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System of parallel forces:
A rigid body is subjected to system of parallel forces as shown in the figure. Reduce this system to,
(i) A single force system
(ii) A single force moment system at point B
Sol.: It is equivalent force system
(Acting downward)
R = 15 - 60 + 10 -25 = -60 N
Now taking moment about A, apply varignon's theorem
R.X = 60 × 0.4 - 10 × 0.7 + 25 × 1.4
60.X = 52, × = 0.867 m Where X is distance of resultant from point A. (1) A single force be 60 N which is acting downward
(2) Now a force of 60 N = A force of 60 N (down) at point B
and anticlockwise moment of 60 × (1.4 - 0.866) = 31.98 Nm at B.
60 N force and 31.98 Nm moment anticlockwise .......ANS
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how is varignons theorem useful in engineering mechanics
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