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posted by  Linzy on 11/4/2009 10:34:23 AM  |  status: Live  |  Earned Karma: 50

statics

Course Textbook Chapter Problem Needs by
N/A N/A N/A N/A 11/9/2009 at 4:00:00 PM
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Linzy Franks lcb8@student.uwf.edu
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posted by walken on 11/4/2009 12:09:21 PM  |  status: Live
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creating a free body diagram will assist you in solving the problem
First thing we need to do is sum the forces in the y direction.  (given: FAy = FBy)
ΣFy = 0 = -1800N/m(a) – (600N/m) (4-a) + 2(FAy)

When the forces in the y direction are summed we have 2 unknowns (a, FAy) so we will need a second equation. Let’s sum the moments about point A.

ΣMA= 0 = 1800N/m(a)(2/3)(a) + (600N/m)(4-a)(2/3)(4+a) – 4m(FBy)

Now we have two equations and two unknowns. Solve the system of equations.

Solving these two equations will output two possible answers,

1.      a = 1m, FAy = FBy = 1800 N

2.       a = 2m, FAy = FBy = 2400 N

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