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 0.25, and the center of mass of the 129-lb door is at its geometric center. Neglect the small diameter of the wheels. 41" P 25" B 5" 41"

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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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 0.25, and the center of
mass of the 129-lb door is at its geometric center.
Neglect the small diameter of the wheels.
41"
Answer:
(a) P =
lb
(b) P=
lb
25"
B
41"
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 0.25, and the center of mass of the 129-lb door is at its geometric center. Neglect the small diameter of the wheels. 41" Answer: (a) P = lb (b) P= lb 25" B 41"
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