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Problem 29 Hard Difficulty

A solid aluminum ingot weighs 89 $\mathrm{N}$ in air. (a) What is its volume? (See Table
$13.1 .$ ) (b) The ingot is suspended from a rope and totally immersed in water. What
is the tension in the rope (the apparent weight of the ingot in water)?

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Video Transcript

this problem starts with an aluminum bar that has a weight of 89 new ins. When an air we have two jobs. First we're going to find its volume. And then we're going to figure out what the tension and a rope would be if we held that aluminum bar by that rope so that it was suspended in water. So first, let's figure out the volume. That's pretty straightforward. Now we know that the density of aluminum is equal to 2.7 times 10 to the third kilo grams per cubic meter. That's our density, that we can look up. And from this unit, you know that density is equal to mass over volume. We also should already know that the weight of an object is equal to its mass times acceleration due to gravity. So what we can do with that information right away as we know that the weight of the bar is 89 new ins. We need its massive. We're gonna be able to do anything with that density. And this bar is on Earth, where the acceleration due to gravity is 9.8 meters per second squared. Let's divide by 9.8 on both sides and what we end up with. We know now that the mass of this aluminum bar is 9.8163 kilograms is our mass. Now that we know are mass, we can figure out the volume of this bar. We can start with our density equation. Density equals mass over volume. What we can do is we can rearrange by multiplying by velocity on both sides dividing by density to get sorry our volume on both sides and dividing by density To get that our volume is equal to mass over density will substitute in. That means that our volume is equal to that 9.8163 kilograms over our density to 0.7 times 10 to the third kilogram meters kilogram, meters per meter cubed. Now we can calculate that using our calculator we end up with a volume of 0.36 Sorry 33 64 meters cubed. We can also say that we have 3.3 64 leaders over. We want cubic meters or leaders. We can also both put these with two significant figures were given a weight with two significant figures. So I can say that our volume is 0.34 cubic meters or I can say it's 3.4 leaders. So we have found our volume. That was our step one. And our step to again is now we're going to find the tension in a rope if this bar was suspended in water. So I think the easiest way to go about doing this is to draw a free body diagram with the forces on this bar. So our box will represent our bar. The primary force pulling down on it is going to be its weight. And we luckily no, but it's weak is 89 new ones and then this bar is going to be at rest suspended in water. Um, so we're going to have to force is kind of holding it up. First, we're going to have a buoyant force on this farm calling it F sub B. That's the water pushing up on it, which is going to help push it upward to counter act its weight. We're also going to have the tension in the rope, which is what we're looking for that forced to detention so we can figure out what this buoyancy is. And then if we know that the force up is equal to the force down, which has to be If this bar is suspended at rest, we can figure out what that force of tension is. So first, let's figure out our buoyant force. Now we should know that a buoyant force is equal to the weight of water displaced by your object. And luckily to figure out the weight of our water displaced by our object, we know that weight is equal to mass times acceleration due to gravity. Um, we have this whole bar suspended in water again the bars made of aluminum. But it's going to displace the same volume of water. We know that volume of water being displaced from before is going to be equal to 0.3364 cubic meters. If I want to find the mass of that volume of water that was displaced using our density equation, we can say that mass is equal to our density times our volume. So the mass of our water displaced is equal to the density of water. 1000 kilograms per cubic meter times the volume of the water and the mass of our water that is displaced is going to be 3.364 kilograms. Now that is our mass of our water displaced, not the weight of our water displaced. To get that, we're going to have to multiply 3.364 kilograms times acceleration due to gravity, which is 9.8 meters per second squared and the weight of our water displaced is 32.967 Newton's and again, this is the weight of our water displaced. So it is equal to the buoyant force. Been a highlight that in red and kind of write it down here. So are buoyant. Force is equal to 32.967 Newman's So what we know know is we have a total wait down of 89 new ins, a buoyant force up of 32.96 noons and then an unknown tension force upward. So the force downward and the total force upward both need to be equal for this object to be at rest, so our total force upward needs to be equal to 89 new ins, and that's going to be equal to our buoyant force 32.96 plus this unknown tension force. And it's easy enough to solve for weekend subtract by are buoyant force on both sides so also tracked 32.96 fromthe left, 32.96 from the right. That will leave us with our tension force. And by doing that, we can say that our attention force is equal to 50 6.2 New ends is our tension force. And just like the previous part of the problem, we were given to Sig Figs with our original weight. So I can round that, too, to sink figs and get 56 Newton's as my final answer.

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