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Question

The 20 cm diameter disc in the figure can rotate on the axle through its center. What is the net torque about O (in Nm)?


A
5.06
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B
2.94
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C
0.06
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D
0.94
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Solution

The correct option is D 0.94
As we can see here, 20 N force which is passing through the center O will not produce any torque about point O
For 30 N force which is acting at 45 angle with xaxis, its x component will pass through the origin, hence its horizontal component will also not cause any torque about O.

Let represent vector going into the plane and
out of the plane
Consider direction as positive

Hence, torque because of 20 N force which is acting in tangential direction :
τ1=rF=0.1×30=3 N-m

Torque produced by y component of 30 N force acting at angle 45:
τ2=rF=0.05×30sin45
=0.05×30×12=1.52 Nm

Torque produced by 20 N force acting in negatve y direction:
τ3=r×F=0.05×20
τ=1 Nm

Hence, Net torque produced on O
τnet=τ1+τ2+τ3
τnet=3+1.52+1
τnet=0.94 Nm
or |τnet|=0.94 Nm

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