Gravitational force acting between objects

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The discussion focuses on calculating the gravitational force acting on one object in a rectangular arrangement of four 10kg objects. The user initially calculates the forces between the selected object and the others, but struggles with combining these forces correctly. Guidance is provided to resolve the forces into their x and y components, leading to a final calculation of the resultant force. The user successfully computes the total gravitational force as approximately 7.8 x 10^-9 Newtons and expresses interest in applying the same method to a triangular arrangement of objects. The conversation emphasizes the importance of vector addition in gravitational force calculations.
Abu

Homework Statement


4 10kg objects are located at the corners of a rectangle sides 2 meters and 1 meter. Calculate the magnitude of gravitational force on 1 due to the other 3.

(Same sort of idea)
3 1kg objects are located at the corners of an equilateral triangle of side length 1 meter

Homework Equations


F= Gm*M/r^2

The Attempt at a Solution


So, let's say that I am finding the gravitational force on 1 for the rectangle (labeling the rectangle clockwise starting from the top right). My first idea was to find the gravitational force between 1 and 4 which would be directly to the left.
Then I wanted to find the gravitational force between 1 and 2 which would be directly below the rectangle. Then I wanted to find the gravitational force between 1 and 3, which would be directly diagonal from 1.

So here are my calculations for those, in order:
Force on 1 due to 4: 6.67*10^-11 10*10/2^2 = 1.67*10^-9
Force on 1 due to 2: 6.67*10^-11 10*10/1^2 = 6.67*10^-9
Force on 1 due to 3: 6.67*10^-11 10*10/√5^2 = 1.334*10^-9

But I don't know what to do from here because my answer is wrong. Am I supposed to be adding these 3 forces together, or is there a vertical and horizontal component I am missing or something?

I am assuming that if I understand how to do the rectangle one, I'll be able to do the triangle one. I may need help with that one too though
If my question isn't clear, I'll be happy to clarify it for you. Thank you.
 
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Abu said:
Am I supposed to be adding these 3 forces together,
Yes, but as vectors. Do you know how to add vectors?
 
haruspex said:
Yes, but as vectors. Do you know how to add vectors?
Do you mean you want me to split the forces into x and y components? like 1 to 4 would be in the x direction and 1 to 2 would be in the y direction?
 
Abu said:
Do you mean you want me to split the forces into x and y components? like 1 to 4 would be in the x direction and 1 to 2 would be in the y direction?
Yes, and what about 1 to 3?
 
haruspex said:
Yes, and what about 1 to 3?
Okay so I found the angle with tanΘ=1/2, then Θ = 26.5 degrees
I found the component of 1 to 3 in the X direction to be:
cos26.5= x/1.334*10^-9 and it equals 1.193*10^-9

For the component of 1 to 3 in the Y direction I did:
sin26.5 = y/1.334*10^-9 and it equals 5.95*10^-10

and then adding the x and y components together now would make

X
1.67*10^-9 + 1.193*10^-9 = 2.862*10^-9
Y

6.67*10^-9 + 5.95*10^-10 = 7.3*10^-9

And then imagining this as a triangle with X and Y components, I just use
F = √(2.862*10^-9)^2 +(7.3*10^-9)^2
F = 7.8*10^-9 Newtons

Is this correct? Sorry for the late responses.
If so I think I'm going to try the triangle one next.
 
Abu said:
6.67*10^-9 + 5.95*10^-10 = 7.3*10^-9
I would keep one more digit of precision here. Other than that, this all looks right now.
 
haruspex said:
I would keep one more digit of precision here. Other than that, this all looks right now.
Thank you for your help, I appreciate it.
 

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