The problem shows three pairs of couples, which would require three moments. The resultant moment is then added (according to the rotation) from those three moments.
The moment of each couple can be resolved using:
Moment of the 400N couple:
The moment between this couple requires the force of 400N multiplied by the distance between the forces which is 2m.
=-800N-m "\mathbf{M}_{400N}=-400N\left ( 2m \right )=-800N-m")
The negative sign signifies the counterclockwise rotation of the couple (using the sign convention where +M is for clockwise).
Moment of the 200N couple:
The moment between this couple requires the force of 400N multiplied by the distance between the forces which is 2m.
The negative sign signifies the counterclockwise rotation of the couple (using the sign convention where +M is for clockwise).
Moment of the 200N couple:
Moment of the 300N couple:
Resultant Moment:
Back to 6 COUPLES AND FORCE-COUPLE SYSTEMS
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