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Bar ADB is supported by a thrust bearing at A and a vertical wire at B. Knowing that Fis vertical with a magnitude of F = 130

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Bar ADB is supported by a trust bearing at A and a vertical wire at B. Knowing that F is vertical with a magnitude of F = 130 lb and that the section of bar from D to B is L = 1.9 ft long. determine the tension in the weire and the reactions at A.

Apply conditions pf static equilibrium.

И 60

ΣMy = 0

130 \times 1.5 = T \times (1.5 + 1.9)

Wire Tension, T = 57.342

ΣFx = 0

Ax = 6

ΣFγ = 0

Ay = 0

ΣΕ = 0

A2 = 130 - T

= 130 - 57.342

A2 = 72.658 lbs

Enter the magnitude of the reaction MAX below.

ΣΜ = 0

Max = 130 x 6-T x 6

= 435.948

MAX = 435.948

MAY = 0; MA2 = 0

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