Question

Uniform metre rule of weight $2.0N$ is pivoted at $60cm$ mark. A $4.0N$ weight is suspended from one end, causing the rule to rotate about the pivot.

At the instant when rule is horizontal, what is the resultant turning moment about the pivot?

• A. zero
• B. $1.4Nm$
• C. $1.6Nm$
• D. $1.8Nm$

Answer 1

The correct option is B $1.4Nm$

$\begin{array}{c}\text{Mass of scale on the left portion}\\ \begin{array}{c}{m}_{\text{Left}}=0.12kg\\ \text{weight on left side}=1.2N\end{array}\end{array}$

$\begin{array}{c}\text{Similarly-}\\ M_{\text{Right}}=0.08kg\\ \text{weight on Right side}=0.8N\\ \text{torque moment clock wise -}\\ \begin{array}{cc}{\tau }_{\text{clockwise}}& =(0.8N\times 0.2m)+(4N\times 0.4m)\\ & =1.76Nm\end{array}\end{array}$

$\begin{array}{c}\text{Torque moment anti-clockwise -}\\ \begin{array}{cc}{\tau }_{\text{anticlock}}=& 1.2N\times 0.3m=0.36Nm.\\ \text{Net torque is}& ={\tau }_{\text{clock}}-{\tau }_{\text{anticlock}}\\ {\tau }_{\text{total}}& =1.76Nm-0.36Nm\\ & =1.4Nm\end{array}\\ \text{Hence, option (B) is correct.}\end{array}$