Question

A thin uniform rod AB of mass $1kg$ moves translationally with acceleration $a=2m/s^2$ due to two anti parallel forces $F_1$ & $F_2$. The distance between the points at which these forces are applied is equal to $\ell =0.2m$. If force $F_1=8N$, then find the length of the rod in meter.

• A. $2$
• B. $0.6$
• C. $5$
• D. $4$

Answer 1

The correct option is B $0.6$

$F_1+F_2=ma\Rightarrow 8+F_2=2\Rightarrow F_2=-6N$ here $-ve$ tells $F_2$ is acting in opposite direction to that of $F_1$.

As given, distance between the points of application of forces, $l=0.2m$

Lets assume Force, $F_1$ is acting $x$ (in m ) away from the centre of rod . then $F_2$ will be acting $(0.2-x)$ away from centre.

Torque, $\tau =(\overrightarrow{r}\times \overrightarrow{F})$

As rod is in purely transnational motion, Hence Total torque will be zero.

${\tau }_{total}=(x\times 8)+\left(\right(0.2+x)\times F_2=0$

On putting value of $F_2$, we have:

$8x-1.2-6x=0\Rightarrow x=0.6m$