The question says Suppose that F 49lb Determine the magnitu

The question says:

Suppose that F = 49lb. Determine the magnitude of the resultant force.

2 201b 4 ft 2 . 20 3 ft

Solution

Positon co-ordinates of A

Ax = 0, Ay = -2, Az = 4

Position vector A

A^ = -2y^ +4 z^

Position co-ordinates of B

Bx = 1.5, By = -3, Bz =0

Position vector B

B^ = 1.5x^ -3 y^

Position co-ordinates of C

Cx = -2Sin20⁰ = - 0.68 , Cy = 2Cos20⁰ = 1.88,   Cz =0

Position vector C

C^ = -0.68x^ +1.88y^

Force F is in the direction of AC^

AC^ = C^ -A^ = (-0.68x^ +1.88y^) -(-2y^ +4 z^ )

                   = -0.68x^ + 3.88 y^ - 4z^

|AC| = sqrt(0.68² +3.88² + 4²) = 5.61

Unit vector along AC = AC^/5.61 = -0.03 x^ + 0.69 y^ - 0.71 z^

F^ = 49*(0.03 x^ - 0.69 y^ + 0.71 z^) = -1.47x^ +33.81 y^ - 34.94z^

AB^ = B^ - A^ = (1.5x^ -3 y^) - (-2y^ +4 z^ ) = 1.5x^ - y^ - 4z^

|AB| = sqrt(1.5² + 1² + 4² ) = 4.27

   Force along AB = (20/4.27)( +1.5x^ - y^ - 4z^)

                             = 7.03x^ - 4.68 y^ - 18.74 z^

Resultant of 20 lb and F= 49 lb

= (-1.47x^ +33.81 y^ - 34.94z^) + (7.03x^ - 4.68 y^ - 18.74 z^)

= 5.56x^ -29.13 y^ - 53.68z^

Magnitude of the resultant = sqrt( 5.56² + 29.13² + 53.68² ) = 61.33 lb