'm out of room I will just add a new page thought I could make it but I couldn't and on this new page I am going to have a whole bunch of arithmetic but we're not quite there yet let's say density of water times gravity factored out and everything divided by one minus specific gravity of sea water divided by the specific gravity of steel yeah okay and then we are multiplying by our volume of our displaced wood that's 12 feet minus 18 inches times pi over four times two inches squared times the specific gravity of sea water give myself more room yes yes iPad shrink that parentheses that's what I wanted and then we are subtracting this stuff cool hey look we made our lives a little bit easier by factoring stuff out is there anything else that I can factor out you're right I can do that I can factor out the pi over four and the two inches squared so let's go density of water at standard temperature and pressure times gravity times pi over four times two inches squared divided by one minus specific gravity proportion of sea water and steel now we have 12 feet minus 18 inches times specific gravity of sea water minus here let's make that smaller okay here we go we're gonna write real real well I have confidence in us 12 feet minus 18 inches everyone knows that when you write small you have to talk softly times specific gravity of sea water and then we are subtracting 12 feet times the specific gravity of maple and if you are sitting there going John couldn't we factor out 12 feet as well you're right we could we could bring this inside and then factor out 12 feet and then we'd be left with specific gravity of sea water divided by 12 feet and then some other terms but but I'm okay with our current level of optimization it might not be the highest level of optimization possible but I think it's good enough for now now let's start computing so I'm gonna start plugging in numbers for real here we have the density of water at standard temperature and pressure we're gonna go look that up in table a1 are you ready here we go table a1 it is at 20 degrees celsius and one atmosphere density water is 1.937 slugs per cubic foot 1.937 slugs per cubic foot and then we are multiplying by 32.2 feet per second squared and then we are multiplying by pi over four and then we are multiplying by two squared inches squared and then we are dividing by this cool unitless proportion specific gravity of sea water I'm writing small so I'm talking small specific gravity of steel and then we are taking that quantity and multiplying by a length so I'm going to write that length in feet so I'm breaking my cardinal rule I'm doing some mental math hopefully you guys can forgive me I guess I don't have to I can write this as 18 over 12 yeah I'll do that instead don't want to break my cardinal rule that's like the most important one the prime directive of calculating quantities 12 minus 18 over 12 times specific gravity of sea water hey I'm plugging in numbers why don't I plug in numbers down here John what are you doing specific gravity of sea water was 1.025 is 1.025 and then specific gravity of steel was 7.85 7.85 and then back over here multiplying by the specific gravity of sea water that's 1.025 and then we are subtracting 12 times the specific gravity of maple which was 0.6 and that entire quantity is in feet okay now we're going to have to continue because we want an answer in pounds so I will start by recognizing that one pound you'll notice I'm starting out my destination and working backwards one pound of force is one slug multiplied by one feet per second squared and so far slugs cancel slugs one foot cancels one foot second squared cancel second squared I have feet in the numerator square inches in the numerator and feet cubed in the denominator which means I need 12 inches in one foot square everything square everything square everything square everything one square is boring inches squared cancels inches squared and then square feet and feet cancel cubic feet aha and that will yield an answer in pounds of force okay calculator come back I know it's been like 45 minutes here we go 1.937 I'll add some parentheses just for a good measure 1.937 times 32.2 times pi times two squared what do you think do we take the time to find the carrot let's find the carrot two carrot that's pi come on calculator carrot two aha we did it multiplied by 12 minus 18 over 12 you guys didn't break the card and rule times 1.025 minus 12 times 0.6 close parentheses and that is divided by four times the quantity one minus the quantity 1.025 divided by 7.85 was it 7.85 let me just double check good 7.85 and then closing parentheses times 12 on the quest for the carrot got it squared now how many closing parentheses do we need apparently that number is sufficient the double check that doesn't look like it's missing anything important I think those parentheticals make sense okay here we go answer is 5.575 pounds of force that is the weight of the seal again we got there by figuring out how much volume there would be if we attached a steel mass to the bottom of the wood now out of curiosity how much would our answer have been affected if we had assumed that the volume of the steel was insignificantly small that is so small relative to the volume of the buoy that we could ignore it all together well I'm glad you asked if we had assumed that our volume of our steel was insignificant relative to the volume of the wood then we would have had a much easier calculation because this would have been just the volume of the wood displaced and that was pi over four times two inches squared times the height which was 12 feet minus 18 inches which I'm going to write as 12 minus 18 over 12 feet times the specific gravity of seawater which was 1.025 1.025 times the density of water which I'm going to choose to factor out times gravity which I'm going to assume or which I'm going to choose to factor out minus the volume of the buoy as a whole which was 12 feet times pi over four times two inches quantity squared times the specific gravity of the maple which is 0.6 and at that point we have factored density of the water which was 998 kilograms per cubic meter nope nope john metric is not what we want right now back to the table it was 1.937 1.937 yeah 1.937 I can remember that 1.937 it's like a different number but not slugs or cubic foot times the page turn apparently times 32.2 feet per second squared uh-huh okay and I'm going to be as efficient I eat lazy as possible I'm going to factor out two inches squared and pi over four again so now I have 12 minus 18 over 12 feet times 1.025 minus 12 feet times 0.6 and then I'm multiplying by the quantity 1.937 slugs per cubic foot times 32.2 feet per second squared times pi over four times two squared inches squared and then again in order to get pounds to recognize that a pound of force is one slug times one feet per second squared and then 12 inches in the denominator is one foot in the numerator square everything so I have feet inside of this quantity here and that feet is going to cancel this feet and then second squared cancel second squared slugs cancel slugs square inches cancel square inches feet and square feet cancel cubic feet leaving me with pounds of force therefore I can take 12 minus 18 over 12 times 1.025 minus 12 times 0.6 and I think that's the appropriate number of parentheses let's hope 1.937 times 32.2 times pi and just for funsies here I'm gonna write times two times two and then we're dividing that hoping that we had a leading parentheses and we had four in the denominator and 12 squared use a carrot again just you know keeping you guys on your toes we had a syntax error which means we forgot our leading parentheses he said hoping he had forgotten the leading parentheses now I don't want to retive this all so how do I switch my calculator into insert mode is it no it's not that is that no this week on fluid mechanics John tries to figure out how to run his calculator okay no so let's just hope for the best and get rid of one of these I think no no that didn't work okay I gotta start over this time with a leading parentheses not all because my calculator is in overwrite mode as opposed to insert mode and now it's an alpha mode calculator okay you know what I'm just gonna reset you emulator that's what you get by the way I'm hoping that I'm going to remember to cut all this out but if you're seeing it that's a sign that I did not remember to cut all this out so here we go calculating that quantity for the first time totally the first time I'm going to add three parentheses this time at 12 minus 18 over 12 times 1.025 and then we are multiplying that by nothing we are subtracting 12 times 0.6 and then we are multiplying again by 1.937 times 32.2 times the pi symbol times two carrot two quantity divided by four times 12 squared and we get 4.84 pounds of force 4.8476 pounds of force so if we had neglected the volume occupied by the steel we would have been off by what like 12 percent is that significant I don't know depends on the circumstances minus 5.575 divided by 5.575 yeah we had about a 13 percent error that's fun and then just to not confuse people who only look at the solution and don't watch this video I'm going to get rid of this there we go look that was a fun aside we did and then again we can have a conversation like how about if this were floating in regular water instead of seawater would you need to add more steel or less steel in order for it to still protrude 18 inches if you said less steel you're right regular water is not going to be as dense which means that the same buoy would sit lower if it had the same amount of steel so in order to get back up to 18 inches we need to get rid of some of the steel