How Does Newton's Law Apply to Multiple Gravitational Forces?

[fredrick08]

[SOLVED] Newton theory of gravity

Homework Statement

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.

Homework Equations

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

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 don't know where i went wrong, can someone please help me?

 

[fredrick08]

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

 

[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..

 

[tiny-tim]

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 don't know where i went wrong, can someone please help me?

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!

 

[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 don't 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 don't see how R can be anything else but R=0.2m... if its not could you please show me why??

 

[fredrick08]

hey, and anyone tell me how to paste pictures in the thread? do i need to upload it the net or something? i can't just give the directory of the file on my comp??

 

[tiny-tim]

yes … but i don't understand why you would use pythagoras since the 2 10kg balls, is the same as one 20kg ball... since the i cancel out? …

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.

hey, and anyone tell me how to paste pictures in the thread?

I think you have to upload onto something like photobucket, and then quote the url (leaving out the http://) .

 

[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 I am giving you the wrong picture.

 

[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...

 

[fredrick08]

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

 

[fredrick08]

heres a pic hope it works = )

 

[tiny-tim]

Hi fredrick08!

nice pic!

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

mystery …

 

[fredrick08]

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

 

[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:
[tex] r= \sqrt{0.2^2 + 0.05^2} = 0.206 m[/tex]

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

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:
[tex]F = G \frac{m_1 m_2}{r^2} = 6.67 \times 10^{-11} \frac{200}{0.206^2} = 3.14 \times 10^{-7}[/tex]

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:
[tex]F_y = F \cos \phi = 3.14 \times 10^{-7} \cos(0.25 rad) = 3.04 \times 10^{-7}[/tex]

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

[tex]F_{total} = 6.08 \times 10^{-7}[/tex]

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

 

[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.

 

[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.

 

[fredrick08]

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

 

[fredrick08]

can anyone explain to me why nick used Fy=FCos(theta) because isn't sin(theta) in the y direction?

 

[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

 

[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.

 

[fredrick08]

hi everyone, I am still going at it, the lecturer said today that along the line of wat nick said... but I am still not 100% sure my new effort is

for 1 ball.
r=pythag of .05 and .2= .206
and theta equal arctan(.05/.2)=.25rad

F(on m3)=(2GMm/r^2)*Cos(theta) or F(on m3)=2GMm/(r^2*Cos(theta))
this is wat the lecturer give me the impression how it was meant to be done?? can anyone make any sense of what he has tried to do?

 

[fredrick08]

btw my lecturer also said that the answer in the book 3x10^7jN was wrong by a factor of 2... what ever that means...

 

[fredrick08]

ive been trying for so many hours and i think what nick did on page 1, is what the lecturer told me to do... can anyone concur that this is correct?

 

[jimvoit]

Common Error

I believe I managed to get the same result as the book by making the same error..That is by computing the forces between the wrong masses.

 

[tiny-tim]

ive been trying for so many hours and i think what nick did on page 1, is what the lecturer told me to do... can anyone concur that this is correct?

Hi fredrick08!

I don't think much of your teacher … telling you the book answer is wrong after you've had all the worry.

It's arguable whether putting answers in the book at all is a good idea … but it's definitely bad to put the wrong answers in, or to leave them with you without telling you!

Yes … Nick89's method is the right one.

The only difference I'd make is that you needn't find the angle to find its cosine … cos = adjacent/hypotenuse, so it's just 0.05/0.206

 

[fredrick08]

ok tiny tim, can i just ask why cosine(.05/.206) instead of cosine(.2/.206) coz isn't .2 the adj, 0.05 the opp and .206 the hyp?

 

[tiny-tim]

oops!

ooh yes … you're right …

I didn't check the diagram! ​

 

[fredrick08]

ok well I am just gunna call it solved. my final answer is

F(on m1)=(2GMm/r^2)Cos(theta)jN
=(2*6.67x10^-11*20*10/.206^2)Cos(.2/.206)=3.55x10^-7jN!

thanks everyone for your help very much appreciated, i don't really care if its right or wrong now, way too much effort has gone into this question lol = )