Sum of 3-D forces in a cube

  • #1

Homework Statement


A cube has an edge length of 2.0m. At each corner there is a different spherical mass. The masses are 1kg, 2kg, 3kg, 4kg, etc. u to 8kg. They Coordinate system: 1kg is at the origin and the 4kg, 2kg, and 8kg masses lie on the positive x, y, and z axis respectively. What is the net gravitational force acting on the 1kg mass? (magnitude and direction)

(I know I don't have an image, but it won't let me insert one and at the point I'm at, I think I'm past needing the image)

Homework Equations


To find the force each mass is putting on the 1kg mass, I used equation:
F = G m1m2/r2
where G is 6.67 * 10-11 and r is the distance between 2 objects.

I substituted 1 for m1 and 2 for r, coming up with simplified equation:
F = (6.67*10-11) (m2) / 4

The Attempt at a Solution


I used that equation to find the x, y, and z components of every force.
I then added to find [tex]\Sigma[/tex]x, [tex]\Sigma[/tex]y, and [tex]\Sigma[/tex]z.
I got:
[tex]\Sigma[/tex]x = 3.0015 * 10-10
[tex]\Sigma[/tex]y = 2.668 * 10-10
[tex]\Sigma[/tex]z = 4.3355 * 10-10

My question is, what do I do now to find the total magnitude of the force?
Once I get that, how do I find the angle / direction of the force?

If I were to use the 3-d distance equation, how would I find the angle?
I'm completely stuck, so any help is greatly appreciated!

Thanks so much guys, this is my first day on the site so I don't quite know what I'm doing!! (:
 
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Answers and Replies

  • #2
is no one answering because my wording is way off or is it something else? just wondering!
 
  • #3
tiny-tim
Science Advisor
Homework Helper
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welcome to pf!

hi harrysawizard! welcome to pf! :smile:

(have a sigma: ∑ :wink:)
My question is, what do I do now to find the total magnitude of the force?
Once I get that, how do I find the angle / direction of the force?
you've found Fx Fy and Fz

so the force is the vector (Fx,Fy,Fz) …

its magnitude is √(Fx2 + Fy2 + Fz2), and for the unit vector of the direction, jsut divide by the magnitude! :wink:

(btw, you can't generally expect quick answers here … people are asleep, or doing something … you should give it 24 hours before worrying)
 
  • #4
thanks so much for all your help!
i used the distance formula for 3 dimensions and get 5.9096 * 10-10 to be the final magnitude.

We've only covered a bit of unit vectors in math, and we haven't touched them at all in physics, so how would I use them to find the direction?

When it's just 2 dimensions that I'm finding the net force of, I'll break the vector down into x and y and use the inverse tangent to find [tex]\Theta[/tex], but I just don't understand how to approach it in 3 dimensions.
 
  • #5
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi harrysawizard! :smile:

(have a theta: θ :wink:)
… When it's just 2 dimensions that I'm finding the net force of, I'll break the vector down into x and y and use the inverse tangent to find [tex]\Theta[/tex], but I just don't understand how to approach it in 3 dimensions.
You could use Euler angles (see http://en.wikipedia.org/wiki/Euler_angles" [Broken]), but i don't think anybody does for something like this …

you either specify the unit vector, or you just leave it as (Fx,Fy,Fz)
 
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