Answer
$73.97\;N$
Work Step by Step
Applying the Bernoulli’s equation between two points: one is on the water surface and another is at the pipe's opening, we obtain
$p_1+\frac{1}{2}ρv_1^2+ρgh_1=p_2+\frac{1}{2}ρv_2^2+ρgh_2$
where, we take the level of the hole as our reference level.
Therefore, $h_1=d=6\;m$, $h_2=0\;m$ and $v_1=v_2=0\;m/s$
or, $p_1+ρgh_1=p_2$
or, $p_2-p_1=ρgh_1$
or, $\Delta p=ρgd$
or, $\Delta p=1000\times9.81\times6\;Pa$
or, $\Delta p=58860\;Pa$
The magnitude of the force acting on the plug due to pressure difference is
$F=\Delta pA$
or, $F=\Delta p\frac{\pi d^2}{4}$
Substituting the given values
$F=58860\times\frac{\pi \times 0.04^2}{4}\;N$
or, $\boxed{F=73.97\;N}$
At static condition, the magnitude of the frictional force between plug and pipe wall is equals to the the magnitude of the force acting on the plug due to pressure difference.
Therefore, the magnitude of the frictional force between plug and pipe wall is $73.97\;N$
