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

Explanation

Consider a mass `Deltam` of liquid flowing across the corner in time `Delta t`. We will apply Newton's second 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 `=Delta vec(P)=vec(P)_(f)-vec(P)_(i)`
`Delta vec(P)` can be calculated by te vector subtraction, geometrically.
As `P_(i)=p_(f)implies theta=45^(@)`
`Delta P= sqrt((P_(i)^(2)+p_(f)^(2)))=sqrt(2(Deltam)^(2)V^(2))=sqrt(2)Delta mV`
Force exerted on the liquid `=(Delta P)/(Delta t)=(sqrt(2)Delta mV)/(Delta t) = sqrt(2)Vrho AV`. ltbr. Hence, the pipe must be pushed at the corners as shown in figure. with force `F=sqrt(2)rho AV^(2)` at an angle of `45^(@)`.