# Please give me the formula (Newon's law)

1. Nov 27, 2008

### victorx3

1. The problem statement, all variables and given/known data

1.Two identical spheres whose centers are 2.0m apart attracts each other with a force of 1.0x10 to the -5 power N. Find the mass of each sphere.

2.A pipe whose inside diameter is 30mm is connected to three smaller pipes whose diameters are 15mm each. If the liquid speed in the larger pipe is 1m/s, what is its speed in the smaller pipes.

2. Relevant equations

None?

3. The attempt at a solution

F = [G(Mm)]/R
Can anyone please change this formula into M=???

And I don't really know the formula for the second problem.

And future thanks.

2. Nov 27, 2008

### redargon

equation manipulation is a basic requirement and should be known well before attempting these sorts of problems. mulitply and divide both sides by some of the variables until you're left with M on one side and everything else on the other side.

The second questions is about flows. what do you know about flowrates and velocities of fluids in a pipe?

3. Nov 27, 2008

### HallsofIvy

Staff Emeritus

4. Nov 27, 2008

### victorx3

5. Nov 27, 2008

### andrewm

You are not yet ready, mathematically, for physics. And I don't mean formulas; I mean your logic and intuition.

6. Nov 27, 2008

### victorx3

Can anyone just give me the formula? And yes, I've read the "Why hasn't anyone answered my question?" thread. The problem with me is that I'm not familiar with the formula to use. I can't derive them.

Last edited: Nov 28, 2008
7. Nov 28, 2008

### redargon

I refer back to post #2.

If we just give you the formula, then we might as well just do your homework for you and give you the answer too. This is not the point. So, instead we will try to help you on your weakness and figure it out on your own and we'll help as you go along. Little steps.

Ok, so, you have F= (GmM)/R^2 and you want M to be the subject of the formula (ie. everything on one side and just M on the other). First thing I would do is move the R^2 across to the left hand side. To do this, multiply both sides by R^2. you will see that on the right hand side (RHS) the R^2/R^2 will become 1 and the left hand side (LHS) will be F*R^2. There is still a term on the RHS that you need to get to the LHS (the G), how would we do this?

Just keep at it and you'll get there and we can go step by step. So, next step, your turn.

8. Nov 28, 2008

### victorx3

Okay, now I know how it works. Just tell me if I'm wrong, I know you can solve this in a heart beat.

M^2=(r^2)(F)/G

M^2= (2m)^2 (1.0x10^-5N)/6.67x10^-11 N-m^2/kg^2

M^2 = (4m^2) (1.0x10^-5N)/6.67x10^-11 N-m^2/kg^2
Cancelation
M^2 = (4) (1.0x10^-5)/6.67x10^-11 kg^2
This is what the calculator shows:
M^2= 5.997001499E-17kg^2
By getting the square root we get:
M=0.000000007kg
M=7x10^-9kg

Am I right? Because I really want to ace this one.

9. Nov 28, 2008

### redargon

the formula is correct, but my calculator shows a different answer:
it gives M^2=599700.1499
and therefore M=774.4kg

for what course are these questions asked, ie. what level is your education at the moment?

10. Nov 28, 2008

### victorx3

Well that was my original answer (but I thought the value is too high - I forgot to root it.)

Right now, I'm in my 4th and final year as high school. Yes, my skills are way below the average right now. But, obviously, I'm the one who's trying to change. And I need PF to help me achieve that.

As for the second problem, I don't know the general formula for it, so I can't really derive from any, and yes, I know I should use Bernoulli's.

11. Nov 28, 2008

### andrewm

M=0.000000007kg

This mass is very small considering the spheres are 2m apart.

I always try to notice these things for myself. This helps me be surprised when I get a surprising answer! And often, a surprising answer to a sundry question is a wrong answer.

12. Nov 28, 2008

### redargon

whooooah... easy tiger, i'm not pointing fingers, i'm just trying to establish at what level and accuracy I should answer your questions and what sort of assumptions we can make based on the level of the work.

Ok, so number 1 down, 1 to go.

the second question is simpler than having to use bernoulli's. Do you know a formula that relates velocity, flowrate and pipe area? What is special about flowrate through a pipe compared to velocity for example?

13. Nov 28, 2008

### victorx3

Sorry but I really don't understand, can you please clarify it?

PS: I'll look at your answers tom. Right now I really need some sleep.

14. Nov 28, 2008

### redargon

i probably won't be online over the weekend, but on monday again.

For now: draw a sketch of the problem and write down all the stuff you know: diameters, velocities.
Try find a formula for velocity of a fluid in a pipe that is related to the pipe's cross sectional area or diameter. Read through your class notes and see if there is any mention of flowrate. remember flowrate and velocity are different things but are related.

15. Nov 28, 2008

### victorx3

I found this formula R=vA.

If I derive it, it will become v=R/A. v=1m/s / 0.015m(converted)

I have a hunch that the formula is correct. But I think my approach is wrong.

16. Nov 29, 2008

### Sjorris

First try to understand the situation on a qualitive level, there is one large pipe through which passes a certain discharge (flow) of the liquid. Now, the large pipe is connected to three smaller pipes. The discharge must stay the same, the amount of fluid that passes through the large pipe per second must also flow through the smaller pipes per second, because it is a conserved quanitity, meaning, no liquid can suddenly dissappear. Also, the liquid doesn't pile up at the point where the large pipe meets the small pipes or some other weird effect which would change the discharge.

So, now you know that the discharge is the same for the large pipe as it is for the three smaller pipes combined. Now, if some amount of liquid has to pass through a pipe, it has to pass through it faster if the pipe has a smaller area. This suggests that the velocity of the liquid is inverse proptionally to the area of the pipe(s).
The discharge increases when the area of the pipe increases, and it also increases when the velocity increases, so discharge = area * velocity, or R=v*a, as you have found.
Now, simply calculate the discharge through the large pipe. We know that the discharge for the three smaller pipes is the same, so you can plug the discharge for the large pipe into R=v*a equation again. You also know the area of the three pipes (three pipes, don't forget to multiply by three!), so your only unknown is v, which you can now solve for algebraicly (try to do this first, and then plug in the numbers, it will keep you from confusing numbers later on) and then calculate the value of v. Good luck!

PS. Don't forgot to add the correct units for each value you find.

17. Dec 1, 2008

### redargon

Sjorris is correct, read his post carefully.

also, just side note: it doesn't matter what variables you use as long as you define them, but we often use Q for volumetric flowrate ie. Q=vA.