Gravitational force acting between objects

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Homework Help Overview

The discussion revolves around calculating the gravitational force acting on one object due to multiple other objects positioned at specific coordinates, specifically in a rectangular and triangular arrangement. The subject area includes gravitational forces and vector addition in physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the method of calculating gravitational forces between objects and the need to consider vector components for accurate results. Questions arise regarding the addition of forces and the correct approach to handle diagonal components.

Discussion Status

Participants are actively engaging with the problem, exploring the calculation of forces and their components. Some guidance has been provided regarding the vector addition of forces, and there is an ongoing exploration of how to apply this to both the rectangular and triangular configurations.

Contextual Notes

There is a mention of potential confusion regarding the addition of forces and the need for clarity on vector components. Participants express a desire to understand the calculations better before moving on to the next problem.

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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