Question

A tube of uniform cross-section $A$ is bent to form a circular arc of radius $R$ forming three quarters of a circle. A liquid of density $\rho$ is formed through the tube with a linear speed $v$. Find the net force exerted by the liquid on the tube.

Answer 1

To solve this question, first we will rewrite the given numeric data and then first we will find the change in momentum to solve for the Net Force exerted by the liquid on the tube. And then finally we will conclude the Net Force when change in momentum and time is found respectively.

Complete step by step solution:
Given that:
Cross-section, $A$ ;
Circular radius $= R$
Liquid of density $= \rho$
Net force exerted by the liquid on the tube $= ?$
Change in momentum of the liquid:
$= m{v^2}$
$= density \times volume$
$\therefore m = d \times A \times {H_m}$
$= (\rho )A\dfrac{3}{4}(2\pi R)$
Therefore,
$F = \dfrac{{\text{change in momentum}}}{{\text{change in time}}}$
$\Rightarrow F = \dfrac{{\sqrt {2\pi R + \dfrac{3}{4}(\rho )A{V^2}} }}{{\dfrac{3}{4}2\pi R/V}}$
$\Rightarrow F = \dfrac{1}{{\sqrt 3 }}\rho A{V^2}$
Hence, the net force exerted by the liquid on the tube is $\dfrac{1}{{\sqrt 3 }}\rho A{V^2}$.

Note:
Remember that force is a vector, so when more than one charge exerts a force on another charge, the net force on that charge is the vector sum of the individual forces. Remember, too, that charges of the same sign exert repulsive forces on one another, while charges of opposite sign attract.