Sorry to repost this, again, but I still don't understand.

Newton’s Law of Gravity specifies the magnitude of the interaction force between two point masses, m1 and m2, separated by the distance r as F(r) = Gm1m2/r^2. The gravitational constant G can be determined by directly measuring the interaction force in the late 18th century by the English scientis Henry Cavendish. This apparatus was a torsion balance consisting of a 6.00-ft wooden rod suspended from a torsion wire, with a lead sphere having a diameter of 2.00 in and a weight of 1.61 lb attached to each end. Two 12.0-in, 348-lb lead balls were located near the smaller balls, about 9.00 in away, and held in place with a separate suspension system. Today’s accepted value for G is 6.674E-11 m^3 kg^-1s^-2.
a) Determine the force of attraction between the larger and smaller balls that had to be measured by this balance.
b) Compare this force to the weight of the small balls.

Ok... so for the first one I have to use the equation given and change the values to kilograms and meters, and use the radii instead of the diameters, this is what I know, but I don't know how to do it when there are three spheres (the big one and the 2 small ones)... I don't get what they are asking for the second part...

Someone please help, it will be deeply appreciated...

Physics Urgent!! - bobpursley, Saturday, April 30, 2011 at 3:31pm
Double the force: you have two big ones, each pulling on one small

Sorry to repost this again but I'm still not clear... Ok, so I have r1= 0.0254m, m1=0.730283 kg, r2= 0.152m and m2=157.850 kg... I calculated
(6.674E-11)(0.730283)(157.850)/(0.152m-0.0254m)^2 and I got 4.77E-7 and multiplyed it by 2 and then I got 9.54 E-7... Does this looks right?

Is that for part A?

Physics Urgent!!!!! - bobpursley, Saturday, April 30, 2011 at 8:30pm
I didn't punch it on the calculator. OK, on A) it asks (I think) for each force, do don't multipy by 2 .

But, I just have to plug in one number for part A that must be E-7 and for part B I have to plug only one number too that must be E-8... Is my equation right?? If I multiply the answer by two, I still don't get the answer to E-8.



Physics (please help!!!!) - bobpursley, Sunday, May 1, 2011 at 11:42pm
show me your work.

Physics (please help!!!!) - Abi, Monday, May 2, 2011 at 12:11am
I showed it....

r1= 0.0254m, m1=0.730283 kg, r2= 0.152m and m2=157.850 kg... I calculated
(6.674E-11)(0.730283)(157.850)/(0.152m-0.0254m)^2 and I got 4.77E-7 and multiplyed it by 2 and then I got 9.54 E-7...

To find the force of attraction between the larger and smaller balls, you need to use the equation given: F(r) = Gm1m2/r^2.

First, convert the values given to the appropriate units:
- Diameter of the lead sphere attached to each end of the wooden rod is 2.00 in, which is 0.0508 m in radius.
- The weight of each lead sphere is 1.61 lb, which is 0.730283 kg in mass.
- The radius of the larger lead balls is 0.152 m and their weight is 348 lb, which is 157.850 kg.

Now you can plug the values into the equation to calculate the force of attraction between the larger and smaller balls.

F = (6.674E-11)(0.730283)(157.850)/(0.152+0.0508)^2
F = 9.54E-7 N

So the force of attraction between the larger and smaller balls that had to be measured by this balance is approximately 9.54E-7 N.

For part b) of the question, you need to compare this force to the weight of the small balls. The weight of the small balls can be calculated using the formula W = mg, where m is the mass in kg and g is the acceleration due to gravity (which is approximately 9.8 m/s^2).

W = (0.730283)(9.8)
W ≈ 7.16 N

So the weight of the small balls is approximately 7.16 N.

Now you can compare the force of attraction (9.54E-7 N) to the weight of the small balls (7.16 N) to see how they relate to each other.