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

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Homework Help Overview

The discussion revolves around the gravitational attraction between two 100kg lead spheres suspended from massless cables, which are anchored 1 meter apart. Participants are exploring how the gravitational force affects the distance between the centers of the spheres.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating the gravitational force between the spheres and the tension in the cables. There are attempts to derive the angle of the cables using force components and to understand the implications of gravitational attraction on their positions.

Discussion Status

Some participants have provided insights into the forces acting on the spheres and the relationship between tension and gravitational force. There is ongoing exploration of how to apply trigonometry to find the angle of the cables, with various interpretations of the forces involved.

Contextual Notes

Participants note that the gravitational force is very small compared to the tension in the cables, leading to discussions about the insignificance of the gravitational attraction in altering the distance between the spheres. There is also mention of the effect of the Earth's gravity on the cables.

bcjochim07
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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}}
which 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!
 
Last edited:
  • #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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