Gravitational Force between three forces

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SUMMARY

The discussion centers on calculating the net gravitational force acting on a 20kg ball due to two 10kg balls positioned 20cm above it, with one 10kg ball 5cm to the right and the other 5cm to the left of center. The gravitational force formula used is F=GMm/R², where G is the gravitational constant (6.67E-11). The calculated distance R is approximately 0.206m, leading to a net force in the y-direction of 6.1E-7 N. Participants concluded that the expected answer of 3E-7 N may be incorrect, suggesting potential errors in the reference material.

PREREQUISITES
  • Understanding of Newton's law of universal gravitation
  • Familiarity with gravitational constant (G=6.67E-11)
  • Basic algebra for manipulating equations
  • Knowledge of vector components in physics
NEXT STEPS
  • Review gravitational force calculations in multi-body systems
  • Learn about vector addition in physics
  • Study the implications of gravitational force discrepancies in academic materials
  • Explore advanced gravitational concepts, such as gravitational potential energy
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Students studying physics, educators teaching gravitational concepts, and anyone interested in solving complex gravitational force problems.

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1. Homework Statement [/b]
There's a 20kg ball and two 10kg balls 20cm above it. One of the 10kg balls is 5cm to the right of center and the other is 5cm to the left of center. What is the net gravitational force on the 20kg ball

2. Homework Equations [/b]
F=GMm/R23. The Attempt at a Solution [/b]
M=20kg, m=10kg, G=6.67E-11 and R=√(.202+.052)=.206m
\SigmaFx=0; because the 10kg ball are equal distances away from the 20kg ball along the x-axis
\SigmaFy=2*(((6.67*10-11*10*20)/(.2062))*(.2/.206))=6.1*10-7
I know the answers supposed to be 3*10-7
 
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Your work looks correct to me.
 
All right thanks I'm just going to assume the books wrong it wouldn't be the first time
 

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