Question

The design of boats is based on Archimedes' Principle, which states that the buoyant force on an object in water is equal to the weight of the water displaced. Suppose you want to build a sailboat whose hull is parabolic with cross section y ax2, where a is a constant. Your boat will have length L and its maximum draft (the maximum vertical depth of any point of the boat beneath the water line) will be H. Every cubic meter of water weighs 10.000 newtons.

Required:
What is the maximum possible weight for your boat and cargo?

Answer 1

Answer:

Maximum Possible weight  = $\frac{4}{3}$ HL$[\sqrt{\frac{H}{a} }$] x 10,000 Newtons.

Explanation:

Solution:

Note: This question is incomplete, it lacks the sketch of the boat which is necessary to draw the volume equation. But I have found similar question and drawn the equation accordingly and solved the question.

In order to find the maximum possible weight for your boat and cargo,  we need to solve this definite integral equation of volume.

V = 2L$[Hx - \frac{ax^{3} }{3} ]^{\sqrt{\frac{H}{a} } }_{0}$

With limits 0 and $\sqrt{\frac{H}{a} }$

Putting the Limits into the variable x, we will get.

V = 2L$[H\sqrt{\frac{H}{a} } - \frac{a(\sqrt{\frac{H}{a} } ^{3} }{3} ]$

Solving the above equation further,

V = 2HL$[\sqrt{\frac{H}{a} } {\frac{3}{3} - \frac{1}{3} }$]

V = 2HL$[\sqrt{\frac{H}{a} } {\frac{2}{3} }$]

So, final equation of the volume after the integration is:

V = $\frac{4}{3}$ HL$[\sqrt{\frac{H}{a} }$]

Now, we know that Every cubic meter weighs 10,000 newtons. So,

Maximum Possible weight  = $\frac{4}{3}$ HL$[\sqrt{\frac{H}{a} }$] x 10,000 Newtons.

Answer 2

We have that the maximum possible weight for your boat and cargo  is mathematically given as

$Maximum\ weight = \frac{4}{3}HL(\sqrt{H/a}) x 10,000 N$

Maximum possible weight for your boat and cargo

Question Parameters:

Every cubic meter of water weighs 10.000 newtons.

Generally the equation for the definite integral equation of volume  is mathematically given as

$V = 2LHx-\frac{ax^3}{3}_0^{\sqrt{h/\alpha}}\\lim-->0 ,\sqrt{h/ \alpha}\\\\V = 2LH\sqrt{h/ \alpha}-\frac{a(\sqrtr{h/ \alpha})^3}{3}\\\\V = 2LH(\sqrt{h/ \alpha}*\frac{2}{3})\\\\V=\frac{4}{3}HL(\sqrt{H/a})$

$Maximum\ weight = \frac{4}{3}HL(\sqrt{H/a}) x 10,000 N$

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