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Q.
The total torque about pivot $A$ provided by the forces shown in the figure, for $0L = 3.0\, m$, is
AMUAMU 2012System of Particles and Rotational Motion
Solution:
Moment of force about pivot $A$ ( $80\, N$ force)
$=80 \times \frac{3}{2} \times \sin 30^{\circ}=60\, N - m$ (anticlockwise)
Moment of force about pivot $A$ (70 $N$ force)
$=70 \times 3 \times \sin 30^{\circ}=70 \times 3 \times \frac{1}{2}=105\, N - m$ (anticlockwise)
Moment of force about pivot $A (60\, N$ force $)$
$=60 \times \frac{3}{2} \times \sin 90^{\circ}=90 N - m$ (clockwise)
Moment of force about pivot $A (90 N$ force $)$
$=90 \times 0 \times \sin 60^{\circ}=0$
Moment of force about pivot $A (50\, N )$
$=50 \times 3 \times \sin 180^{\circ}=0$
The total torque about pivot
$A T=(60+105-90)=75\,N - m$
