Three forces at on a wheel of radius 20cm as shown in figure.

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The net torque produced by the forces is 1.8Nm clockwise

The torque produced by the force of 4N is 0.8Nm anticlockwise

The torque produced by the force of 9N is zero

The torque produced by the force of 8N is 0.8Nm clockwise

answer is A, B, D.

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Detailed Solution

The torques produced by forces 8N,4N and 9N respectively are

$τ_{1} = 8 N \times 0 .2 m \times \sin 30^{\circ} = 0 .8 Nm (clockwise)$
$\begin{array}{l} τ_{2} = 4 N \times 0 .2 m \times \sin 90^{\circ} = 0 .8 Nm (anticlockwise) \\ τ_{3} = 9 N \times 0 .2 m \times \sin 90^{\circ} = 1 .8 Nm (clockwise) \end{array}$
$Net torque = 0 .8 - 0 .8 + 1 .8 = 1 .8 Nm clockwise$

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