Chapter 11, Problem 20A

You hold a meterstick with the same suspended masses at the angles shown. Rank the torque needed to keep the stick steady from largest to smallest.

To determine

The rank of the torque needed to keep the stick steady from largest to smallest.

Answer to Problem 20A

  ${\tau }_B>{\tau }_D>{\tau }_A>{\tau }_C$

Explanation of Solution

Introduction:

The torque acting about an axis of rotation due to the force $F$ is given by,

  $\begin{array}{c}\tau =F\times \left(r\right)\\ =F\text{r}\sin \theta \end{array}$

Where,

  $r=$ denotes the shortest distance of the point of application of force from the axis of rotation.

  $\theta =$ angle between the force and the distance

For the given cases, weight will be same for all cases and acts in downward direction. Hence, the angle between the distance and the force for the case C and D are $90°$ .

Then, the torque can be written as,

  ${\tau }_A=F{r}_A\sin 135°=0.707\left(1\right)F=0.707F$

  $\begin{array}{l}{\tau }_B=F{r}_B\sin 125°=0.866\left(1\right)F=0.866F\\ {\tau }_C=F{r}_C\sin 90°=F\left(0.5\right)\left(1\right)=0.5F\\ {\tau }_D=F{r}_D\sin 90°=F\left(0.75\right)\left(1\right)=0.75F\end{array}$

Conclusion:

Hence, the rank is,

  ${\tau }_B>{\tau }_D>{\tau }_A>{\tau }_C$