Is this correct?Solved: Magnitude of Net Gravitational Force on Mass in Square

In summary, the law of gravitation states that the magnitude of the net gravitational force on one mass due to three other masses can be calculated using the equation (GMm)/r^2 and superposition of forces. By breaking the forces into x and y-axis components and adding them using Pythagoras, the net force can be determined. The net force for the specific scenario given is 2.35 x 10^-3.
  • #1
huh
15
0
law of gravitation- is this better?

Could someone check this please?

Four identical masses of mass 600 kg each are placed at the corners of a square whose side lengths are 14.0cm (.14m).

What is the magnitude of the net gravitational force on one of the masses, due to the other three?

It's asking for the absolute value of the force.

Ok so the equation to use is (GMm)/r^2

And superposition of forces should be used, correct?

This is what I am trying: for the mass in the upper left hand corner
G= 6.67 x 10^-11

(G*600*600)/(.14^2)= 1.23 x 10^-3 for the masses beside and below

(G*600*600)/(.07^2)= 4.90 x 10^-3 for the mass diagonal from the one used
(.07 because the attraction would decrease by half, right?)

So adding these together:
(2)(1.23 x 10^-3) + 4.90 x 10^-3= 7.35 x 10^-3

or
(1.23 x 10^-3)+(-1.23 x 10^-3)+ 4.90 x 10^-3 = 4.90 x 10^-3
 
Last edited:
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  • #2
You have to break them up into x and y-axis components, add those, then use Pythagoras to add them back up.
 
  • #3
huh said:
(G*600*600)/(.07^2)= 4.90 x 10^-3 for the mass diagonal from the one used
(.07 because the attraction would decrease by half, right?)

I'm not sure of your reasoning here.

I would calculate the new distance by pythagoras and use that.

You're right - the attaraction DOES half, but you've used that fact to half your distance, squared your distance and substitued it into the equation which has given you a GREATER force for that diagonal one... which can't be right!

Also remember when you add, you're adding vectors.
 
  • #4
Okay let's try again:

(G*600*600)/(.14^2)= 1.23 x 10^-3 for the masses beside and below

use cos(0) and sin(0)
x-component: 1.23 x 10^-3 y-component: 0
use cos(90) and sin(90)
x-comp: 0 y-comp: 1.23 x 10^-3

(G*600*600)/[(.14^2)+(.14^2)]= 6.13 x 10^-4 for the mass diagonal from the one used
(I saw something like this denominator in an example I found)
use cosine and sine of 45 degrees
x and y-comp: 4.33 x 10^-4

So adding these together:
F(x)=0 + 1.23 x 10^-3 + 4.33 x 10^-4 = 1.66 x 10^-3
F(y)=4.33 x 10^-4 + 1.23 x 10^-3

using pythagorean theorem:
2.35 x 10^-3
 
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Related to Is this correct?Solved: Magnitude of Net Gravitational Force on Mass in Square

1. What is the equation for calculating the magnitude of net gravitational force?

The equation for calculating the magnitude of net gravitational force is F = G * ((m1 * m2)/r^2), where F is the magnitude of the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

2. How is the magnitude of net gravitational force affected by the masses of the objects?

The magnitude of net gravitational force is directly proportional to the masses of the objects. This means that as the masses increase, the force also increases.

3. What is the significance of the distance between the objects in calculating the magnitude of net gravitational force?

The distance between the objects has an inverse relationship with the magnitude of net gravitational force. This means that as the distance between the objects decreases, the force increases and vice versa.

4. How does the gravitational constant affect the magnitude of net gravitational force?

The gravitational constant, denoted by G, is a universal constant that determines the strength of the gravitational force. It is a very small constant, and as it remains constant, it does not affect the magnitude of net gravitational force.

5. Can the magnitude of net gravitational force be negative?

No, the magnitude of net gravitational force is always a positive value. Negative values can only arise when considering the direction of the force, but the magnitude itself will always be positive.

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