Question

A horizontal jet of water is made to hit a vertical wall with a negligible rebound. If the speed of water from the jet is 'v', the diameter of the jet is 'd' and the density of water is 'p' , then what is the force exerted on the wall by the jet of water?

Answer 1

Answer:

F = π/4 ρ d² v²

Explanation:

Force is mass times acceleration:

F = ma

Acceleration is change in velocity over change in time:

F = m Δv / Δt

Since there's no rebound, Δv = v.

F = m v / Δt

Mass is density times volume:

F = ρ V v / Δt

Volume per time is flow rate:

F = ρ Q v

Flow rate is velocity times cross sectional area:

F = ρ (v A) v

F = ρ A v²

Area of a circle is pi times radius squared, or pi/4 times diameter squared:

F = ρ (π/4 d²) v²

F = π/4 ρ d² v²