# Newton theory of gravity

1. Apr 29, 2008

### fredrick08

[SOLVED] newton theory of gravity

1. The problem statement, all variables and given/known data
what are the magnitude and direction of the net gravitational force on the 20kg mass.
Now this is kinda tricky but there is an 20kg ball on the origin, and two 10kg balls 20cm above the x axis, and one is 5cm right and the other is 5cm left of the y axis.

2. Relevant equations
F=GMm/R^2
G=6.67x10^-11

3. The attempt at a solution
F(on m1)=F(m2 on m1)+F(m3 on m1)
F(m2 on m1)=GMm/R^2=(6.67x10^-11*20*10)/0.2^2=3x10^-7jN
F(m3 on m1)=" "=" "=" ' "
Thus F(on m1)=3x10^-7j+3x10^-7j=6.67x10^-7jN

Now i thought this question was kinda straight forward, but in the answers in the book the answer is 3x10^-7jN, half of what i got.... and i dont know where i went wrong, can someone plz help me?

2. Apr 29, 2008

### fredrick08

anyone have any idea??? i know that -.05i and .05i cancel out, but maybe do you double the .2j value???

3. Apr 29, 2008

### malawi_glenn

Please wait a bit more. One hour is nothing.

the majority of PF users are sleeping at the moment, so please wait.

Also, this is NOT advanced physics..

4. Apr 29, 2008

### tiny-tim

Hi fredrick08!

Yes, it is straightforward … but not if you don't draw a proper diagram, and label it carefully.

R isn't .02, is it?

Use Pythagoras, and try again!

5. Apr 29, 2008

### fredrick08

yes, i tried pythagoras, but then your r values would be, -.05i+.2j and .05i+.2j, which would give a magnitude of sqrt(.05^2+.2^)=.206m, but i dont understand why you would use pythagoras since the 2 10kg balls, is the same as one 20kg ball...... since the i cancel out???? and even if you use this u still get

F(on m1)=(6.67x10^-11*20*20)/.2^2=6.67x10^-7???? how do they half that value???

im srry tiny-tim, but i dont see how R can be anything else but R=0.2m...... if its not could you plz show me why??

6. Apr 29, 2008

### fredrick08

hey, and anyone tell me how to paste pictures in the thread??? do i need to upload it the net or something? i cant just give the directory of the file on my comp??

7. Apr 29, 2008

### tiny-tim

Hi fredrick08!

ok, you obviously did draw a diagram … my apologies!

But you're trying to make physics too simple … and unfortunately it isn't!

Think of the gravitation force as a vector along the hypotenuse.

(Force always is a vector, of course.)

Then that force is equal to the sum of two forces along the two sides, isn't it?

The x-direction forces are opposite, so they cancel.

It is only the y-direction forces which add up!

But the value of each y-direction force has to be calculated using R for the hypotenuse direction.

Use .206m for R, but then only use the j component of the result.
I think you have to upload onto something like photobucket, and then quote the url (leaving out the http://) [Broken].

Last edited by a moderator: Apr 23, 2017 at 12:33 PM
8. Apr 29, 2008

### fredrick08

ok, well if i understand you correctly...
F(m2 on m1)=6.67x10^-11*20*10/.206^2=3.14x10^-7??? but then if you multiply that by 2, well i dunno.... srry

have you tried doing it yourself? did it work out? i will try and upload a picture, in case im giving you the wrong picture.

9. Apr 29, 2008

### fredrick08

i did a similar problem in class, but that was with m1 and m2 had different i and j values. where you added the component vectors, and then did pythag to find the resultant magnitude.... but this one has just confused me, coz it looks simple.. but somehow not as i thought...

10. Apr 29, 2008

### fredrick08

btw, thanks everyone who has helped, very much appreciated = )

11. Apr 29, 2008

### fredrick08

heres a pic hope it works = )

12. Apr 29, 2008

### tiny-tim

Hi fredrick08!

nice pic!

erm … yes, I have tried it, and I can't make it 3x10^-7.

mystery …

13. Apr 29, 2008

### fredrick08

did u get my answer? or is it possible that the book is wrong? i might ask tomoz at uni... thanks heaps anyways = )

14. Apr 29, 2008

### Nick89

I'll give it a go:

From the symmetry of the problem we can see directly that the net force on the 20 kg ball will be in the y direction.
Also, the force will be exactly twice the y-component of the force of one ball.

The distance between one of the 10kg balls and the 20kg ball is according to pythagoras:
$$r= \sqrt{0.2^2 + 0.05^2} = 0.206 m$$

The angle between the line from one of the 10kg balls and the 20 kg balls and the y-axis is:
$$\phi = \arcsin \frac{0.05}{0.2} = 0.25 rad$$

Now, the force, directed along the line between the two balls (note that it therefor has both an x and y component) is given by:
$$F = G \frac{m_1 m_2}{r^2} = 6.67 \times 10^{-11} \frac{200}{0.206^2} = 3.14 \times 10^{-7}$$

The y-component of this force can be calculated using the angle in the right triangle that is created by this force vector and the y-axis:
$$F_y = F \cos \phi = 3.14 \times 10^{-7} \cos(0.25 rad) = 3.04 \times 10^{-7}$$

The force by both balls (in the y-direction) will ofcourse be just twice this:

$$F_{total} = 6.08 \times 10^{-7}$$

I hope I didn't make any mistakes but it's definately not 3 *10^-7...

15. Apr 29, 2008

### fredrick08

oh ok i see... thanks nick, i didnt realize that there was still x and y components... i gunna ask today anyways, knowing my physics course, they prob got the wrong answer from the book lol.

16. Apr 29, 2008

### Nick89

No, the net force is ONLY in the y-direction!
The force of only one of the 10kg balls (acting on the 20kg ball) has a x and y direction.

However, the x-component of the force of one ball will be exactly the same as the x-component of the force of the other ball, but in the opposite direction, and the effects of these two forces will therefore cancel out.

17. Apr 29, 2008

### fredrick08

yes thats what i meant to say.... srry, but thanks for your help = )

18. Apr 29, 2008

### fredrick08

can anyone explain to me why nick used Fy=FCos(theta) because isnt sin(theta) in the y direction?????

19. Apr 29, 2008

### jimvoit

Here is an approach that appears to give the answer in the book. It’s a bit brute force but it works for a more general case also.

The position vectors p1 and p2 from the 20 kg mass m3 to the 10 kg masses m1 and m2 are:
p1=-0.05 i+0.20 j p2=+0.05 i+0.20 j

Their magnitudes are:

r1=√(p1.p1) r2=√(p2.p2) where the period indicates the dot product.

The values of the gravitational field at mass m3 (20 kg) due to the masses m1 and m2 (10 kg) are:
g1= (g m1 m2)/〖r1〗^2 g2=(g m1 m2)/〖r2〗^2

The net vector result of the field due to m1 and m2 is:

g1 p1/r1+g2 p2/r2

Note that p1/r1 and p2/r2 are unit vectors

The answer obtained in this way is 0 i +3.05 〖x 10〗^(-7) j

20. Apr 29, 2008

### jimvoit

Sum of field vectors

Here is an approach that appears to give the answer in the book. It’s a bit brute force but it works for a more general case also.

The position vectors p1 and p2 from the 20 kg mass m3 to the 10 kg masses m1 and m2 are:
p1=-0.05 i+0.20 j p2=+0.05 i+0.20 j

Their magnitudes are:

r1=√(p1.p1) r2=√(p2.p2) where the period indicates the dot product.

The values of the gravitational field at mass m3 (20 kg) due to the masses m1 and m2 (10 kg) are:
g1= (g m1 m2)/〖r1〗^2 g2= (g m1 m2)/〖r2〗^2

The net vector result of the field due to m1 and m2 is:

g1 p1/r1+g2 p2/r2
Note that p1/r1 and p2/r2 are unit vectors

The answer obtained in this way is 0 i+3.05 〖x 10〗^(-7) j

PS. The key idea here is to set up the two field vectors by multiplying the magnitude of the field by a unit vector in the direction of the field, then adding the vectors to get the resultant.