The sliding glass door rolls on the two small lower wheels A and B. Under normal conditions the upper wheels do not touch their horizontal guide. (a) Compute the force P required to slide the door at a steady speed if wheel A becomes "frozen" and does not turn in its bearing. (b) Rework the problem if wheel B becomes frozen instead of wheel A. The coefficient of kinetic friction between a frozen wheel and the supporting surface is 0.33, and the center of mass of the 118-lb door is at its geometric center. Neglect the small diameter of the wheels. Answer: A (a) P = i (b) P- i 28" B 6" P lb lb 38" 38"

The sliding glass door rolls on the two small lower wheels A and B. Under normal conditions the upper wheels do not touch their horizontal guide. (a) Compute the force P required to slide the door at a steady speed if wheel A becomes "frozen" and does not turn in its bearing. (b) Rework the problem if wheel B becomes frozen instead of wheel A. The coefficient of kinetic friction between a frozen wheel and the supporting surface is 0.33, and the center of mass of the 118-lb door is at its geometric center. Neglect the small diameter of the wheels. Answer: A (a) P = i (b) P- i 28" B 6" P lb lb 38" 38"

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

b.5

The sliding glass door rolls on the two small lower wheels A and B. Under normal conditions the upper wheels do not touch their
horizontal guide. (a) Compute the force P required to slide the door at a steady speed if wheel A becomes "frozen" and does not turn in
its bearing. (b) Rework the problem if wheel B becomes frozen instead of wheel A. The coefficient of kinetic friction between a frozen
wheel and the supporting surface is 0.33, and the center of mass of the 118-lb door is at its geometric center. Neglect the small
diameter of the wheels.
Answer:
(a) P =
(b) P =
A
i
i
28"
B
6"
P
lb
lb
38"
38"

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