# Force Exerted by Jet an Inclined Plate - Ishwaranand

## Force Exerted by a Jet an Inclined Flat Plate.

Let
θ = Angle inside between the jet and plate,
A = Area of a cross-section of the jet. Image Source ~ Crafted With ©Ishwaranand - 2020 ~ Image by ©Ishwaranand

Fig. Jet striking stationary sloping plate
• If the plate is smooth and if it is assumed that there is no loss of energy due to impact of the jet, then the jet will move over the plate after striking with a velocity equal to initial velocity i.e., with a velocity v.
• Let us find the force exerted by the jet on the plate in the direction normal to the plate.
• Let this force is represented by Fn
Then
Fn = Mass of jet striking per sec. x [ Initial velocity of the jet before striking in the way from n - The final velocity of jet after striking in the way from n]
= ρav [v sinθ - 0]
= ρav^2 sinθ
Then we have,
Fx = Parts of Fn in the direction of flow.
= Fn cos (90° - θ)
= Fn sinθ
= ρav2 sinθ x sinθ
( ∵ Fn = ρav^2 sinθ )
= ρav^2 sin^2θ
= Fn sin(90° - θ)
= Fn cosθ
= ρav^2 sinθ cosθ
Q. An jet of water of diameter 75 mm moving by a velocity of 25 m/s strikes a fixed plate in such a direction that the angle within the jet and plate is 60°. Determine the force exerted by the jet toward the plate
Given
A diameter of a jet, d = 75 mm = 0.075 m
Area, a = (π/4)d2
= (0.075)^2
= 0.004417 m^2
Velocity of jet, v = 25 m/s
The angle between jet and plate θ = 60°
Solution
Fn= ρav^2 sinθ
= 1000 x 0.004417 x 25^2 x sin(60)
= 2390.7 N
=2.39 kN
Fx = ρav^2 sin^2θ
= 1000 x 0.004417 x 25^2 x sin^2(60)
= 2070.4 N
= 2.070 kN