Question

A steel pipe is being carried down a hallway 9 ft wide. At the end of the hall there is a right-angled turn into a narrower hallway 6 ft wide. What is the length of the longest pipe that can be carried horizontally around the corner

Answer 1

Answer:

21.62m

Step-by-step explanation:

first draw the the pipe

second you need know wich is the angle between the pipe and the corner

β=Tan^-1(6/9)=33.7

find the components using tow triangles

a=9/cos(33.7)=10.81m

b=6/sen(33.7)=10.81m

finally sum the leghts

L=10.81+10.81=21.62mm

attached procedure

Answer 2

The allowable length of the pipe can be found by finding the smallest angle made between the pipe and 6 ft. hallway.

The length of the pipe is approximately 21.07 feet.

Reasons:

Length of the pipe, L = x + y

$sin(\theta) = \dfrac{6}{x}$

$cos(\theta) = \dfrac{9}{y}$

Therefore;

x = 6·cosec(θ)

y = 9·sec(θ)

Length of pipe, L = 6·cosec(θ) + 9·sec(θ)

At the minimum length, we have;

$\dfrac{dL}{d\theta} = \dfrac{d}{d\theta} \left(6 \cdot cosec(\theta) + 9 \cdot sec(\theta) \right) = \dfrac{9 \cdot sin^3(\theta) - 6 \cdot cos^3(\theta)}{sin^2(\theta) \cdot cos^2(\theta)} = 0$

9·sin³(θ) - 6·cos³(θ) = 0

9·sin³(θ) = 6·cos³(θ)

$\dfrac{sin^3(\theta)}{cos^3(\theta)} = \dfrac{6}{9}$

$tan(\theta) = \sqrt[3]{\dfrac{6}{9} }$

$\theta = arctan \left(\sqrt[3]{\dfrac{6}{9} } \right) \approx 41.14^{\circ}$

x = 6 × cosec(41.14°) ≈ 9.12

y = 9 × sec(41.14°) ≈ 11.95

L ≈ 11.95 + 9.12 = 21.07

The length of the pipe, L ≈ 21.07 feet

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