SSC JE Civil Engineering 30-Oct-2020 Shift 2 Paper

Concept:
${\mathrm{V}}_{\mathrm{r}}$ = relative velocity of the jet with respect to the plate
● When the plate is moving (u) in the direction of the jet (V) i.e. away from the jet: ${\mathrm{V}}_{\mathrm{r}}$ = V - U
● When the plate is moving (U) in the opposite direction of the jet (V) i.e. towards the jet: ${\mathrm{V}}_{\mathrm{r}}$ = V + U
Force exerted by the jet on the plate in the direction normal to the plate:
${\mathrm{F}}_{\mathrm{n}}=\mathrm{\rho }\mathrm{A}{\mathrm{V}}_{\mathrm{r}}^2\sin \mathrm{\theta }$
Force exerted by the jet on the plate in the direction of the jet ( ${\mathrm{F}}_{\mathrm{x}}$ ) and is perpendicular to the direction of flow ( ${\mathrm{F}}_{\mathrm{y}}$ )
${\mathrm{F}}_{\mathrm{x}}$ $={\mathrm{F}}_{\mathrm{n}}\sin \mathrm{\theta }$ $=\mathrm{\rho }{\mathrm{AV}}_{\mathrm{r}}^2{\sin }^2\mathrm{\theta }$
${\mathrm{F}}_{\mathrm{y}}$ $={\mathrm{F}}_{\mathrm{n}}\cos \mathrm{\theta }$ $=\mathrm{\rho }{\mathrm{AV}}_{\mathrm{r}}^2\sin \mathrm{\theta }\cos \mathrm{\theta }$
Calculation:
Given, velocity of jet = 20 m/s
Angle between jet and plate = 60° and area of cross-section of jet = 0.01 ${\mathrm{m}}^2$
Flate plate is stationary, Then
Relative velocity of the jet with respect to the plate( ${\mathrm{V}}_{\mathrm{r}}$ ) = 20 - 0 = 20 m/s
The perpendicular force to flow exerted by a jet on a stationary inclined flat plate,
${\mathrm{F}}_{\mathrm{y}}$ $=$ $\mathrm{\rho }\mathrm{A}{\mathrm{V}}_{\mathrm{r}}^2\sin \mathrm{\theta }\cos \mathrm{\theta }$
${\mathrm{F}}_{\mathrm{y}}$ $=$ $1000\times 0.01\times 202\times \sin 60\times \cos 60$
${\mathrm{F}}_{\mathrm{y}}$ = 1732.05 N.