Calculating Force on Windscreen from Water Spray

[kalupahana]

A pipe with a mouth 125cm² of area, spraying water with a speed of 25cm s^-1 in a Service Station. This water perpendicularly fall on to a wind screen of a car.
Find the Force which act on wind screen by water.
[Note:- consider that velocity of water become 0 after it hits the wind screen, that is mv₂=0]

The answer is given as 0.75N, I cannot to reach to this, please help me.

 

[collinsmark]

Hello Kalupahana,

Welcome to Physics Forums!

You'll have to show your attempted solution, and any relevant equations before we can help. For starters, in your relevant equations, how is momentum defined? What is the relationship between force and momentum?

Also, could you double check the problem (and given answer) and make sure you wrote it in correctly? A few things about your problem statement seem suspicious. Water moving at 25 cm/sec isn't really "spraying." That's more of a trickle. A pipe that has an area of 125 cm² means that it has a diameter of around a foot. That's a huge pipe. So you've described a huge pipe that's just trickling out a whole bunch of water (but not really "spraying" it per-se). And the answer, is that in units of Newtons (N) or kN? And are you sure the numerical value is 0.75 instead of 0.78?

[Edit: Oh, and are there any changes in height involved (such as differences between the height of the pipe and the height of the windscreen) that are mentioned in the problem?]

 

[kalupahana]

I also have a problem it practically can be happen. But question is given like that. No height between pipe & wind screen. The answer come in cubic meters (Dimensions L³T^-1, Units m³s^-1).
The answer also given like 0.75N. It's in Newtons.

How to get a relation with those things with force.

 

[collinsmark]

I also have a problem it practically can be happen. But question is given like that. No height between pipe & wind screen. The answer come in cubic meters (Dimensions L³T^-1, Units m³s^-1).
The answer also given like 0.75N. It's in Newtons.

Well, it is possible that your textbook is giving you the wrong answer. It's rare, but I have seen it happen before.

Anyway, go ahead and list your relevant equations, and show us your attempted solution. Then I(we) might be able to point you in the right direction and/or compare your answer with mine(ours).

How to get a relation with those things with force.

Try looking in your textbook for a relationship between force and momentum. This relationship is the key to solving this problem.

 

[kalupahana]

kk i'll try

 

[kalupahana]

The area of mouth of pipe = 0.012 m².
The speed of moving water = 0.25 ms^-1
So flow rate = 0.012 x 0.25 = 0.003 m³s^-1

I don't how to combine this with force. It;s not coming to my memory.

The mass can be taken by eq of density, bt it's bot given in the question.
[density of water = 1000 kg m^-3]
d = m/v
so m = 3kg

Then momentum can be taken by as
mv = 3 x 0.25 = 0.75 kg ms^-1

NOw what to do?

F=d(mv)/dt
F=ma

The water beam become 0 after stuck on wind screen.

F = 0.75/1
= 0.75 N

Can I use time as 1sec in here.

I was reached to answer like this.

 

[collinsmark]

The area of mouth of pipe = 0.012 m².
The speed of moving water = 0.25 ms^-1

Okay, but earlier you mentioned the area of the mouth of the pipe was 125 cm² = 0.0125 m². It makes a significant difference in this problem.

So flow rate = 0.012 x 0.25 = 0.003 m³s^-1

Okay, but if we go with the original 125 cm², the flow rate is 0.003125 m³/s.

I don't how to combine this with force. It;s not coming to my memory.

The mass can be taken by eq of density, bt it's bot given in the question.
[density of water = 1000 kg m^-3]
d = m/v
so m = 3kg

So close! But what you've calculated is

(1000 [kg/m³])(0.003 [m³/sec]) = 3 [kg/s]

The units are in kg/s, not kg.

That is the change in mass per unit time. In other words,

dm/dt = 3 kg/s

(If we use the original 125 cm² for the mouth area, the flow rate turns out to be dm/dt = 3.125 kg/sec)

Then momentum can be taken by as
mv = 3 x 0.25 = 0.75 kg ms^-1

Multiplying those is not really the momentum. It's the change in momentum per unit time.

(dm/dt)v = d(mv)/dt = (3 [kg/s])(0.25 [m/s]) = 0.75 [kg(m/s²)]

(Or if we use the original 125 cm² for the mouth area, the flow rate turns out to be dm/dt = (3.125 [kg/s])(0.25 [m/s]) = 0.781 [kg(m/s²)])

NOw what to do?

F=d(mv)/dt
F=ma

The water beam become 0 after stuck on wind screen.

F = 0.75/1
= 0.75 N

Can I use time as 1sec in here.

The equation in red was the one I was looking for; there 'ya go!

But there is no reason you need to divide by 1 sec. The division by unit time is already present in the above equations (you forgot to include it in your rate of mass flow). But you have already found that
(dm/dt)v = d(mv)/dt = (3 [kg/s])(0.25 [m/s]) = 0.75 [kg(m/s²)]

And you know that F = d(mv)/dt, so you've already found the answer!

And take a look at the units. [kg(m/s²)] = [N]

So you answer is 0.75 N.

However, if we use the original 125 cm² for the mouth area, the answer comes out to be 0.781 N. That is what threw me originally.

If your book/coursework actually has 125 cm² as the area (and everything else as is listed in the problem statement), but gives a final answer of 0.75 N, you book/coursework made a mistake in keeping track of significant figures.

 

[kalupahana]

ok, thnx for all I got it

Most of the times I forgot to put correct units.
Thats my mistake, sorry for it.. kk