Newton's Theory of Gravity Problem help

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The problem involves two 100 kg lead spheres suspended from massless cables anchored 1 meter apart, raising questions about the distance between their centers due to gravitational attraction. The initial assumption is that the spheres would remain 1 meter apart, but the gravitational force causes them to move closer together. To solve the problem, trigonometry and free body diagrams (FBDs) are recommended to analyze the angles and tension in the cables. The gravitational force calculated is 6.67x10^-7 N, but further calculations are needed to determine how this force affects the position of the spheres. Understanding the angles involved is crucial for finding the horizontal displacement and ultimately the distance between the centers of the spheres.
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Homework Statement


"Two 100 kg lead spheres are suspended from 100-m-long massless cables. The tops of the cables have been carefully anchored exactly 1 m apart. What is the distance between the centers of the spheres?

Homework Equations


F = (Gm1m2)/r^2 ?

The Attempt at a Solution


I'm not sure how this problem is related to Newton's theory and I am confused at this problem as I think the answer should simply be 1 meter because that is exactly how far the cables are anchored apart. Using the equation for F that I have provided I get 6.67x10^-7 N. This is the only equation we have really learned in class and I am unsure how this applies to the problem or how I use this to calculate the distance between the centers of the spheres. I feel dumb because this problem sounds so easy but I feel like I'm missing something. The book says it's a medium difficulty problem and that it requires material from earlier chapters but I'm not sure what. I would greatly appreciate any help on this to get me going. Thanks!
 
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Welcome to PF!

aarunt1 said:
Two 100 kg lead spheres are suspended from 100-m-long massless cables. The tops of the cables have been carefully anchored exactly 1 m apart. What is the distance between the centers of the spheres?

Hi aarunt1! Welcome to PF! :smile:

The tops are a metre apart, but the cables can swing, so the bottoms can be nearer if the lead spheres attract each other.

Use trigonometry, and FBDs! :smile:
 
thanks for the welcome. I figured that the spheres would probably attract each other somewhat and I calculated the gravitational attraction (I believe) in my attempt at the solution please correct me if I am wrong tho. I am unsure how to calculate how far 6.67x10^-7 N would push each sphere in. Is there an equation I can use for this? Thanks!
 
aarunt1 said:
I am unsure how to calculate how far 6.67x10^-7 N would push each sphere in. Is there an equation I can use for this? Thanks!

The cables are at an angle, so consider the tension in one of the cables … that's what the FBD is for! :wink:
 
I'm all for the FBD idea but I don't know how to get the angles. If they were provided in the equation that would be cake but I can't remember there being a way to calculate them. Sorry for being so noob I haven't used trig in a long time and it's over a year since I've taken a physics class.
 
aarunt1 said:
I'm all for the FBD idea but I don't know how to get the angles. If they were provided in the equation that would be cake but I can't remember there being a way to calculate them. Sorry for being so noob I haven't used trig in a long time and it's over a year since I've taken a physics class.

Hi aarunt1! :smile:

No problem … the sine of the angle (which is equal to the tangent for very small angles) is the horizontal displacement divided by the length of the cable. :smile:
 

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