   Chapter 14, Problem 3P

Chapter
Section
Textbook Problem

A long piece of galvanized sheet metal with width w is to be bent into a symmetric form with three straight sides to make a rain gutter. A cross-section is shown in the figure.(a) Determine the dimensions that allow the maximum possible flow; that is, find the dimensions that give the maximum possible cross-sectional area.(b) Would it be better to bend the moral into a gutter with a semicircular cross-section? (a)

To determine

To find: The dimensions for the maximum cross sectional area which gives maximum flow for given cross sectional area.

Explanation

Given:

The rain gutter having width (w) and three straight sides is shown below in figure 1.

Calculation:

The area of the rain gutter is calculated using the area of trapezoid (A) as,

A=12h(b1+b2)

Here, height of the trapezoid is h and the upper and lower breadth of the trapezoid are b1 and b2 , respectively.

The cross section area of the rain gutter is the function of A(x,θ)=12h(b1+b2) .

Substitute x(sinθ) for h , (w2x) for b1 and w2x+2xcosθ for b2 ,

A(x,θ)=12(xsinθ)[(w2x)+(w2x+2xcosθ)]=12(xsinθ)(w2x+w2x+2xcosθ)=12(xsinθ)(2w4x+2xcosθ)=122(xsinθ)(w2x+xcosθ)

=(xsinθ)(w2x+xcosθ)A(x,θ)=wxsinθ2x2sinθ+x2sinθcosθ (1)

Here, 0<x12w and 0<θπ2

To find the critical point, differentiate the equation (1) with respect to x .

Equate the above equation is equal to zero (0) .

wsinθ4xsinθ+2xsinθcosθ=0sinθ(w4x+2xcosθ)=0w4x+2xcosθ=02xcosθ=4xw

cosθ=4xw2xcosθ=4x2xw2xcosθ=2w2x

Similarly, differentiate the equation (1) with respect to θ .

By chain rule of differentiation (ddx(u.v)=udvdx+vdudx) ,

Apply the formula (cos2θsin2θ=2cos2θ1) ,

Substitute 2w2x for cosθ ,

Substitute 0 for dAdθ ,

3x2wx=0x(3x

(b)

To determine

To find: Whether it would be better to bend the metal into a rain gutter with semicircular cross section.

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