Sum of 3-D forces in a cube

In summary, the problem involves a cube with different spherical masses at each corner and the task is to find the net gravitational force acting on the 1kg mass at the origin. Using the equation F = Gm1m2/r^2 and substituting the values, the x, y, and z components of the force are calculated. To find the total magnitude of the force, the vector (Fx,Fy,Fz) is used and its magnitude is obtained by taking the square root of the sum of its components squared. To find the direction of the force, unit vectors can be used or the vector can be left in its current form.
  • #1
harrysawizard
4
0

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! (:
 
Last edited:
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  • #2
is no one answering because my wording is way off or is it something else? just wondering!
 
  • #3
welcome to pf!

hi harrysawizard! welcome to pf! :smile:

(have a sigma: ∑ :wink:)
harrysawizard said:
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, just 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
hi harrysawizard! :smile:

(have a theta: θ :wink:)
harrysawizard said:
… 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)
 
Last edited by a moderator:

1. What is the concept of "Sum of 3-D forces in a cube"?

The concept of "Sum of 3-D forces in a cube" refers to the total magnitude and direction of all the three-dimensional forces acting on an object within a cube-shaped space.

2. How is the sum of 3-D forces in a cube calculated?

The sum of 3-D forces in a cube is calculated by adding up all the individual forces acting on an object within the cube and taking into account their respective magnitudes and directions.

3. What factors affect the sum of 3-D forces in a cube?

The sum of 3-D forces in a cube is affected by the magnitude and direction of each individual force, as well as the shape, size, and orientation of the cube itself.

4. Why is it important to understand the sum of 3-D forces in a cube?

Understanding the sum of 3-D forces in a cube is important for accurately predicting the motion and behavior of objects within a three-dimensional space, and for designing structures that can withstand various forces acting on them.

5. Are there any real-world applications of the concept of sum of 3-D forces in a cube?

Yes, the concept of sum of 3-D forces in a cube is used in various fields such as engineering, physics, and architecture to analyze and design structures, predict the stability of objects in motion, and understand the forces acting on objects in three-dimensional environments.

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