How can you make a ship from a 35cm by 35cm aluminum foil and hold around 120 pennies?

Question

Please tell me all the steps, and nothing else can be used other than the piece of aluminum foil.

Answer 1

A penny has a mass of 2.5 g for pennies minted after 1980

so 120 have mass of
300 grams
hmmm - coincidence that this is a nice even number?

Assume the aluminum foil has negligible mass, so you have to create a volume greater that the amount of wate rthat will be displaced.

Water has mass 1000 Kg/m^3

A meter is 100 cm
so a m^3 is 100cm^3
or 1 cm^3 has mass of 1 g

so the volume is 300 cm^3

Now, I set up the following equations:

using a square bottom boat with width and length
w

and height of the sides as h

w^2*h>=300
2*w+h=35
or
35*w^2-w^3-600>=0

In order to minimize the strain on the foil, I want to make w as small as possible. So I looked for a minima, or the smallest w that satisfies the equations.

After some calculus and excel graphing, I found that , a boat with a base that is 5 by 5 cm square
will have height of 15 cm and displace
375 g of water before sinking.
This gives you some freeboard.

It also means that you have good overlap at the corners to give the vessel strength.

The last question is, will it hold the pennies?

A penny has a diameter of 19.05 mm and a thickness of 1.5 mm

If the boat bottom is modified to be 1.9*3 cm
or 5.7 cm by 5.7 cm, then a layer of pennies that is 3X3 will fit in the bottom. That's 9 pennies per layer.

That means you need 120/9, or 14 layers of pennies to reach 120 ( there are 13 layers of 9 with the 14th layer having 3)

14 pennies have a height of
14*0.15 cm
=2.1 cm
So back to my ship:

The bottom is 5.7 cm ^2
which makes the sides 14.65 cm tall

It will displace a maximum of 475 g, well over the 300 requred.

When in the water, it will be 9.2 cm from the water surface to the keel


j

Answer 2

A penny weighs about 3 grams.
120 pennies weigh about 360 grams, make it 400g for slop
because you need to be able to float greater than the exact
weight or it will never float.
.
Since water has a density of about 1 gram/cm^3 you would
need to displace about 400 grams of water in mass, or
400 cc.s of water in volume.

Looking at a square of aluminum foil, and neglecting its weight
I would say looking at a simple model of a flat aluminum sheet with a square bottom of lengths "L" and sides of height "l" that
will look like a cookie pan, wide and short with the pennies in the bottom as wide and deep as possible. The shape would
be as if you pressed the aluminum foil to the bottom of a square pan you have some unused material at the corners that reinforce the sides ... probably a good idea to have the pennies as flat and as much against the sides as possible.

(L) + ( 2 * l ) = 35 looking at the sheet.
( ( L )^2 * ( l ) >= 400 looking at the volume

Now you have 2 equations in two unknowns you can solve.

Answer 3

You will have to consider the strength of the Aluminum foil in holding its shape, and the stability of the boat in the water. If your boat folds in half due to the weight of the pennies, or tips over, it doesn't do you much good. I suggest making a square boat with reinforced side-walls. This way you can evenly distribute the weight of the pennies across the bottom of the boat. you can also calculate the height of the sidewalls necessary to float the 120 pennies by using the density of water. Just give yourself plenty of margin on the sidewall height to keep the boat from taking on water. You can keep the top collapsing by rolling the top edge to create a reinforcing loop at the top of the boat. As a guess, I would try making the boat about 5 times as wide as it is tall.

Answer 4

you have to find out how much 120 pennies weigh, then figure out the how much volume an equivalent weight of water would be.

once you know these, then you have to design the vessel so that it displaces that volume of water before the water starts pouring in over the sides.

since the weight of the pennies are a constant and since the area of aluminum you have to work with is a constant (though it will be reduced by the shape you choose), you want to go the other route an maximize water displacement - that's the thing you have control of in determining shape.

I suggest a hemispherical shape - as a sphere is the lowest surface area (you have a limit on the aluminum foil) per volume enclosed of all shapes.

its a balance of maximizing water displacement and lost surface area due to a limit on surface area. its basically an exercise in efficiency.

Answer 5

frezze the aluminum water then put it in water and set the pennies on top because you only using using water