Calculate Magnitude of Gravitational Force on One Sphere

AI Thread Summary
The discussion focuses on calculating the gravitational force exerted on one sphere by three others positioned at the corners of a square. The gravitational force formula is applied, with specific attention to the x and y components of the forces. A participant initially miscalculated the force components, leading to an incorrect total gravitational force. After clarification, the correct calculations were confirmed, revealing that the diagonal distance between spheres must be considered rather than just the side length of the square. The final total gravitational force was adjusted to approximately 3.13e-8, accounting for rounding errors.
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Homework Statement



Four 9.5 kg spheres are located at the corners of a square of side 0.60 m. Calculate the magnitude of the total gravitational force exerted on one sphere by the other three.

Homework Equations



F=G(mass of #1)(mass of #2)/r^2
F=G(mass of #1)(mass of #3)/r^2
F=G(mass of #1)(mass of #2)/r^2
For #3 take Fsin(45)=x value
For #3 x value=y value

x and y components added
F=squareroot(x^2+y^2)


The Attempt at a Solution


r=0.6m
m=9.5kg
F#2x=1.625e-8
F#2y=0
F#3x=0
F#3y=-1.672e-8
F#4x=1.182e-8
F#4y=-1.182e-8
FTotal= 4.037e-8

Correct answer in book for FTotal: 3.2e-8
 
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Can you show how you got

F#2x=1.625e-8
F#4x=1.182e-8
F#4y=-1.182e-8
 
RE:
Can you show how you got

F#2x=1.625e-8
F#4x=1.182e-8
F#4y=-1.182e-8

Begin Reply:
F#2x=(G)(9.5)(9.5)/(0.6^2)
=1.672e-8
Which was a typo, thanks for pointing that out.
F=#4x=[(G)(9.5)(9.5)/(0.6^2)]sin(45)=
=1.182e-8
F=#4y=[(G)(9.5)(9.5)/(0.6^2)]cos(45)=
=1.182e-8
 
The distance between the two masses diagonally across from each other will NOT be 0.6 m. 0.6 m is the length of the sides of the square.
 
Wow thank you so much! I can't believe I missed that. I plugged in my values again and I got 3.13e-8, which I attribute to rounding. Thanks again!
 
You're welcome.
 
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