1 if there is a 1200N force applied at point B pointing in t

1. if there is a 1200N force applied at point B pointing in the negative z direction, in addition to the force shown, calculate the moment about point O.

Solution

>> Let Force, F is acing at point C

Now, tak, g O as reference, writing all the coordinates :

>> A = (280, 350, 0) mm = (0.28, 0.35, 0) m

>> B = (0, 650, 0) mm = (0, 0.65, 0) m

>> C = (0, 650, 530) mm = (0, 0.65, 0.53) m

>> Now, as F = 3000 N isacting along CA

As,CA = (0.28 i + 0.35 j ) - (0.65 j + 0.53 k)

=> CA = 0.28 i - 0.30 j - 0.53 k

Magnitude = [ 0.28² + 0.30² + 0.53² ]^1/2 = 0.6703

So, Unit Vector along CA = (0.28 i - 0.30 j - 0.53 k)/0.6703 = 0.418 i - 0.448 j - 0.791 k

So, Force, F = 3000(0.418 i - 0.448 j - 0.791 k) = 1254 i - 1344 j - 2373 k

>> Also, Force, F1 = 1200 N and is along negative z - direction

So, F1 = - 1200 k

>> So, Moment about O, Mo = ( OA X F ) + ( OB X F1 )

=> Mo = [ (0.28 i + 0.35 j) X (1254 i - 1344 j - 2373 k) ] + [ 0.65 j X - 1200 k ]

=> Mo = - 376.32 k + 664.44 j - 438.9 k - 830.55 i - 780 i

=> Mo = Required Moment = - 1530.55 i + 664.44 j - 815.22 k