The forces F1 11204 75100k F2 100 I 50 j 250k lb are both

The forces F_1 = ()-11204+ 75]-100k) F_2 =(100 I + 50 j - 250k) lb, are both applied at point A in the pipe arrangement shown below. Determine the resultant moment about point O.

Solution

>> F1 = - 120 i + 75 j - 100 k {lb}

and, F2 = 100 i + 50 j - 250 k {lb}

As, Coordinate of Point A, where both forces are acting = (5,3,4) ft = 5 i + 3 j + 4 k {ft}

>> So, Moment about O of these two forces = (OA) X F1 + [ (OA) X F2 ] = (OA) X (F1 + F2)

=> Mo = (5 i + 3 j + 4 k) X [ (- 120 i + 75 j - 100 k) + ( 100 i + 50 j - 250 k) ]

=> Mo =  (5 i + 3 j + 4 k) X [ - 20 i + 125 j - 350 k ]

=> Mo = 625 k + 1750 j + 60 k - 1050 i - 80 j - 500 i

=> Mo = - 1550 i + 1680 j + 685 k .........REQUIRED MOMENT ......