Determine the moment of forces:
The side of a square ABCD is 1.60 m long .Four forces equal to 6, 5, 4 and 8 N acts respectively along with the line CB, BA, DA, DB. Determine the moment of these forces w.r.t. O, the point of intersection of the diagonals of the square.
Solution
Taking moments w.r.t. O,
Resultant moment M_{o} = - 6 x_{2} - 5 x_{3} + 4 x_{1} + 8 × 0
Where x_{1}, x_{2} and x_{3} are the perpendicular distances of the forces 4, 6, and 5 N, respectively from O, and force of 8 N has zero moment around O as its line of action passes out through this point.
Here, x_{1} = x_{2} = x_{3} = 1.6/2 = 0.8 m
∴ M _{o} = - 6 (0.8) - 5 (0.8) + 4 (0.8) + 0
= - 4.8 - 4.0 + 3.2
= - 5.6 N-m or 5.6 N-m (clockwise)