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 0.23, and the center of mass of the 136-lb door is at its geometric center. Neglect the small diameter of the wheels. 5" A 29" B 5" Answer: P (a) P = i lb (b) P = i lb 44" 44"

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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5.5

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 0.23, and the center of mass of the 136-lb door is at its geometric center. Neglect the small
diameter of the wheels.
5"
A
29"
B
5"
Answer:
P
(a) P =
i
lb
(b) P = i
lb
44"
44"
Transcribed Image Text: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 0.23, and the center of mass of the 136-lb door is at its geometric center. Neglect the small diameter of the wheels. 5" A 29" B 5" Answer: P (a) P = i lb (b) P = i lb 44" 44"
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