Question

A wheel of radius $20cm$ has forces applied to it as shown in the figure. The net torque produced by the forces $4N$ at $A,8N$ at $B,6N$ at $C$ and $9N$ at $D$ at angles indicated is :-

• A. $5.4N-m$ anticlockwise
• B. $1.80N-m$ clockwise
• C. $2.0N-m$ clockwise
• D. $5.4N-m$ clockwise

Answer 1

Answer: (B)

Moment of $4N$ force
$=4\times \frac{20}{100}N-m$ (anticlockwise)
Moment of $8N$ force
$=8\times \frac{20}{100}\times sin30^{\circ }N-m$ (clockwise)
Moment of $9N$ force $=9\times \frac{20}{100}N-m$ (clockwise)
Moment of $6N$ force $=6\times \frac{20}{100}\times sin{0}^{\circ }=0$
$\therefore \tau =\left(4\times 0.2-8\times 0.2\times \frac{1}{2}-9\times 0.2\right)=-1.8N-m$
$=1.8N-m$ (clockwise)