A 4 metre long bar rotates freely around a central pivot. 3 forces act upon it: 7N down, 2m to the left of the pivot; 8N up, 1m to the left of the pivot; 4N up, 1m to the right of the pivot. Apply one additional force to place the bar in equilibrium.
Begin by calculating the current force and moment of the bar.Force: 8+4-7=5 (or vice versa)Moment: 72= 14 Nm anticlockwise 81= 8 Nm clockwise 4*1= 4 Nm anticlockwiseSum calculated moments: 14+4-8=10 (or vice versa, again)Therefore a force of 5N down, placed to produce a 10Nm clockwise moment is required; a force of 5N down, located 2m to the right of the pivot.
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