Determine the moment about the origin O 077i 332k 299i 077j

Determine the moment about the origin O. 0.77i -33.2k 29.9i + 0.77j 28.3i - 2.49j 28.3i + 249j -0.77i + 33.2k

Assume the Tails of the Force Vectors is the Point F

Head of Upward Force is Point E Head o Downward orce is Point C.

Point D is the intersection of X axis and the dotted line

Position Vector along FE is FE = OD i – DB k + BE j = 5i + 7 j -3k

Unit vector along FE is = (5i + 7 j -3k )/sqrt(5^2+7^2+(-3)^2 = (5i + 7 j -3k )/9.11

And along FC Vector FC = OD i – OF j +DC k = 5i -6j +5k

Unit vector along FC is = (5i -6j +5k)/sqrt(5^2+(-6)^2+5^2 = (5i -6j +5k)/9.27

Now Moment about Origin can be calculated by adding each moments of forces present.

M1= Position Vector of OF X Force Vector along FE

M2 = Position Vector of OF X Force Vector along FC

Now we have

Position Vector of OF = 6j

Force Vector along FE = Magnitude of Force along FE .Unit Vector along FE

Force Vector along FE = 4x(5i + 7 j -3k )/9.11 N = 2.19i+3.07j-1.31k

Force Vector along FC = Magnitude of Force along FC .Unit Vector along FC

Force Vector along FC = 4x(5i -6j +5k)/9.27 N = 2.15i-2.58j+2.15k

Then M1= 6j X (2.19i+3.07j-1.31k)

M1= -13.14k – 7.8i

M2 = 6j X (2.15i-2.58j+2.15k)

M2= -12.9k+12.9i

Then Resultant Moment = M1+M2 = = -13.14k – 7.8i +(-12.9k+12.9i)

M= -26.04k +5.1i

M= 5.1i – 26.04 k

Magnitude of the resultant Momentum = sqrt(5.1^2+(-26.04)^2 = 26.53 Nm