How Do You Calculate the Holding Force of a Water Nozzle?

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To calculate the holding force of a water nozzle with a radius of 25mm and a flow rate of 750L/min, the fluid velocity is determined to be approximately 6.366 m/s. The challenge lies in calculating the pressure needed to derive the force, which is ultimately given as 79N, corresponding to a pressure of about 40,000 pascals. The discussion highlights the use of various fluid dynamics equations, including the impulse-momentum equation, to solve for the required force. Despite initial difficulties with variable management in the equations, clarity is achieved through community input. Understanding these principles is crucial for accurately calculating the holding force of a water nozzle.
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Homework Statement


A nozzle has a radius of 25mm. Water emerges at the rate of 750L/min. Find the force with which the nozzle must be held.

Homework Equations


F = m \times a
\Delta m / \Delta t = pAv
A_1 \times v_1 = A_2 \times v_2
R = vA

The Attempt at a Solution


I've so far been able to calculate fluid velocity: v = r/A = 6.366m/s. However, finding pressure in order to find force eludes me. I've tried looking into different units, trying the above formulae (but there's always one variable somehow dangling somewhere), and so on.

The answer is given to be 79N, so I do know that pressure is equal to about 40000 pascals. This, of course, doesn't help much.

Any help is very much appreciated. :redface:
 
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Try looking at a variation of the impulse-momentum equation, Ft = m delta v
 
Thank you verily much for your input. I completely understand now. (:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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