Question

The total torque about pivot $A$ provided by the forces shown in the figure, for $0L=3.0m$, is

• A. 210 N-m
• B. 140 N-m
• C. 95 N-m
• D. 75 N-m

Answer 1

Answer: (D)

Moment of force about pivot $A$ ( $80N$ force)
$=80\times \frac{3}{2}\times \sin 30^{\circ }=60N-m$ (anticlockwise)
Moment of force about pivot $A$ (70 $N$ force)
$=70\times 3\times \sin 30^{\circ }=70\times 3\times \frac{1}{2}=105N-m$ (anticlockwise)
Moment of force about pivot $A(60N$ force $)$
$=60\times \frac{3}{2}\times \sin 90^{\circ }=90N-m$ (clockwise)
Moment of force about pivot $A(90N$ force $)$
$=90\times 0\times \sin 60^{\circ }=0$
Moment of force about pivot $A(50N)$
$=50\times 3\times \sin 180^{\circ }=0$
The total torque about pivot
$AT=(60+105-90)=75N-m$