# Homework Help: Force and momentum

1. Dec 2, 2007

### myworldisorange

[SOLVED] force and momentum

this is my homework problem and me and my freinds are kinda stuck
The great limestone caverns were formed by dripping water.
The acceleration of gravity is 9:81 m=s^2:
If water droplets of 0:73 mL fall from a height of 16 m at a rate of 17 per minute, what is the average force exerted on the limestone floor by the droplets of water during a 3 min
period? Answer in units of N.

now we are on the subject of momentum and we know we have to find
F = dp/dt
however we dont know how to take the derivative of the momentum equation because we dont know where the rate of dropping fits in
any help would be greatly appreciated. to get us on the track

2. Dec 2, 2007

### dotman

Hello,

You've got all the info you need to figure out how much force a single drop is going to exert, right?

After that you need to figure out how many drops are going to fall during a 3 minute period, and you're going to need to use the rate that they drop at to figure that out.

Hope this helps.

3. Dec 2, 2007

### myworldisorange

my friend keeps trying to find velocity "how do you find velocity for m1v1 + m2v2 = v'(m1+m2)"

what I got is this
f=ma
1ml = .001kg
.73ml = 7.3e-4kg
a = 9.81
so
f = 7.3e-4 * 9.81
f = .00073
thats the force per drop
and since there are 17 drops per min
for 3 minutes there are 51 drops
so 51 * .00073
should equal the force applied by the water

srry im getting confused by rate of change and the volume =\

Last edited: Dec 2, 2007
4. Dec 2, 2007

### dotman

Hello,

Let me see if I can get you moving a little closer, you've got the right kinds of ideas here. You're on the momentum section, so you should probably be using momentum somehow to figure this out.

Firstly, you're using the volume exactly in the manner they intended-- to calculate mass to use in your equation. So that's fine. Next, you've used the rate exactly as intended, to figure out how many total drops are going to fall. Great.

Now you've got to figure out the average force applied to the floor of the cavern. What do they mean, exactly, by average force? Well, say you've got something thats not moving, and you want to accelerate it to some set speed. You've got to apply a force to do that, right? But how much? Clearly you could apply a really large force for a short amount of time (think spiking a volleyball), and that will get it moving. Or, you could apply a smaller force for a longer period of time-- it will accelerate more slowly, but the end result will be the same. If you knew what final amount of speed you wanted, you could calculate exactly how much force you would need for any given time period. Or, conversely, if you knew what force you wanted, you could figure out how long you needed to apply it.

Now, you've got some water drops hitting a floor. Clearly each one is going to impart some force on the floor. The average force applied, then, would be the sum of these small forces, over the time it takes them all to hit the floor. See, you can have the water drip down like this, or you could pour it all at once less often, or whatever; as long as its the same amount of water, from the same height, etc etc, you're going to get the same result.

Now here's the hint. There's an equation which relates momentum, force, and time. Do you know what it is? Before you even do that, consider this: what is the momentum of a single drop when it hits the floor?

5. Dec 2, 2007

### myworldisorange

yes got it
thank you very much

6. May 4, 2010

### velvet

Re: [SOLVED] force and momentum

Hello there, I have a similar question, I wonder if you can help me.

I am using scales to measure the volume of liquid coming from a bottle, and then differentiating this volume to give flow rate. The bottle emits a constant flow rate and I hold it so that the stream falls vertically. When the stream of liquid initially hits the scales, it creates a momentum artefact, i.e. a 'spike' in the flow rate which then settles down to the correct flow rate. Obviously this spike also appears at the end of the flow, but this time negative.

I am investigating the size of this artefact when the bottle is held at different heights from the scales. The area under the spike, i.e. the 'volume artefact' seems to increase with the square root of the height. (It looks like that sort of relationship but I am not sure. I believe if I plotted 'height of bottle' against 'area under artefact', the curve would tend towards becoming flat, which a √height curve does not.) I would like to come up with an equation to describe the relationship but I am not sure where to begin.

Many thanks :)