Please answer!! 7.7.In Examples 7.3 and 7.4 the analysis was simple becausethe jet was at right angles to the solid surface. You canobserve with a garden hose that such jets, perpendicularto walls or sidewalks, go off radially in all directions,with more or less circular symmetry. You will alsoobserve that there is a region near the jet that is muchshallower than the rest, as shown in Fig. 7.23A more interesting and complexproblem is the flow of a jet against a flat surface that isnot perpendicular to it; you can also observe this flowwith a garden hose. The flow is more or less circularlysymmetrical, but much more goes away in the directionaway from the hose than in the direction toward the hose.But why does any of it flow back in the direction towardthe hose?We can understand this if we replace the three-dimensionalproblem (circular jet, moving in x, y, and z) directions with atwo-dimensional jet (as might issue from a rectangular slot) thatis constrained to move only in the x and y directions (bydirecting it into an open rectangular channel, which preventsflow in the z direction). This flow is sketched in Fig. 7.27.Friction is assumed to be negligible, so from B.E. (ignoringgravity) we see that both streams flowing along the wall musthave the same velocity as that in the jet, V₁. In the figure thestream going off to the upper right, (2), is larger than that goingto the lower left, (3), in accord with the observation describedabove. If the flow is frictionless, then there can be no shear stresson the wall, so the resisting force must act normal to the surfaceas shown. We could attempt to solve for these flows by writingthe x and y components of the steady flow momentum balance,but that adds more terms and only makes the analysis harder (tryit!).Instead, we choose a new set of axes for our momentumbalance, with one axis, the s direction, parallel to the plate andthe other, the r direction, perpendicular to the plate, as sketchedon Fig. 7.27. We now apply Eq. 7.17 in the s direction, finding m1, V₁0 m₂V₁ cos 0 - m₂V₂ - m3V3 + F₁m₂, V₂SystemboundaryFIGURE 7.27A jet impinging on a surface not perpendicular to it. We solve inthe r-s coordinate system, instead of the x-y system.(7.65)From the assumption of frictionless flow, we can seethat F = 0, that the absolute magnitudes of V₁, V2, andV3 are the same, but that V3 is in the -s direction, so that it isequal to - V₁. Making these substitutions and dividing by V₁ wefind0 m₁ cos 0 m₂ + m3(a) Using the material balance to eliminate m3 show theequation for m2 /m₁.(b) Show the equation for the force exerted on the wall in the rdirection.

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