The force required to hold a water hose

Calculate the force required to hold a water hose of radius 0.04 m when it emits a jet of water at a speed of 20 m/s.

Solution.

During time Δt the hose emits a tube of water and the tube length can be calculated using

l = vΔt,

where l – the length, v – the velocity of the jet of water and Δt – the time interval.

Also, the tube’s area of cross-section A is given by equation

A = πr^2.

The mass m of water in the tube is

m = ρV,

where ρ – the density of water (1000 kg/m^3) and V is the volume of the tube of water

which equals A*l or πr^2*vΔt.

Hence, the mass m of water in the tube is ρ*πr^2*vΔt

The mass loss per second is

dm/dt = ρ*πr^2*vΔt/Δt = ρ*πr^2*v.

Since the force F = d(mv)/dt

or

F = m*dv/dt + v*dm/dt.

The velocity v is constant then dv/dt = 0 and

F = v*dm/dt.

Therefore,

F = v*dm/dt = v*ρ*πr^2*v = ρπ*(vr)^2.

Now we can sabstitute the values and calculate the force

F = 1000*π*(20*0.04)^2 ≈ 2010.62 N

 
 

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