Problem 268 | Resultant of Non-Concurrent Force System

Problem 268
The resultant of four forces, of which three are shown in Fig. P-268, is a couple of 480 lb·ft clockwise in sense. If each square is 1 ft on a side, determine the fourth force completely.

Three forces in Planar Space


Solution 268


i have a correction sir,,, my answer is, the fourth force is 200 lb, 1 ft above O.
The four forces in the force system produces a couple of 480 lb-ft clockwise, meaning the resulatant of the four forces is 0 but it has a resultant moment of 480 lb-ft which is a couple.
Considering the figure P-268 taking moments at O clockwise positive, and the equation should be written in this way,
-200x+120(2)+110(4)=480. (assuming the missing fourth force which is 200lb left, is above O.)
solving for x.
x= 1ft positive, meaning the assumption is correct, therefore 200 lb force is 1 ft above O.


The solution is incorrect. Assuming the missing fourth force which is 200 lb left is above O, taking moments at O and taking clockwise moments as positive, the equation should be written in this way
M = -200d + 120(2) + 110(4) = 480

solving for d,,, we get d= 1ft above O. The answer is positive because the assumptions are correct

Jhun Vert
Jhun Vert's picture

Yes you are correct. The solution above will yield a 480 ft·lb couple in "counterclockwise" direction. This page is noted for revision. Thank you.

MATHalino's picture

This page has been corrected. Thank you very much kainur for calling our attention.

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