One day when you come into physics lab you find several plastic h...

Question

One day when you come into physics lab you find several plastic hemispheres floating like boats in a tank of fresh water. Each lab group is challenged to determine the heaviest rock that can be placed in the bottom of a plastic boat without sinking it. You get one try. Sinking the boat gets you no points, and the maximum number of points goes to the group that can place the heaviest rock without sinking. You begin by measuring one of the hemispheres, finding that it has a mass of 21 g and a diameter of 8.0 cm. What is the mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat?

Accepted Answer

Hi, everyone in this practice problem, we are being asked to determine the maximum mass of uh monitoring equipment. We will have a floating device holding a monitoring equipment for offshore environmental assessments where the device consists of a spherical buoy made out of lightweight plastic material. We're being asked to determine the maximum mass of the monitoring equipment that can be placed inside of the buoy without causing it to sink where it is given that the spherical buoy has a mass of two times 10 to the power of three kg and a diameter of two m. The options given for the maximum mass are a 3.46 times 10 to the power of four kg. B 3.23 times 10 to the power of four kg C 2.58 times 10 to the power of three kg and D 2.19 times 10 to the power of three kg. So for this particular practice problem, the device will not sync as long as the buoyant force is equal to the gravitational force, namely it must have neutral buoyancy. So according to that F B or the buoyant force will then equals to the gravitational force F G of the buoy plus the gravitational force F G of the equipment E Q. So in this case, we wanna recall again that boy and force F B can be call by multiplying row of the fluid multiplied by the volume of the buoy or volume submerge multiplied that again by the gravitational acceleration G. And we want to uh substitute that into our equation that we have. So F B will be row of the water multiplied by the volume of the buoy multiplied then again by the gravitational acceleration which will be equals to F G of the, which is just the weight of the buoy. So M of the buoy multiplied by G plus the M of the equipment. E Q multiplied by G. We can cross out the G S because we have a gravitational acceleration in all three terms here. So then we will get an equation where row of the water multiplied by volume of the Buie will equals to M Buie plus M equipment. Uh Once again, row, uh W is the density of water. Three of the buoy is the volume of the spherical buoy. M buoy is the mass of the spherical buoy. And M equipment is the mass of the monitoring equipment. So then what we are interested to find is the M of the equipment. So we can rearrange this equation to get M E Q equation which will then be row W multiplied by V buoy minus M buoy. So then we can actually start substituting a lot of the information into this equation right here. So then we will have first M E Q to be equals to row W which is just going to be 1000 kg per meter cube. And then P buoy is given uh by the equation because it is a spherical buoy. We are looking for the volume of a sphere which is going to be represented by the equation of four divided by three multiplied by pi R cube minus Ambu. And Bui is uh given to be two times 10 to the power of three kg. So then we can um substitute our R value for the here so that our m equipment will then equals to kg divided by a meter cube multiplied by 4/3 or four divided by three, multiplied by a pi multiplied by R cube, which is given in the problem statement to be um two m divided by two or half of two m. So that will be one m. So one m to the power of three minus two times 10 to the power of three kg. So the diameter is two m. So the R is half of that, which is one m. So then calculating all of this, we will get the mass of the equipment maximum to then be 4186.67 kg minus 2000 kg, which will resulted in 2186.67 kg or rounding it up mass, maximum mass of the equipment will be close to be 2.19 times 10 to the power of three kg. So this will be the answer to this practice problem where the maximum mass of the monitoring equipment that can be placed inside of the buoy without causing it to sink is 2.19 times 10 to the power of three kg, which will correspond to option D in our answer choices. So option D will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our other videos on similar topics and that'll be it for this one. Thank you.

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