Calculate the magnitude of the gravitational force

[mathcrzy]

1. Homework Statement

Four 9.0 kg spheres are located at the corners of a square of side 0.72 m. Calculate the magnitude of the gravitational force on one sphere due to the other three.


2. Homework Equations

Fg=G(M1)(M2)/r^2

3. The Attempt at a Solution

I figured out the force the each of the two spheres perpendicular to the corner one by doing Fg=[(6.67e-11)(9)(9)]/[.72^2]=1.04e-8. then i used the pathageon theorum to find the radius to the sphere in the opposite corner of the square by doing square root of .72^2+.72^2=1.02. I am now having trouble finding the force of the two corner balls pulling at a 45 degree angle towards the middle of the square.

 

[learningphysics]

For the 3rd force at a distance of 1.02 m... apply the force formula to get the magnitude of the force... ie Gm1m2/(1.02)^2

then what is the x-component of this force... what is the y-component of this force...

add up the x-compoents of all 3 forces.

add up the y-components of all 3 force.

then you have the x-component and y-component of the net force... get the magnitude...

 

[m2003]

The magnitude of the gravitational force on one sphere due to the other three can be calculated by using the equation Fg=G(M1)(M2)/r^2, where G is the gravitational constant, M1 and M2 are the masses of the two objects, and r is the distance between them.

In this case, M1 and M2 are both 9.0 kg, and the distance between them is the diagonal of the square, which can be found using the Pythagorean theorem as √(0.72^2 + 0.72^2) = 1.02 m.

Plugging in these values, we get:

Fg = (6.67e-11)(9.0)(9.0)/(1.02^2) = 6.11e-9 N

Therefore, the magnitude of the gravitational force on one sphere due to the other three is 6.11e-9 N.