The uniform panel door weighs 67 Ib and is prevented from opening by the strut C, which is a light two-force member whose upper end is secured under the door knob 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 force in the vertical z-direction. Calculate the compression C in the strut and the horizontal components of the forces supported by hinges A and B when a horizontal force P=57 Ib is applied normal to the plane of the door as shown.
The uniform panel door weighs 67 Ib and is prevented from opening by the strut C, which is a light two-force member whose upper end is secured under the door knob 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 force in the vertical z-direction. Calculate the compression C in the strut and the horizontal components of the forces supported by hinges A and B when a horizontal force P=57 Ib is applied normal to the plane of the door as shown.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Could you please solve the problem below, with work, in the same method as the solution posted below? Thank you!
Transcribed Image Text:The uniform panel door weighs 67 Ib and is prevented from opening by the strut C, which is a light two-force member whose upper end
a rubber cup which does not slip on the floor. Of the door hinges A and B, only B can support force in the vertical z-direction. Calculate the compression C in the strut and the horizontal components of the
forces supported by hinges A and B when a horizontal force P = 57 Ib is applied normal to the plane of the door as shown.
secured under the door knob and whose lower end is attached to
43"
P
46"
35"
8"
24"
Answers:
C =
1 Ib
Axy
-2 lb
%3D
Byy =
3 Ib
Transcribed Image Text:3/89
E Mno.
=0: 50 (48) - o.6 C (48) :
Ax
C = 100 Ib
P=
50 lb |
48'
60 lb
Ay
I Ma,=0:-50(8) - (0.6'100) 32
+64 By =0
By EMay-
B.
By = 36.25 lb
60 (4)-(0.8-10) 40
+ 64BX =0,
Bx = 27.5 lb
%3D
45.5 l6
=
%3D
EMB, =0:-50(72) + (0.6'100) (32) + 64Ay = 0
Ay
= 26.25 Ib
E MBy =0: 60(24) - 80 (48) +64 Ax=0
Ax = 27.5 1b
Axy=
= 38.0 Ib
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