Testing for Gravitational constant

[Zaros]

Homework Statement

A student attempts to measure the gravitational constant G by suspending two 100.0kg spherical masses from a high ceiling in a shopping center and measuring the deflection of the cables from the vertical.
The cables are 45.0m long and attached to the ceiling 1.00m apart. What is the distance between of the two masses (center of mass two center of mass) at the bottom?

Homework Equations

F = (Gm1m2)/(r^2)
W=mg

The Attempt at a Solution

I worked out that the force between the two when they are 1m apart is 6.67*10^-7 and the weight of the spheres is 980N. I know that the weight is a vertical component and the force is a horizontal component but I'm unsure how I can use the force to calculate the distance between them.

Thanks for the help
Zaros

 

[diazona]

I'd start by thinking about it this way: what is the reason that the distance between the masses would not be 1m?

 

[Zaros]

I understand the concept that your getting at just don't know how i would do it mathematically. i.e. The force exerted by gravity is an attractive one and so the two spheres would be pulled together by this attractive force. But how would i be able to use the force calculated earlier to evaluate how far they move?

Thanks for such a speedy reply :)
Zaros

 

[diazona]

OK, let me pose a different question: if there is an attractive force between the two spheres, why don't they just come together and touch each other?

 

[Zaros]

is it because of the weight of the spheres and the tension in the cables?

 

[diazona]

Right. Anyway, now that you've identified the forces acting on the spheres, can you draw a free-body diagram and then write Newton's second law for each of the spheres? (Actually you could get away with doing just one of them, since it should be a symmetrical situation)

 

[Zaros]

Okay iv drawn a free body diagram with all the forces labeled. From this I'm able to see that as there is no motion vertically then the vertical component of the tension must be equal to that of the weight and the horizontal component should be equal to the gravitational force. How will Newtons second law help as F=ma will just let me get the acceleration but does not help in finding the distance. I'm probably missing something that's dead easy to see but I can't quite figure it out

 

[diazona]

No, ΣF=ma will not help you get the acceleration. You should already know what the acceleration is, then you can use it to find something about the forces.

It's often the case that when doing a force problem, you don't see at first how Newton's law will help. Just write it out anyway. Nearly all force problems are solved that way: draw one or more free-body diagrams and then write Newton's second law for each of them, so I'd advise you to get used to that procedure.

 

[Zaros]

As the sphere is not falling there is no acceleration in the vertical plane so Tension is equal to the weight. The sphere has moved towards the other one so there should be some form of acceleration going on but as you said earlier the spheres stop moving so therefore at the end the tension must be equal to the gravitational force. Is it something to do with the angle i.e. looking for the angle when the horizontal component of the tension is equal to the gravitational force and from that working out the distance related with this angle?

 

[diazona]

Is it something to do with the angle i.e. looking for the angle when the horizontal component of the tension is equal to the gravitational force and from that working out the distance related with this angle?

Horizontal? But yes, that's the idea. Try it and see how much you can figure out.

 

[Zaros]

okay doing this i got the total distance between the two spheres to be 0.9999999387m so they barely moved. Should this be about right?

 

[diazona]

You know about how much 100kg is, right? Would you expect spheres of this mass separated by an ordinary distance like 1m to displace each other by a large amount?

 

[Zaros]

No I didn't expect them to move by much so thought this value would be about right.
Thanks for all the help.
Zaros