KE_rms = (1/2)*M*(v_rms)^2 (2-10)

PD

I believe your equation (I numbered it 2-10) may apply to
linear acceleration (1/2 of peak). We have sinusoidal
acceleration in this case. I believe this holds:

KE_rms = (v RMS)^2 * M (sqrt 0.5) = 0.000008 (2-12)

jc

Dirk Van de moortel wrote:

"PD" <TheDraperFamily@gmail.com> wrote in message news:1140033005.040981.180240@g43g2000cwa.googlegroups.com...

Dirk Van de moortel wrote:

"Samantha" <no@email.nope> wrote in message news:Xns976B8F56DDAA5itsucksnotld@65.32.5.122...

Hello,

I'm faced with a problem that I've been trying to solve, let alone
understand. Let's say we have two masses on an x-axis, where m_1 = (m)
and m_2 = (3*m), and they are separated by a distance d. If we place a
third mass m_3 = (m) between m_1 and m_2, where is the point such that
the resultant (or net) gravitational force on m_3 is equal to zero?

I know that in this case, F = (G*m_a*m_b)/d^2.

If I assign F_1 = (G*m*m)/x^2, and F_2 = 3(G*m*m)/(d-x)^2, I must set F_1
= F_2, for F_1 - F_2 should equal 0 in the problem above.

I solve for x, but in doing so, I can never seem to get x untangled from
d. Nevertheless, the answer is 0.366 * d, but I cannot figure out where
to begin the journey to this answer. Am I even setting the problem up
correctly to begin with? Is it merely a proportional problem?

Please help me in the right direction! Thanks!

So you must solve the equation
1/x^2 = 3/(x-d)^2
for x.
You can easily manipulate it into the standard form
2 x^2 + 2 d x - d^2 = 0

Surely you have learned how to solve an equation like
a x^2 + b x + c = 0
?

You will get two values for x, one of which will be between
0 and d, probably 0.366 d, like you suggested.
The result must obviously depend on d - look at this
number as some kind of scaling factor of the situation.
The other solution will probably be negative and can be
ruled out.

Actually, you have to be careful about ruling out solutions. In this
case, there is nothing wrong with a negative value of x -- it indicates
a point on the line that passes through m and 3m, and there's nothing
obviously unphysical about it. In fact, it's worth worrying about what
that other solution means.

Yes, I had spend a few moments on this and had a
similar thought :-)
I reasoned that since we chose x = 0 for one of the
masses, and since we use the magnitudes of the forces
as opposed to the components, that x was conceived
as a distance in the model, so a negative solution could
be ruled out without even looking.
Likewise, had we taken x1 and x2 as coordinates of the
masses, with x1 < x2, then any solution x < x1 or x > x2
could be ruled out just as well.
So in this case I would insist looking at the other solution
as unphysical.

Dirk Vdm

Since you are comparing magnitudes of forces, the second solution has
an obvious interpretation. It's the place on the side of the smaller
mass opposite the larger mass where the two forces are equal.

However,
in that case they are pointing in the same direction, not opposite
direction, and so they don't cancel there; that information about
direction is lost by setting the magnitudes equal.

Dirk Van de moortel wrote:

"PD" <TheDraperFamily@gmail.com> wrote in message news:1140033005.040981.180240@g43g2000cwa.googlegroups.com...

Dirk Van de moortel wrote:

"Samantha" <no@email.nope> wrote in message news:Xns976B8F56DDAA5itsucksnotld@65.32.5.122...

Hello,

I'm faced with a problem that I've been trying to solve, let alone
understand. Let's say we have two masses on an x-axis, where m_1 = (m)
and m_2 = (3*m), and they are separated by a distance d. If we place a
third mass m_3 = (m) between m_1 and m_2, where is the point such that
the resultant (or net) gravitational force on m_3 is equal to zero?

I know that in this case, F = (G*m_a*m_b)/d^2.

If I assign F_1 = (G*m*m)/x^2, and F_2 = 3(G*m*m)/(d-x)^2, I must set F_1
= F_2, for F_1 - F_2 should equal 0 in the problem above.

I solve for x, but in doing so, I can never seem to get x untangled from
d. Nevertheless, the answer is 0.366 * d, but I cannot figure out where
to begin the journey to this answer. Am I even setting the problem up
correctly to begin with? Is it merely a proportional problem?

Please help me in the right direction! Thanks!

So you must solve the equation
1/x^2 = 3/(x-d)^2
for x.
You can easily manipulate it into the standard form
2 x^2 + 2 d x - d^2 = 0

Surely you have learned how to solve an equation like
a x^2 + b x + c = 0
?

You will get two values for x, one of which will be between
0 and d, probably 0.366 d, like you suggested.
The result must obviously depend on d - look at this
number as some kind of scaling factor of the situation.
The other solution will probably be negative and can be
ruled out.

Actually, you have to be careful about ruling out solutions. In this
case, there is nothing wrong with a negative value of x -- it indicates
a point on the line that passes through m and 3m, and there's nothing
obviously unphysical about it. In fact, it's worth worrying about what
that other solution means.

Yes, I had spend a few moments on this and had a
similar thought :-)
I reasoned that since we chose x = 0 for one of the
masses, and since we use the magnitudes of the forces
as opposed to the components, that x was conceived
as a distance in the model, so a negative solution could
be ruled out without even looking.
Likewise, had we taken x1 and x2 as coordinates of the
masses, with x1 < x2, then any solution x < x1 or x > x2
could be ruled out just as well.
So in this case I would insist looking at the other solution
as unphysical.

Dirk Vdm

Since you are comparing magnitudes of forces, the second solution has
an obvious interpretation. It's the place on the side of the smaller
mass opposite the larger mass where the two forces are equal. However,
in that case they are pointing in the same direction, not opposite
direction, and so they don't cancel there; that information about
direction is lost by setting the magnitudes equal.

PD

"Henry Lemington-Wholeflavors" <cwocwocwo@yahoo.co.uk> wrote in message
news:1140037061.622548.125770@g43g2000cwa.googlegroups.com...
|
| Spaceman wrote:
| > "Henry Lemington-Wholeflavors" <cwocwocwo@yahoo.co.uk> wrote in message
| > news:1140034454.871980.283980@f14g2000cwb.googlegroups.com...
| > |
| > | Samantha wrote:
| > | > Hello,
| > | >
| > | > I'm faced with a problem that I've been trying to solve, let alone
| > | > understand. Let's say we have two masses on an x-axis, where m_1 =
(m)
| > | > and m_2 = (3*m), and they are separated by a distance d. If we place
a
| > | > third mass m_3 = (m) between m_1 and m_2, where is the point such
that
| > | > the resultant (or net) gravitational force on m_3 is equal to zero?
| > | >
| > | > I know that in this case, F = (G*m_a*m_b)/d^2.
| > | >
| > | > If I assign F_1 = (G*m*m)/x^2, and F_2 = 3(G*m*m)/(d-x)^2, I must
set
| > F_1
| > | > = F_2, for F_1 - F_2 should equal 0 in the problem above.
| > | >
| > | > I solve for x, but in doing so, I can never seem to get x untangled
from
| > | > d. Nevertheless, the answer is 0.366 * d, but I cannot figure out
where
| > | > to begin the journey to this answer. Am I even setting the problem
up
| > | > correctly to begin with? Is it merely a proportional problem?
| > | >
| > | > Please help me in the right direction! Thanks!
| > |
| > | This is really simple. I did it in my head in a second.
| > |
| > | 3m/x1^2 = m/x2^2
| > |
| > | IOW, one length is root(3) times the other.
| > |
| > | Therefore 1/(1 + root(3))*d is one length, and root(3)/(1 + root(3))*d
| > | is the other, where d = x1 + x2.
| > |
| > | 1/(1 + root(3)) is 0.366
| > |
| > | No need to get your knickers in a twist.
| >
| > Dear Henry,
| > 0.366 what?
| > The question asks for a point in between m_1 and m_2.
| > where is this 0.366 point you are talking about?
|
| 0.366 multiplied by d, you illiterate hopeless fucking moron. Read my
| post again.

Dear Henry,
I read your post,
the problem is I read the Question, and your answer
does not answer it.
0.366 * d is not a point between mass 1 and mass 2
You need more of an answer for a point between 2 masses
like that.
Your answer is wrong.
(and your insults are funny)