How Does Position Affect Gravitational Force Between Three Objects?

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The discussion revolves around calculating the net gravitational force on a 63.5 kg mass positioned between two larger masses of 125 kg and 548 kg. The first part of the problem was successfully solved, yielding a force of 0.00004836 N. However, the second part, which involves determining the distance from the 548 kg mass where the net force on the 63.5 kg mass is zero, presents challenges. Participants suggest setting up the equation symbolically and emphasize that the distances must be correctly represented to balance the forces. The conversation highlights the importance of understanding force directions and the correct application of gravitational equations.
GiantSheeps
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Homework Statement


PART 1: Objects with masses of 125 kg and 548 kg are separated by 0.385 m. A 63.5 kg mass is placed midway between them.
Screenshot%2B2015-03-20%2Bat%2B6.20.08%2BPM%2B-%2BEdited.png

Find the magnitude of the net gravitational force exerted by the two larger masses on the 63.5 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m2 /kg2 . Answer in units of N.

PART 2: Leaving the distance between the 125 kg and the 548 kg masses fixed, at what distance from the 548 kg mass (other than infinitely remote ones) does the 63.5 kg mass experience a net force of zero? Answer in units of m.

Homework Equations


[/B]
F = Gmm/(d^2)

The Attempt at a Solution



So i got part 1 no problem, the answer is .00004836 N, but part 2 is a real tough one. For my first attempt, i set up the formula ( (548)(63.5)(6.67E-11)/x^2 ) + ( (125)(63.5)(6.672E-11)/(x+.385)^2 ) = 0

x being distance, and (x + .385) distance from smaller mass since it remains fixed

I got 0.8165837351598118 as an answer, but that was wrong.

Where did I go wrong? How can I solve this? Any help would be greatly appreciated

Thanks!
 
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I would recommend setting up the equation in symbols first. Then you can simplify by cancelling common factors on each side of the equation.

The expression x + .385 does not represent the correct distance.

Your equation reads ( (548)(63.5)(6.67E-11)/x^2 ) + ( (125)(63.5)(6.672E-11)/(x+.385)^2 ) = 0. Note that both terms on the left are positive. So, they can't add to zero.
 
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How would I do that? Gmm/d^2 - Gmm/d^2 = 0?
I used - because as you said if it were addition it could not add to zero, but am I setting that up right at all? Wouldn't the gravity of the two objects on the smaller one have to end up being equal to zero? Here's my reasoning behind the x + .385 distance
(its really large image, sorry, i have no idea why)
IMG_20150320_195707066.jpg
 
In order for the total force on the 63.5 kg mass to be zero, does the 63.5 kg mass need to be between the other two masses or somewhere else? Think about the direction of the forces on the 63.5 kg mass.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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