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November 23, 2014

November 23, 2014

Posted by **Abi** on Monday, May 2, 2011 at 12:59pm.

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...

Physics Urgent!!!!! I really need help with this - bobpursley, Monday, May 2, 2011 at 11:21am

The issue I have is the distance apart.

you wrote .152-.0254. The problem stated "about 9 inches away". Well, that is definitily vague, but I assume it is surface to surface distance, which means then distance from center to center is .2286 (9") plus .1524 (6inches) plus .0254 (1") which is .406meters.

Now, if one takes the "about 9 inches" to mean center to center, then it is .2286m.

In any event, I don't see how you arrived at your distance.

Physics Urgent!!!!! I really need help with this - Abi, Monday, May 2, 2011 at 12:59pm

I was substracting both radii, now I tryed

(6.674E-11)(0.730283)(157.850)/(0.2286)^2 and I got 1.47E-7 and tryed 6.674E-11)(0.730283)(157.850)/(0.2286 +.1524 +.0254)^2 and I got 4.66E-8... but niether is right....

- Physics Urgent!!!!! I really need help with this -
**bobpursley**, Monday, May 2, 2011 at 2:13pmI dont see an error. The only other thing is the problem. You calculated the force between one large, and one small ball. I am still wondering if the problem wants the sum of both forces, but it is hardly clear.

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