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# Buoyant force example problems edited

## Video transcript

let's say the window and some I have some object and when it's outside of water its weight is so wait outside of water is I don't know 10 Newtons and let's say well then I submerge it in water I put on a weighing machine in water it's weight so let's call it weight in water is I don't know it's two Newtons so what must be going on here well the water must be exerting some type of upward force to counteract at least eight Newtons of the person's original of not the person the objects originally this would be a very small person most people's weights in Newtons are in the hundreds but anyway so so and that difference is the point force right so a way to think about it is once you put the object in the water so let's say it's a you know it could be a cube but now to be a cube could be anything it's in the water so let me draw the water this is the water up here we know that we have a downward there's a downward weight right that is ten Newtons but we know that once it's in the water the net weight is two Newtons so there must be some force acting upwards on the object of eight Newtons right and that's the buoyant force that we learned about in the previous video and the video about Archimedes principle this is the buoyant force so the full the buoyant force is equal to 10 minus 2 is equal to eight right that's that's how much the water is pushing up and what is that also equal to that equals the volume about the weight of the water displaced so eight Newtons is equal to weight of water displaced and what is the weight of the water displaced that's the volume of the water displaced times the density of water times gravity right and so what is the volume of water displaced well it's just the volume of water divided eight Newtons by the density of water is 1,000 kilograms per meter cubed this is 8 Newton so a Newton is what kilogram kilogram meter per second squared and then what's gravity it's 9 point 8 meter per second squared and if we look at all the units they actually do turn out with just you end up having just meters cubed but let's do the maths equals 8.2 times 10 to the minus 4 meters cubic meters so just knowing the difference in the weight of an object I can figure out the value the difference when I put it in water I can figure out the volume and so this could be a fun game to do next time your friends come over is way yourself outside of water then get some type of spring or waterproof waterproof weighing machine put it at the bottom of your pool stand on it and figure out what your weight is assuming that you're you're you're dense enough to go all the way into the water but you could figure out somehow your weight in water and then you would know your volume there's other ways you could just figure out how much the surface of both though the water increases and take that water away but anyway this was interesting just knowing how much the buoyant force of the water was or how much lighter we are when we go the object goes into the water we can figure out the volume of the object and this might seem like a very small volume but just keep in mind in a meter cubed in a meter cubed you have let's see you have 27 square feet right you have 27 square feet so if we multiply that number times 27 it equals 0.02 0.02 square feet roughly and point oh two square feet how many in a square foot there's actually c12 to the third power times 12 times 12 is equal to 17 28 times point no to so this is actually 34 square inches so the object isn't as as small as you may have thought it to be it's actually you know it's like maybe a little bit bigger than three inches by three inches by three inches so it's a reasonably sized object anyway let's do another problem let's say I have some I have some balsa wood and I know that the density of balsa wood is 130 kilograms per meter cubed meter cubed that's the density of balsa wood and I have some you know big cube of balsa wood and what I want to know is if I put that so let me draw the water that's the water and I have some big cube of balsa wood which I'll do in brown so that that big cube of balsa wood and the water should go on top of it just so you see that it's submerged in the water all right I want to know what percentage of the cube goes below the surface of the water interesting question so how do we do that well for the object to be at rest for this big cube to be addressed there must be zero net forces on this object right so in that situation the buoyant force must completely equal the weight or you know the force of gravity well what's the force of of gravity going to be well the force of gravity is just the weight of the object and that's the volume of let's just say the wood volume of the balsa wood times the density of the balsa wood times the density of balsa wood times gravity and what's the buoyant force the buoyant force is equal to the volume of the displaced water right volume of displaced water but that's also well it's the volume of displaced water and it's the volume of the of the cube that's been submerged the part of the cube that submerged that's volume that's also equal to the amount of volume of water displaced right so you could say that's the volume of the block submerged which is the same thing remember is the volume of the water displaced times the density of water times gravity remember this density of water so remember the buoyant force is just equal to the weight of the water displaced and that's just the volume of the water displaced times the density of water times gravity and of course the volume of the water displaced is the exact same thing as the volume of the block that's actually submerged and since the block is stationary it's not accelerating upwards or downwards we know that these two quantities must equal each other so the volume of the would the entire volume not just the amount the submerge times the density of the would times gravity must equal the volume of the would submerged which is equal to the volume of the water displaced times the density of water times gravity well we have the acceleration of gravity we have that on both sides so we can cross it out let me switch colors to use a mop monotony and then let's see what happens if we if we divide both sides by the volume of the balsa wood you get let's put this over here so you get let's divide both so I'm just rearranging this equation I think you'll figure it out we can get put divide both sides by that you get the volume submerged divided by the volume of the balsa wood right I just divided both sides by VB and divide and switch sides is equal to and Alice is equal to the density of the balsa wood divided by the density of water that makes sense I just I just did a couple of quick algebraic operations but hopefully that you know get rid of the G and that should make sense to you and now we're ready to solve our problem because my original question is what percentage of of the object is submerged and so that's exactly this number right if we say what this is the volume submerged over the total volume this is the percent submerged and so that equals the density of balsa wood which is 130 kilograms per meter cubed divided by the density of water which is 1,000 kg/m^3 so 130 divided by 1,000 is 0.13 so vs over VB is equal to 0.1 3 which is the same thing as 13% so exactly 13% of this object will be of this balsa wood block will be submerged in the water that's that's pretty neat to me and it actually didn't have to be a block it could have been shaped like a horse I'll see in the next video