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# please provide help for the following screenshot, only an answer for letter B is needed src="/qa/attachment/10754528/" alt="Screen Shot 2019-10-14 at 4.34.21 PM.png" /> ATTACHMENT PREVIEW Download attachment Screen Shot 2019-10-14 at 4.34.21 PM.png The uniform panel door weighs 56 lb 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 = 48 lb is applied normal to the plane of the door as shown. Answers: C = 110.557 lb Axy= 57.2766 lb Bxy= ﬂ 64.0632 lb

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I found approximately the same results (enclosed doc) NB, location of... View the full answer

Equilibrium
. ZF = 0 47
A + B = 0
. 2 Fy = 0 ()
Ay + By + B + Fay = 0
Axy = VAitAy = 58, 57 lb
where Fey = Fc. d = Fc. mc'd
So By = - Ax = - 51.25 lb
is the unit vector on strut c.
M = - 227+ 38 k...

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