Superposition net gravitational force

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SUMMARY

The discussion focuses on calculating the net gravitational force acting on a central sphere (mass m5 = 0.00424 kg) due to four surrounding spheres with masses m1 = 0.00372 kg, m2 = 0.00958 kg, m3 = 0.00408 kg, and m4 = 0.00835 kg, arranged in a square formation with an edge length of 0.232 m. The gravitational force is calculated using the formula F = G*Ma*Mb/r^2, where G is the gravitational constant. The participant emphasizes the importance of considering the vector components of the forces and the cancellation effects of the forces from opposite spheres. Trigonometric functions are necessary for resolving the forces into their respective x and y components.

PREREQUISITES
  • Understanding of Newton's law of universal gravitation
  • Familiarity with vector notation and components
  • Knowledge of trigonometric functions for resolving forces
  • Basic principles of gravitational interactions between multiple bodies
NEXT STEPS
  • Study the application of Newton's law of gravitation in multi-body systems
  • Learn how to resolve forces into components using trigonometry
  • Explore gravitational force calculations in two-dimensional arrangements
  • Investigate the concept of gravitational force cancellation in symmetric configurations
USEFUL FOR

Physics students, educators, and anyone interested in gravitational force calculations in multi-body systems will benefit from this discussion.

sophzilla
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I would appreciate any help on this problem:

In Fig. 14-18a, a square of edge length .232m is formed by four spheres of masses m1 = .00372 kg, m2 = .00958 kg, m3 = .00408 kg, and m4 = .00835 kg. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = .00424 kg?

1499s9d.gif


I started out by using F = G*Ma*Mb/r^2. So for the x-direction, it would be something like:

Fi = G*M3*M4/r^2 + G*M1*M2/r^2, and similar things for the y-direction.

But I don't know how to include the m5 into my equations. Do I have to involve angles and trig functions?
 
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Okay, here's the first thing that needs to happen. Just as a forwarning, trig will be used to the fullest extent. Remeber that when you want to find the net gravitational force, you don't include the particle that you are ovbserving. The only part that it plays is the distance from the other particles. Also, remeber that particles on either side of a central particle have cancelling effects, depending on the masses and the radial distances.
 
Be careful with your components in the vector notaion. Remeber that vertical components of force vectors have nothing to do with the horizontal with tregards to the algebraic setup.
 

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