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

 
kainur

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.

kainur

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
MATHalino's picture

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

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