# Force of gravity between 2 bars

1. Aug 5, 2006

### NotMrX

Suppose you had two bars with mass m and length L and there were parallel to each other, along the same line and a distance d apart. How do we calculate the force between them?
______________ ______________
|-----L--------| |------d-------| |-------L-------|

I figured out how to do it if it is a bar and point mass. Then I thought maybe if add up the particles forces along the line but I couldn't get it to work out.
Density = E = M/L; dm = E dr
F = G(massofparticle)E dr/r^2 = G(mass of particle)(1/r)
mass of partlcle = dm = E dr
And then evaluate the integral again to get the force for the sum of particles but I don't think that method works.

2. Aug 5, 2006

### NotMrX

the drawing messed up

____________ empty space ___________
|------L------||----d-----||-----L------|

3. Aug 7, 2006

### mukundpa

I think the method is correct but first you have to find the force on a small partical of length dx at distance x from the end of second rod due to all particels of the first rod and then integrate again for all particles of second.

The best way to find field due to first rod as a function of distance r from ond end of first rod and then integrate for the different force on the second rod.

4. Aug 10, 2006

### NotMrX

F= GMdm/r^2 where M is the mass of the particle
dm = Density dx = (m/L) dx
F = G M m/L *(1/x^2)dx where x can be from d to d + L
F = G*M*m/L/x +C

F = G*m/L*dm/x = G*(m/L)^2 Ln X where is x from L+d to 2L+D
F= G*(m/L)^2 Ln [(2L+D)/(L+D)]

Something doesn't seem right with the bounds of the integral?

5. Aug 10, 2006

### mukundpa

$$F = \int [\int \frac{G(m/L)dx}{(r-x)^2}] (m/L)dr$$

The inner integral is for first rod from x=0 to x=L and the outer integral is for second rod from r= L+d to r=2L+d

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6. Aug 12, 2006

### NotMrX

Thank you for your help. I think I figured out where that integral came from so I tried to work some similar examples.

for two bars parrelel in a rectangle

_________

_________

I ended up with the following integral where lambda is density and both bars has same mass and length with a vertical distance Y between them.
I hope I got latex to work correctly.

$$/{int}/{int}/frac{G/{lambda}Y}{[Y^2 + (x-r)^2]^(3/2)}dx/{lambda}dx$$

7. Aug 12, 2006

### NotMrX

If I is the integral sign

I I [G*(m/L)*Y/(Y^2+(x-r)^2)^(3/2)*dx*m/L*dr

Where both integrals are from 0 to L

Did anyone else get the same answer?

Last edited: Aug 12, 2006
8. Aug 12, 2006

### NotMrX

$$\int\int\frac{G\lambdaY}{\sqrt[3]{Y^2+(x-r)^2}}dx\lamda\dr$$
fingers crossed

9. Aug 12, 2006

### Staff: Mentor

I could be wrong, but can't you just calculate the gravitational attraction with the simplification of putting all the mass at the center of mass for each object? Isn't that how we deal with gravitational attraction for objects and the Earth?

10. Aug 13, 2006

### lapo3399

Well... If you're talking about using the mass and the radius of gyration of the two bars, I think that might work.