Calculating Average Force of a Fire Hose

[chukie]

A fire hose sprays 450 liters of water a minute onto a fire with a velocity of 17.0 m/s. Water has a density of 1 kg/liter. What is the average force acting on the fireman holding the hose due to the water?

Is it just:
F=pt
F=450*17.0*60
F=459000N

The force seems kind of big?

 

[cepheid]

Your equation is wrong (F is not equal to p*t). Let's go through this step by step:

450 L/min = 450/60 L/s = 7.5 L/s = 7.5 kg/s

If you think about it, in one second, 7.5 kg of water is pushed out at 17.0 m/s, meaning that it exits the hose with a final momentum of (7.5 kg)(17.0 m/s) = 127.5 kg*m/s = 127.5 N*s

This momentum change on that water (from 0 to 127.5) can be considered to occur over one second, and therefore, whatever force propels that 7.5 kg "chunk" of water can be considered to occur over one second (since we're just looking for the average force). The force is therefore just equal to the momentum change divided by the time over which it occurs:

(127.5 N*s) / (1 s) = 127.5 N

Since this is the force by which the rest of the water in the hose (due to the water pressure) pushes on our 7.5 kg bit of water that is exiting, it is also (by Newton's third law) the force by which that 7.5 kg bit of water pushes BACK on the hose.

I dunno...that's the only way I can think of to simplify what seems to be a rather complex problem for introductory physics.

 

[chukie]

thanks so much! 127.5 N makes a lot more sense than the big number i came up with =)