Question

If a rod of mass 0.2 kg is in equilibria in given condition then friction force acting on rod in newtons is. [Force F is acting horizontally and $g=10m/s^2$] (Take $\sqrt{3}=1.73$). Write upto two digits after the decimal point.

Answer 1

By horizontal equilibrium,
$f=F\dots \left(1\right)$

By ${\tau }_p=0\Rightarrow mg\frac{L}{2}cos30=FLsin30$

$F=f=mg\frac{\sqrt{3}}{2}=1.73$