Question

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

• A. $5.06$
• B. $2.94$
• C. $0.06$
• D. $0.94$

Answer 1

The correct option is D $0.94$

As we can see here, $20\text{N}$ force which is passing through the center $\text{O}$ will not produce any torque about point $\text{O}$
For $30\text{N}$ force which is acting at ${45}^{\circ }$ angle with $x-$axis, its $x-$ component will pass through the origin, hence its horizontal component will also not cause any torque about $\text{O}$.

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

Hence, torque because of $20\text{N}$ force which is acting in tangential direction :
${\tau }_1=r_{\perp }F=0.1\times 30=3\text{N-m}$ $⨂$

Torque produced by y component of $30\text{N}$ force acting at angle ${45}^{\circ }$:
${\tau }_2=r_{\perp }F=0.05\times 30\sin {45}^{\circ }$
$=0.05\times 30\times \frac{1}{\sqrt{2}}=\frac{1.5}{\sqrt{2}}\text{Nm}$ $⨀$

Torque produced by $20\text{N}$ force acting in negatve y direction:
${\tau }_3=r_{\perp }\times F=0.05\times 20$
$\tau =1\text{Nm}$$⨀$

Hence, Net torque produced on $\text{O}$
${\tau }_{\text{net}}={\tau }_1+{\tau }_2+{\tau }_3$
${\tau }_{\text{net}}=-3+\frac{1.5}{\sqrt{2}}+1$
${\tau }_{\text{net}}=-0.94\text{Nm}$
or $\left|{\tau }_{\text{net}}\right|=0.94\text{Nm}$