Finding smallest forces required to produce a given moment

[Leo M]

The Given Moment at P is Mp = 10 lb-ft.

a) Corners A and D are 10” apart = 0.833 ft.

The equivalent moment is Force X Distance = Fad X 0.833 ft = 10 lb-ft.

So, smallest F = 10 lb-ft/0.833 ft. = 12 lb, applied perpendicularly to side AD.

b) Corners B and C are 11.66 inches apart = 0.97 ft.

The equivalent moment is Fbc X 0.97 ft = 10 lb-ft

So, smallest Fbc = 10/0.97 = 10.3 lb., applied perpendicularly to side BC.

c) The greatest distance between two points on the block is from A to C, 18.867 inches = 1.572 ft.

The equivalent moment is Fac X 1.572 ft = 10 lb-ft

So smallest Fac = 10/1.572 = 6.36 lb, applied perpendicularly to the line AC.

Hope this helps and is in time for you - dashed it off quickly this AM - have to run.

Good luck.