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hi guys, i was wondering if you could help, I'm really stuck on this question, a water hose ejects water in a fine jet at a rate of 250 litres per minute and at a speed of 26m/s. The density of water is 1000kg per metre cubed. what is the force required to hold the hose steady? any help would be very much appreciated

Forward Momentum (Water) = Mass * Velocity

Volume Ejected In m^-3 (Per Second) = (250 * 0.001)/60

Per Second: Mass Ejected (Water) = Density * Volume = 1000 * [(250*0.001)/60] = 0.0041666666 kg

Therefore: Forward Momentum (Water) = 0.0041666666 * 26 = 0.1083 kgms^-2

As initial Momentum was zero (stationary event), this is now a recoil situation. Hence:

Magnitude Of Momentum (Hose) = 0.1083 kgms^-2 (Acting In Opposite Direction To That Of Water Flow).

To keep the hose steady, the resultant momentum must be zero, hence a momentum of 0.1083 kgms^-2 acting in the same direction as that of which the water flows (forward), is required. This is to keep the hose steady by equally matching the backwards momentum that the hose otherwise experiences.

So: The force (momentum) required to keep the hose steady is 0.1083 kgms^-2, or 1.083 Ns. (Whichever unit you prefer).

Hope that helps, if it is correct.

Volume Ejected In m^-3 (Per Second) = (250 * 0.001)/60

Per Second: Mass Ejected (Water) = Density * Volume = 1000 * [(250*0.001)/60] = 0.0041666666 kg

Therefore: Forward Momentum (Water) = 0.0041666666 * 26 = 0.1083 kgms^-2

As initial Momentum was zero (stationary event), this is now a recoil situation. Hence:

Magnitude Of Momentum (Hose) = 0.1083 kgms^-2 (Acting In Opposite Direction To That Of Water Flow).

To keep the hose steady, the resultant momentum must be zero, hence a momentum of 0.1083 kgms^-2 acting in the same direction as that of which the water flows (forward), is required. This is to keep the hose steady by equally matching the backwards momentum that the hose otherwise experiences.

So: The force (momentum) required to keep the hose steady is 0.1083 kgms^-2, or 1.083 Ns. (Whichever unit you prefer).

Hope that helps, if it is correct.

thanks a lot, when I'd previously done it i got 108.3, so I'm guessing yours is right

I got 108.3 as well :\

250 litres is 250 000 ml, which is 250 000 cm^3, which is 0.25 m^3, which is 250 kg.

250 kg of water per minute is 25/6 kg of water per second

and this water is being shot out at 26 ms^-1, so the rate of change of momentum is (25/6)(26) kg m s^-2 - this is the force (force = rate of change of momentum)

250 litres is 250 000 ml, which is 250 000 cm^3, which is 0.25 m^3, which is 250 kg.

250 kg of water per minute is 25/6 kg of water per second

and this water is being shot out at 26 ms^-1, so the rate of change of momentum is (25/6)(26) kg m s^-2 - this is the force (force = rate of change of momentum)

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