Question

Determine the following: (A) The angle formed by cables AB and AD (B) The perpendicular distance from point B to the line AD (C) If the force in cable AC is 500N, find the component of that force in the direction of line AB If the force in cable AC is 500N, find the moment of this force about a line joining BD (D) What is the perpendicular distance from point B and cable AC? (E) 0.36 m 0.56 m 0.48 m 0.52 mーーッ 0.9 m

Answer 1

Tension along A B The force is directed from A to B whose coordinates are as follows: 0 0 B0.56, 0 Therefore, the position vectors are Hence, A B 0.56 i + 0.9 j + 0 k The magnitude of the vector is obtained as The unit vector along the direction of this vector is: A BA B 0.5280849 i + 0.000 k Tension along A C The force is directed from C whose coordinates are as follows: A to 0, 0.9 0.0 .-0.48 Therefore, the position vectors are: 0.90 k - 0.48 k Hence, 0 +0.9 0.48 k The magnitude of the vector is obtained as o2 +0.9 2 A C -0.48.- 1.020 m The unit vector along the direction of this vector is: AC/A C 0.000 0.882 j - A C 0.471 k Tension along A D The force is directed from A to D whose coordinates are as follows: 0 -0.9 D 0.52, 00.36 m Therefore, the position vectors are 0.9 0 k 0+0.36 k O D-0.52 Hence -0.52 0.9 0.36 k The magnitude of the vector is obtained as

A D0.522 + o.9 2 + +0.361.100 m The unit vector along the direction of this vector is: AD: AD/AD 0.473 0.819 j + 0.328k

- cost[o-446] -63.51。 : WAR o.528 0849 - 0473 .819 0.328 c) L + 441 J-235.5 k : 374. 409 N 0.56 - 0 48 44-235.5 = 21レ68→_ 131.88j.一246.96k BDI = - 1.138 m 122 845

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