Question

Calculate the net torque on the system about the point $\text{O}$ as shown in figure if $F_1=11\text{N},F_2=9\text{N},F_3=10\text{N},a=10\text{cm}$ and $b=20\text{cm}$ (All the forces are along the tangents)

• A. $2.2\text{Nm}$ out of the plane
• B. $3.0\text{Nm}$ into the plane
• C. $2.2\text{Nm}$ into the plane
• D. $3.0\text{Nm}$ out of the plane

Answer 1

The correct option is B $3.0\text{Nm}$ into the plane


As we know, $\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}=r_{\perp }F$
Torque because of force $F_1$ at origin,
$\overrightarrow{{\tau }_1}=r_{\perp }\overrightarrow{F_1}=b.F_1=0.2\times 11=2.2\text{Nm}$ (direction into the plane)

Torque because of force $F_2$ at origin,
$\overrightarrow{{\tau }_2}=r_{\perp }.F_2=b.F_2=0.2\times 9=1.8\text{Nm}$ (direction into the plane)

Torque because of force $F_3$ at origin,
$\overrightarrow{{\tau }_3}=r_{\perp }.F_3=a.F_3=0.1\times 10=1\text{Nm}$ (direction out of the plane)

Hence net torque (assume direction of torque into the plane as positive)
$\overrightarrow{\tau }=\overrightarrow{{\tau }_1}+\overrightarrow{{\tau }_2}+\overrightarrow{{\tau }_3}$
i.e $\tau =2.2+1.8+(-1)=3.0\text{Nm}$ (into the plane)