Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 9 - Exercises and Problems: 51

Answer

a) (0,0,13) b) (0,0,11)

Work Step by Step

We know that the sides are perfectly symmetrical, so the center of mass would normally be in the center of the silo if there were no base. We now must consider the base. The base is at the bottom, so we find the percentage of the mass it takes up and multiply this value by 15, which would normally be the center of mass: $15 \times ( \frac{6000}{44000})=2.04$ Thus, the height of the center of mass is: $15 - 2.04 \approx 13$ Thus, the center of mass is: a) (0,0,13) b) We solve this by adding the mass to the overall silo. We know that the center of mass of the first 44,000 kg is at 13 meters high. We know that the center of mass of the $\pi(2^2)(30)(800)(2/3)=201,061 \ kg$ remaining is 10. Thus, we find the overall center of mass: $h_{cm} = \frac{(201,061\times 10) + (44,000\times 13)}{44,000+201,061}\approx 11$ Thus, the center of mass is: b) (0,0,11)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.