# Physics Urgent!!

posted by
**Abi** on
.

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?