A rock has mass 1.80 kg. When the rock is suspended from the lower end of a string and totally immersed in water, the tension in the string is 12.8 N. What is the smallest density of a liquid in which the rock will float?

Video transcript

Alright. So for this problem, we are working with the buoyancy force. Let's see what they're asking us, students perform an experiment using a two kg metal block. They tie the block and immerse it into water. They way the stone while immersed and find its way to be 11.3 Newtons. We need to determine the least density of a liquid that will make the block float. So this is a little bit of a tricky question. The first key is going to be recognizing that we have two different scenarios. We have one scenario where we have a block. I'm gonna do our free body diagrams as I talked with this. Okay, so one scenario where we have a block and we have it tied to a string. So there is a tension force when they put it on, when they immerse it into the water, there's going to be a buoyancy force just because it's submerged and then there is the weight of the block. Alright, that scenario one and the liquid, the fluid for this is water. The second scenario, what they're asking us to solve for is a different case where we just have the buoyancy force always have the weight of the block and those people each other so that the block is floating. Okay, so once we have that kind of in mind and this second case is in this unknown fluid and we're finding the density of this fluid. So once we have this in mind, let's go back to what are known as are we know the mass of the block is two kg. We also know that the weight when it is submerged when it is tied to the block is 11. newtons. So it might be tempting to say, oh the weight is 11.3 newtons. But actually what that's telling us is that the tension on the string is 11.3 newtons when the sloth is submerged in water. We also need to recall our buoyancy force equation. So that's the row G V where rho is the density of the fluid that you are submerged in JIA's gravity and then V is volume. So between the two scenarios, this road term is going to change, but the gravity term Is a constant. Right? We know that that's just 9.8 m/s squared. And volume is going to be the same because the block is the same. So let's recall Newton's law where the sum of the forces equals mass times acceleration. We don't have any acceleration. So this term goes to zero and we're going to do this for both of our free body diagrams and then kind of see where we end up with what we need to know. So for this first one we have the buoyancy force and the tension equal the weight of the block. For this second free body diagram starts out the same. We don't we're not moving. And we have the buoyancy force is just equal to the weight. So as far as the weight goes. We need to recall that the weight equals the mass times gravity. We know the mass of the block. We know gravity is a constant. And so for this case The weight is going to equal two kg Times 9.8 m/s squared for both of these situations. Okay, so Plug this into our calculator and we come out with the weight of the block is 19. new ones. So this is good for this first equation because we know tension. Now we know wait. So we can get our buoyancy force, buoyancy force is weight minus tension. So that equals 19.6 newtons -11.3 Newtons from our attention. That gives us 8.3 Newtons. That's our total buoyancy force in this first problem from there, we know the density of water. We know gravity and we can solve for volume and that's helpful to us because the only unknown in this second equation other than what we are solving for is volume. So let's see what that looks like. So we know it's the density of this unknown fluid times gravity times volume equals weight. So we already sell for weight. We know we are solving for this, this density term. So we no, wait, Is 19.6 Newtons divided by gravity times volume. So gravity The gravitational force 9.8 m/s squared and then volume is the only term that we need to be able to solve for this density. So, let's go back up here and use this first scenario to solve for volume. So we have row GV equals 8. newtons. We know that in water the density is kg per meters cubed density is 9.8 m per second, gravity is 9.8 m per second squared volume is what we're solving for. And that is all equal to 8.3. Yes. So if we're gonna solve for volume, we are going to um divide out the this half of the equation. We plug that into our calculators and we get eight points 47 times 10 to the negative. four m. Cute. So now that we know that volume, we can like this into our second scenario to solve for this density of this unknown fluid. Okay, so we have 9.19.6 Newtons 9. meters per second squared times 8.47 times 10 to the negative fourth meters. Cute, plug that in. And we get 23 60 kg for a huge And that is the density of this unknown fluid. That where the ice would just, you know, be equal to the buoyant buoyancy force. It would be perfectly level in this ice. All right, look at our answers answer. C is 23 60 kg per meter cute. And that's the answer for this problem. We'll see you in the next video