A horizontal water jet with a velocity of 10 m/s and cross-sectional area of 10 mm2 strikes a fixed plate held normal to the flow direction. Find the total force acting on the plate. (Take density of water, =1000 kg/m3).
Answers
Answer:
The velocity of water jet, v = 10 m/s
Area of cross-section of the water jet, A = 10 mm² = 10⁻⁵ m²
Since the plate is held fixed normal to the flow direction, therefore, the velocity of the plate “u” = 0.
Density of water is given as = 1000 kg/m³
Now,
The force exerted by the water jet on to the plate is given as,
F = ρ * A * [v – u]²
where ρ = density of water
Substituting the given values, we get
F = 1000 * 10⁻⁵ * [10 – 0]² = 1000 * 10⁻⁵ * 100 = 1 N
Thus, the total force acting on the plate is 1 N.
The total force acting on the plate is 1 newton.
Explanation:
The total force acting on the plate is given by the formula:
F = ρ × A × [v – u]²
Where,
ρ = Density of water = 1000 kg/m³
A = Cross-sectional area of water jet = 10 mm² = 10 × 10⁻⁶ m²
v = Velocity of water = 10 m/s
u = Velocity of plate = 0 m/s (fixed plate)
On substituting the values, we get,
F = 1000 × 10 × 10⁻⁶ × (10 - 0)²
F = 1000 × 10 × 10⁻⁶ × 100
∴ F = 1 N