How Close Do Two 100kg Spheres Hang When Suspended 1m Apart?

AI Thread Summary
The discussion revolves around calculating the distance between the centers of two 100kg lead spheres suspended 1m apart by massless cables. Participants analyze the gravitational attraction using Newton's law of universal gravitation and explore the forces acting on the spheres. They conclude that the gravitational force is extremely weak, leading to negligible movement between the spheres. The tension in the cables and the angles they form with the vertical are also considered, with trigonometric relationships used to find the angle of inclination. Ultimately, the gravitational force is confirmed to be significantly smaller than the forces due to tension, emphasizing the minimal effect of gravitational attraction in this scenario.
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Gravitational Attraction-Please Help!

Homework Statement


Two 100kg lead sphers are suspended from 100m long massless cables. The tops of the cables have been anchored 1 m apart. What is the distance between the center of the spheres?

Homework Equations


Fg= Gm1m2/r^2


The Attempt at a Solution



Would I say find the force between them when they are just anchored, using 1m as my value for r? Then I thought I might use a force body diagram to find the angle that the cables are hanging and then the distance between the spheres. But the problem is, isn't the force going to increase as they get closer, so wouldn't they just keep moving toward each other?
 
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Don't worry about it. Do what you said you were going to do. The movement toward each other is 'tiny', the correction to that is 'tiny'^2, or maybe even 'tiny'^3. Ignore it. It's completely insignificant.
 
Ok, I summed up the forces on one sphere

Fx= GM^2/(1m)^2 In this case it is 6.67 *10^-7 N
Fy= T-mg T= 980 N

But how do I get an angle from that?
 
Any thoughts on this question anybody??
 
Fy is just mg. Now call theta the angle the cable makes with the vertical. Split T into x and y components. The total force on the mass is Fx+Fy+T=0.
 
The change in the distance because the cables will both point to the center of the earth, which is equal to:
\frac {d l}{R_{earth}}
wich is 0.016 mm is much bigger than the change because of the mutual attraction of the lead balls.
 
Ty=980 and then does tension on x equal the force of gravity. I guess I don't understand what you mean by Fx + Fy + T = 0
 
bcjochim07 said:
Ty=980 and then does tension on x equal the force of gravity. I guess I don't understand what you mean by Fx + Fy + T = 0

I mean that the sum of all of the forces on a stationary object is 0. Ty does equal 980N. What does Tx equal? Ty also equals T*cos(theta) and Tx equals T*sin(theta). Which you would know if you'd split T into x and y components. Does that suggest a way to find theta?
 
I can't think of anything other than gravity between the spheresthat Tx would equal. When I draw my force body diagram, I have gravity from the other sphere along x and mg along y. But that doesn't seem to make the forces sum to zero. Once I figure Tx out though, I can draw a triangle and use trigonometry to find that angle.
 
  • #10
Tx does equal the gravitational attraction between the spheres! If you can use trig from there then you are good to go. They do sum to zero Ty=-Fy and Tx=-Fx.
 
  • #11
arctan(6.67E-10/980) = 6.8 * 10-10 degrees that's very tiny!
 
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  • #12
bcjochim07 said:
arctan(6.67E-10/980) = 6.8 * 10-10 degrees that's very tiny!

You meant 6.67*10^(-7), right? And your final answer is correct but in radians, not degrees. But yes. Still very tiny. Gravitational force is extremely weak.
 
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