A) 210 N-m
B) 140 N-m
C) 95 N-m
D) 75 N-m
Correct Answer: D
Solution :
Moment of force about pivot A (80 N force) \[=80\times \frac{3}{2}\times \sin {{30}^{o}}\]= 60 N-m (anticlockwise) Moment of force about pivot A (70 N force) \[=70\times 3\times \sin {{30}^{o}}=70\times 3\times \frac{1}{2}\] = 105 N-m (anticlockwise) Moment of force about pivot A (60 N force) \[=60\times \frac{3}{2}\times \sin {{90}^{o}}\] = 90 N-m (clockwise) Moment of force about pivot A (90 N force) \[=90\times 0\times \sin \,{{60}^{o}}=0\] Moment of force about pivot A (50 N) \[=50\times 3\times \sin \,{{180}^{o}}=0\] The total torque about pivot A \[T=(60+105-90)\] = 75 N-m
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