Question

4 m 8. Two cables AB &AC hold the 6 m shown pole in the figure Determine the resultant force & the angles and ф F2-35 N Im 3 m 2 m
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Answer #1

Coordinate of O (0,0,0)

Coordinate of A (0,4,6)
Coordinate of B (4,5,0)
Coordinate of C (-2,8,0)

vector AB = 4i+j-6k (direction of F1)

unit vector AB^ = 0.5494i+0.13736j-0.82416k

F1 = 35(0.5494i+0.13736j-0.82416k) = 19.229i+4.807j-28.844k

vector AC = -2i+4j-6k (direction of F2)

unit vector AC^ = -0.26726i+0.5345j-0.80178k

F2 = 50(-0.26726i+0.5345j-0.80178k) = -13.363i+26.725j-40.089k

Resultant force, F = F1 + F2 = 5.866i+31.532j-68.933k
magnitude = 76.029
AO = -4j-6k
AO.AB = AO AB cos theta
32 = 7.211*7.28 cos theta
theta = 52.44 degrees

AO.AC = AO AC cos phi
20 = 7.211*7.4833 cos phi
phi = 68.245 degrees

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