A liquid of density `rho` is flowing inside 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 in the two cases shown?

Explanation

(a) Consider a mass `Deltam` of liquid flowing across the corner in time `Deltat`.
We will apply Newton.s Iind law of motion to the mass `Deltam`.
`"Magnitude of initial momentum "=P_(i)=(Deltam)V`
`"Magnitude of final momentum"=P_(f)=(Deltam)V`
`"Change in momentum"=Deltabar(P)=barP_(f)-barP_(i)`
`DeltabarP` can be calculated by vector subtraction geometrically.

`"As "P_(i)=P_(f) rArr theta =45^(@)`
`DeltaP=sqrt((P_(i)^(2)+P_(f)^(2)))=sqrt(2(Deltam)^(2)V^(2))=sqrt2(Deltam)V`
The force on liquid has same direction as the direction of change in momentum of liquid. The force by lilquid on the pipe is equal and opposite to this force.
Hence the pipe must be pushed at the corners with force `sqrt2AV^(2)rho` an angle of `45^(@)` with horizontal.
The force by liquid on the pipe is equal and opposite to this force.
(b) Force exerted on the liquid
`=(Deltavecp)/(Deltat)=(2DeltamV)/(Deltat)=2Vrho" (volume per sec)"`
`=2V rho AV=2AV^(2)rho`