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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

A 160-lb man carries a 25-lb can of paint up a helical staircase that encircles a silo with a radius of 20 ft. If the silo is 90 ft high and the man makes exactly three complete revolutions climbing to the top, how much work is done by the man against gravity?

To determine

To find: The work done by the man against the gravity.

Explanation

Given data:

The weight of the man is 165-lb.

Weight of the paint can is 25-lb.

The staircase encircles a silo, the radius of the silo is 20 ft, and height of the silo is 90 ft.

The man needs to make three complete revolutions to reach the top the staircase.

Formula used:

Write the expression to find the work done on the object with the force vector F .

W=abFr(t)dt (1)

Here,

F(t) is the force vector exerted by the man,

r(t) is the velocity vector of the object, which is the derivative of position vector r(t) ,

a is the lower limit of the scalar parameter and

b is the upper limit of the scalar parameter.

Write the expression to find the vector r(t) .

r(t)=ddt[r(t)] (2)

Here,

r(t) is the position vector.

When the man moves to up, the gravity of the earth tries to pull the combined weight of man and the paint towards downside. Therefore, the force exerted by man is equal and opposite to the force exerted by the gravity.

The combined weight of the man and paint is 185-lb. As the direction of force exerted by the man is up side (z-direction), the force is written as follows.

F=(0)i+(0)j+185k

As the radius of silo is 20 ft, height is 90 ft, and man needs 3 revolutions to reach top, parameterize the staircase as follows.

Consider the parameter as t and write the x, y, and z parameters.

One revolution means 2π and 3 complete revolutions means 6π .

x=20cost,y=20sint,z=906πt,0t6πx=20cost,y=20sint,z=15πt,0t6π

Write the position vector r(t) from the parametric equations as follows.

r(t)=20costi+20sintj+15πtk

Calculation of r(t) :

Substitute 20costi+20sintj+15πtk for r(t) in equation (2),

r(t)=ddt[20costi+20sintj+15πtk]=ddt(20cost)i+ddt(20sint)j+ddt(15πt)k=(20sint)i+(20cost)j+(15π)k

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Chapter 16 Solutions

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Sect-16.1 P-11ESect-16.1 P-12ESect-16.1 P-13ESect-16.1 P-14ESect-16.1 P-15ESect-16.1 P-16ESect-16.1 P-17ESect-16.1 P-18ESect-16.1 P-21ESect-16.1 P-22ESect-16.1 P-23ESect-16.1 P-24ESect-16.1 P-25ESect-16.1 P-26ESect-16.1 P-29ESect-16.1 P-30ESect-16.1 P-31ESect-16.1 P-32ESect-16.1 P-33ESect-16.1 P-34ESect-16.1 P-35ESect-16.1 P-36ESect-16.2 P-1ESect-16.2 P-2ESect-16.2 P-3ESect-16.2 P-4ESect-16.2 P-5ESect-16.2 P-6ESect-16.2 P-7ESect-16.2 P-8ESect-16.2 P-9ESect-16.2 P-10ESect-16.2 P-11ESect-16.2 P-12ESect-16.2 P-13ESect-16.2 P-14ESect-16.2 P-15ESect-16.2 P-16ESect-16.2 P-17ESect-16.2 P-18ESect-16.2 P-19ESect-16.2 P-20ESect-16.2 P-21ESect-16.2 P-22ESect-16.2 P-23ESect-16.2 P-24ESect-16.2 P-25ESect-16.2 P-26ESect-16.2 P-31ESect-16.2 P-32ESect-16.2 P-33ESect-16.2 P-34ESect-16.2 P-35ESect-16.2 P-36ESect-16.2 P-37ESect-16.2 P-38ESect-16.2 P-39ESect-16.2 P-40ESect-16.2 P-41ESect-16.2 P-42ESect-16.2 P-43ESect-16.2 P-44ESect-16.2 P-45ESect-16.2 P-46ESect-16.2 P-47ESect-16.2 P-48ESect-16.2 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P-6RECh-16 P-7RECh-16 P-8RECh-16 P-9RECh-16 P-10RECh-16 P-11RECh-16 P-12RECh-16 P-13RECh-16 P-14RECh-16 P-15RECh-16 P-16RECh-16 P-17RECh-16 P-18RECh-16 P-19RECh-16 P-20RECh-16 P-21RECh-16 P-22RECh-16 P-23RECh-16 P-24RECh-16 P-25RECh-16 P-27RECh-16 P-28RECh-16 P-29RECh-16 P-30RECh-16 P-31RECh-16 P-32RECh-16 P-33RECh-16 P-34RECh-16 P-35RECh-16 P-36RECh-16 P-37RECh-16 P-38RECh-16 P-39RECh-16 P-40RECh-16 P-41RECh-16 P-1PCh-16 P-2PCh-16 P-3PCh-16 P-5PCh-16 P-6P

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