Water flows out of a big tank along a tube bent at right angles, the inside radius of the tube is equal to `r=0.50cm` (figure). The length of the horizontal section of the tube is equal to `l=22cm`. The water flow rate is `Q=0.50` litres per second. Find the moment of reaction forces of flowing water, acting on the tube's walls, relative to the point O.

Explanation

Let the velocity of water flowing through the tube at a certain instant of time be u, then `u=(Q)/(pir^2)`, where Q is the rate of flow of water and `pir^2` is the cross section area of the tube.
From impulse momentum theorem, for the stream of water striking the tube corner, in x-direction in the time interval `dt`,
`F_xdt=-rhoQudt` or `F_x=-rhoQu`
and similarly, `F_y=rhoQu`
Therefore, the force exerted on the water stream by the tube,
`vecF=-rhoQuveci+rhoQuvecj`
According to third law, the reaction force on the tube's wall by the stream equals `(-vecF)`
`=rhoQuveci-rhoQuvecj`.
Hence, the sought moment of force about 0 becomes
`vecN=l(-veci)xx(rhoQuveci-rhoQuvecj)=rhoQu lveck=(rhoQ^2)/(pir^2)lveck`
and `|vecN|=(rhoQ^2l)/(pir^2)=0*70N*m`