Newton's Theory of Gravity Problem help

[aarunt1]

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!

 

[tiny-tim]

Welcome to PF!

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!

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!

 

[aarunt1]

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!

 

[tiny-tim]

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!

 

[aarunt1]

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.

 

[tiny-tim]

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!

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.