The bottom of a steel 'boat' is a 5.0 m x 10 m x 2.0 cm piece of steel (pₛₜₑₑₗ = 7900 kg/m³) . The sides are made of 0.50-cm-thick steel. What minimum height must the sides have for this boat to float in perfectly calm water?

Question

Diagram of a rectangular steel boat with dimensions labeled for buoyancy calculations.

Accepted Answer

Hi everyone in this practice problem. We're being asked to calculate the height necessary for the walls of a platform to remain afloat in a lake. We will have a floating platform possessing a rectangular base with a dimension of four m by eight m by a 1.5 centimeter thickness made from a material with a density of 8100 kg per meter cube. The platform is enclosed by walls constructed from platinum, having a 0.32 centimeter thickness. We are being asked to find the lowest height necessary for the walls to ensure that the platform will remain afloat in the lake. The density of the water is approximately 1000 kg per meter cube. And the options given for for the height necessary for the walls are a 0.11 m b 0.22 m C 0.36 m and D 0.24 m. We are also given the figure to represent the system that we have in this problem statement and um starting to actually solve this problem. We want to notice that for the platform to remain afloat inside of the lake, the weight must be equal to the buoyant force. Namely W must be equals to F B F B is the buoyant force and W is the weight of the floating platform. And we are going to look into first the Buon force and then the weight. So the buoyant force is represented by F B which will equals to row, W multiplied by P P multiplied by G row W is the density of the water P P is the volume of the platform and G is the gravitational acceleration. The fall of the platform is given to B V P which will be equals to a multiplied by B multiplied by. So the height of the uh of the wall itself is going to be H and we are given the thickness which is going to be C which is going to be 1.5 centimeter. So the total height is going to be H plus C. So PB will be equals to a multiplied by B multiplied by open parenthesis H plus C. So we want to substitute that into our boy and force equation so that we will get then a combination of F B to be equals to row W multiplied by a multiplied by B multiplied by open parenthesis H plus C multiplied by G just like. So, so we will keep this as it is for now and then we are going to now look into the weight of the platform. So the weight of the platform will be equals to the weight of the base or F base plus two times the weight of the first part plus two times the weight of the second part. The first part is going to be the weight of the A multiplied by D multiplied by H. So it's going to be the weight of the thickness or of the walls on the A side. And then the F two is going to be the weight on the B side. So in this case, then F base will equals to row multiplied by V base multiplied by G and uh two, F one will equals two multiplied by a row multiplied by a volume, one multiplied by G and two F two will equals to two multiplied by a row multiplied by V two multiplied by G. So in this case, we can actually pull the row times G out so that our W or the weight will that equals to row multiplied by G multiplied by open parenthesis, P base plus two, P one plus two, P two close parenthesis just like that. So now let's actually uh write down what fee base is what P one is and what P two is to actually neglect any sort of confusion. So W will then equals to row multiplied by G multiplied by open parenthesis. Fe base fee base is going to be just a multiplied by B multiplied by C. So I'm just gonna write that down as ABC to make it simpler for all of us. And then P one is going to be the uh volume of the wall on the A side. So that will be a multiplied by D multiplied by H. And then lastly, oh, don't forget the two in front. And lastly, P two is going to be, um, B multiplied by D multiplied by H. So there'll be two B DH just like so close parentheses row is going to be the density of the platinum, which is going to be the material of which the walls are constructed with. So we wanna next substitute the boy and force F B and W together into our W equals F B equation. So that will then give us W equals F B W will be row multiplied by G multiplied by open parent C ABC plus two A DH plus two B DH equals that to row W multiplied by A B multiplied by H plus C. In parentheses, multiplied by G, we can cross out the gravitational acceleration but we can't cross out the density because they are different. One is the density of the water and one is the density of the platinum. So we do have to keep both sites. But after crossing the gravitational acceleration out, we will then left with row multiplied by ABC plus two A DH plus two B DH, close parentheses equals row W multiplied by A B multiplied by in parentheses H plus C. So we actually know all of this information either from the figure given or the problem statement in our problem. So we want to substitute all of these values into this equation and still leave the H to be unknown because that is the one thing that we are interested to find. OK. So uh on the left side, we have the row of the platinum which will be kilograms per meter cube multiplied that by ABC A is four m. So four m B is eight m and C is 1.5 centimeters. So that'll be 1.5 times 10 to the power of negative two m plus two ad. So two multiplied by four m multiplied by D which is going to be 0.32 centimeter. So that will be 0.32 times 10 to the power of negative two m multiplied that again by H which is unknown. And then we wanna plus that with two B DH. So two multiplied by eight m multiplied by 0.32 times 10 to the power of negative two m for the D multiplied that with H close parenthesis equals to the row of water which is 1000 kg per meter cube multiplied by A B which is four m multiplied by eight m multiplied that by H plus C which will be H plus C is going to be 1.5 times 10 to the power of negative two m just like. So, so that will be our um equation after substituting all of the known values into our equation. So simplifying this by actually doing the multiplication, we will then get um on the left side, I am going to just drop all the units to avoid any sort of confusion because we know that everything is in si so the H is going to just come out in meter. So let's just uh drop all the units. So then the left side will then equals to multiplied by 0.48 plus 0.256 H plus 0.512 H equal that to 1000 multiplied by 32 multiplied by H plus 1.5 times 10 to the power of negative two just like so, so next, we want to actually um do this algebra. So then our left side will then equals to plus 207.36 H plus 414.72 H to be equals to 32,000 H plus 480. You wanna then compile all the terms with ages into one site. So we'll have 32,000 H minus 207.36 H minus 414.72 H to then be equals to 3888 minus which will come out to be 31,377 30. H equals to 3408 which will then give us an H value of 0.1 oh nine. And that will be in meter rounding it up. We will then get 0.11 m or 0.11 m, which will be the answer to this practice problem which will correspond to option A in our answer choices. So option A will be the answer to this particular practice problem with the lowest height necessary for the walls to ensure that the platform remains afloat in the lake to be 0.11 m. 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 lesson videos on similar topics and that'll be it for this one. Thank you.

The bottom of a steel 'boat' is a 5.0 m x 10 m x 2.0 cm piece of steel (pₛₜₑₑₗ = 7900 kg/m³) . The sides are made of 0.50-cm-thick steel. What minimum height must the sides have for this boat to float in perfectly calm water?

Verified Solution

Key Concepts

Buoyancy

Density

Volume Displacement