2.) The uniform panel door weighs 66 lb and is prevented from opening by the strut C, which is a light two- force member whose upper end is secured under the doorknob and whose lower end is attached to a rubber cup which does not slip on the floor. Of the door hinges A and B, only B can support thrust. Calculate the compression force C in the strut and the horizontal components of the reactions (the -resultant of the components of the reaction not along the hinge axis) supported by hinges A and B. Take P=65 lb and applied normal to the plane of the door as shown. [Answer: C 112 lb, An = 46.4 lb, Bn = 55.4 lb] The two force member description of C is a way of telling you that you can assume the force it exerts follows a line of action parallel to the strut. So you can express it in terms of its unknown magnitude times the appropriate unit vector. 44° 38 55°

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
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2.) The uniform panel door weighs 66 lb and is prevented from opening by the strut C, which is a light two-
force member whose upper end is secured under the doorknob and whose lower end is attached to a
rubber cup which does not slip on the floor. Of the door hinges A and B, only B can support thrust.
Calculate the compression force C in the strut and the horizontal components of the reactions (the
resultant of the components of the reaction not along the hinge axis) supported by hinges A and B. Take
P=65 lb and applied normal to the plane of the door as shown.
[Answer: C 112 lb, An = 46.4 lb, Bn = 55.4 lb]
The two force member description of C is a way of telling you that you can assume the force it exerts
follows a line of action parallel to the strut. So you can express it in terms of its unknown magnitude
times the appropriate unit vector.
44"
38°
35°
55°
Transcribed Image Text:2.) The uniform panel door weighs 66 lb and is prevented from opening by the strut C, which is a light two- force member whose upper end is secured under the doorknob and whose lower end is attached to a rubber cup which does not slip on the floor. Of the door hinges A and B, only B can support thrust. Calculate the compression force C in the strut and the horizontal components of the reactions (the resultant of the components of the reaction not along the hinge axis) supported by hinges A and B. Take P=65 lb and applied normal to the plane of the door as shown. [Answer: C 112 lb, An = 46.4 lb, Bn = 55.4 lb] The two force member description of C is a way of telling you that you can assume the force it exerts follows a line of action parallel to the strut. So you can express it in terms of its unknown magnitude times the appropriate unit vector. 44" 38° 35° 55°
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