Question

A liquid of density p is flowing with a speed v through a pipe of cross-sectional area a. The pipe is bent in the shape of a right angle as shown. What force should be exerted on the pipe at the corner to keep it fixed?

Answer 1

Answer:

The force exterted is $\sqrt{2}pav^2$

Explanation:

Density of the liquid = p

Speed = v

Cross-sectional area of the pipe = a

The liquid will exert force on the pipe at the bend, this is the required force we need to calculate

Mass of the liquid flowing per second through the pipe

= density × area × velocity

= p × a × v

= pav

This is in the horizontal direction

Therefore, the momentum will be = $(pav)v\hat i$ = $pav^2\hat i$

After the bend, the same amount of liquid will be flowing in the downwards direction

Therefore the momentum will be =  $-pav^2\hat j$

Force exerted = Change in momentum

or, $\vec F= pav^2\hat i-(-pav^2\hat j)$

$\implies \vec F= pav^2\hat i+pav^2\hat j$

Thus, the magnitude of the force = $\sqrt{2}pav^2$