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 $=6\times \frac{20}{100}\times sin0^\circ =0$
$\therefore \tau=\left(4 \times 0 . 2 - 8 \times 0 . 2 \times \frac{1}{2} - 9 \times 0 . 2\right)=-1.8N-m$
$=1.8N-m$ (clockwise)