Chapter 5: Applications of Integration
Section 5.8: Work
Example 5.8.3
With a rope weighing 12 lb/ft, roofing material weighing 50 lbs is hoisted 45 ft to the top of a building. Find the total amount of work done.
Solution
Mathematical Solution
Since the force of gravity on the roofing material is constant, the work done raising just this material to the top of the building is the product of force and the distance through which the force moves. Hence, this work is 50 lbs × 45 ft = 2250 ft-lbs.
The force exerted by gravity on the rope decreases during the lift because the amount of rope hanging from the roof varies from 45 to 0 ft. So, with y measured positive from the ground, an infinitesimal segment of the rope of length dy weighs 1/2 dy lbs. This segment must be lifted through a distance 45−y ft. Hence, the work done in lifting just the rope is given by the integral
∫0451245−y ⅆy=110254 ft-lbs
Maple Solution
Work done in hoisting just the building material to the roof:
Type in the product and press the Enter key.
50⋅45
2250
Work done in hoisting just the rope to the top of the building:
Expression palette: Definite-integral template Write the appropriate integral and press the Enter key.
∫0451245−y ⅆy
20254
Sum to get total work done
Reference summands by their equation labels.
Context Panel: Evaluate and Display Inline
+ = 110254
<< Previous Example Section 5.8 Next Example >>
© Maplesoft, a division of Waterloo Maple Inc., 2024. All rights reserved. This product is protected by copyright and distributed under licenses restricting its use, copying, distribution, and decompilation.
For more information on Maplesoft products and services, visit www.maplesoft.com
Download Help Document