/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 85 The cross-sectional shape of a c... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 85

The cross-sectional shape of a canoe is modeled by the curve \(y=a x^{2},\) where \(a=1.2 \mathrm{ft}^{-1}\) and the coordinates are in feet. Assume the width of the canoe is constant at \(w=2 \mathrm{ft}\) over its entire length \(L=18 \mathrm{ft}\). Set up a general algebraic expression relating the total mass of the canoe and its contents to distance \(d\) between the water surface and the gunwale of the floating canoe. Calculate the maximum total mass allowable without swamping the canoe.

Short Answer

Expert verified
The maximum total mass which will prevent the canoe from swamping is approximately 447 pounds.

Step by step solution

01

Express the Cross-sectional Area

The cross-sectional shape of the canoe is a parabola, given by \(y=a x^{2}\), where \(a\), \(x\) and \(y\) are in feet. The cross-sectional area of the part of the canoe that is underwater is given by integrating the expression for \(y\) over the width that is underwater. This specific width varies from -d to d as only the part that is submerged contributes to buoyancy. So we get \(A = \int_{-d}^{d} a x^{2} dx = [\frac{1}{3}a x^{3}]_{-d}^{d} = \frac{2}{3}ad^{3}\), where \(A\) is in square feet.
02

Obtain Expression for Total Mass

The total volume of the part of the canoe that is underwater is obtained by multiplying \(A\) with \(L\), so we have \(V=AL=\frac{2}{3} a L d^{3}\). Now, using Archimedes' principle, the mass of the water displaced by this volume must be equal to the total mass of the canoe, \(m\), for it to float. The mass of a volume of water is given by multiplying it by the density of water, \(\rho\), which is approximately 62.4 lb/ft鲁 in English units. Hence, we have the equation for the total mass of the canoe as \(m=\rho V = \rho A L = \rho \frac{2}{3}a L d^{3}\). This is still in pound units.
03

Optimize for d and Calculate Maximum Total Mass

This function depicts the total mass for a given depth of canoe submerged in water. The maximum value of mass should occur when \(d\) is equal to \(w/2 = 1\) feet, as the canoe cannot sink more than its own width. You can also confirm this analytically by taking the derivative of mass function with respect to \(d\) and setting it to zero. Either way, we find \(d=w/2\). Substituting this value along with given values of \(a\), \(L\), and \(\rho\) into the mass function yields the maximum total mass that the canoe can support without sinking: \(m= \rho \frac{2}{3}a L (\frac{w}{2})^{3} = 62.4 \times \frac{2}{3} \times 1.2 \times 18 \times 1 = 447.36\) pounds, or approximately \(447\) pounds when rounded.

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