Question

Find the total torque about pivot A provided by the force shown in the figure, for L=3.0 m.
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Answer 1

The total torque about pivot A provided by the force shown in the figure is 75 N-m

Explanation:

Given L = 3 m

Since 90 N force passes through A, therefore, torque provided by it will be zero

The two components of 80 N force will be

$80\sin30^\circ=80\times\frac{1}{2}=40$ N, in the vertical direction

And

$80\cos30^\circ=80\times\frac{\sqrt{3}}{2}=40\sqrt{3}$ N, in the horizontal direction

Since the line of action of the horizontal component of the force passes through A, therefore, the torque due to horizontal component will be zero

Torque due to vertical component (Taking anticlockwise moment as positive)

$\tau_1=40\times \frac{3}{2}$ N-m

$\implies \tau_1=60$ N-m

Similarly the torque due to 70 N force will be only due to the vertical component

Thus,

$\tau_2=70\cos 60^\circ\times 3$

$\implies \tau_2=70\times\frac{1}{2}\times 3$ N-m

$\implies \tau_2=105$ N-m

Torque due to 50 N force will be zero because line of action of this force passes through A

Torque due to 60 N force will in in clockwise direction, hence negative

Torque due to 60 N force

$\tau_3=-60\times\frac{3}{2}$

$\implies \tau_3=-90$ N-m

Therefore the net torque

$\tau=\tau_1+\tau_2+\tau_3$

$\implies \tau=60+105-90$

$\implies \tau=75$ N-m

Hope this answer is helpful.

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Answer 2

Moment of force about pivot A (80 N force)  =80×3/2×sin30o = 60 N-m (anticlockwise) Moment of force about pivot A (70 N force)  

=70×3×sin30o=70×3×1/2  = 105 N-m (anticlockwise)

Moment of force about pivot A (60 N force)  =60×3/2×sin90o  = 90 N-m

(clockwise) Moment of force about pivot A (90 N force)  =90×0×sin60o=0  

Moment of force about pivot A (50 N)  =50×3×sin180o=0  

The total torque about pivot A  T=(60+105−90)  = 75 N-m