OK, I know, I'm postin this way too many times but I'm halfway there now....
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 2:13pm
I don't 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.
Ok, so I took my first answer and multiplyed it by 2 as you previously told me and got it right (2.94E-7)....but I still don't know how to get part B) I multiplyed it by 2 too... since I didn't know what else to do, and is wrong, please help, I have only one attempt left...
weight of the balls: 158kg, or 158*9.8
"this force"= 2.94E-7
I don't know what they mean by "compare". one is about 1.6E10 larger.
yups
To determine the force of attraction between the larger and smaller balls, you need to use the equation F(r) = Gm1m2/r^2.
First, you need to convert the given values to SI units.
- Convert the diameter of the lead sphere from inches to meters:
diameter = 2.00 in = 2.00 * 0.0254 m = 0.0508 m
- Convert the weight of the lead sphere from pounds to kilograms:
weight = 1.61 lb = 1.61 * 0.453592 kg = 0.730283 kg
- Convert the distance between the larger and smaller balls from inches to meters:
distance = 9.00 in = 9.00 * 0.0254 m = 0.2286 m
- Convert the weight of the lead balls from pounds to kilograms:
weight = 348 lb = 348 * 0.453592 kg = 157.850 kg
Now you can calculate the force of attraction between the larger and smaller balls by substituting the values into the equation F(r) = Gm1m2/r^2:
F(r) = (6.674E-11)(0.730283)(157.850)/(0.2286)^2
After evaluating this expression using a calculator, you should get the force of attraction between the larger and smaller balls.
For part B, you are asked to compare this force to the weight of the small balls. The weight of the small balls can be calculated by multiplying the mass of each small ball (0.730283 kg) by the acceleration due to gravity (9.8 m/s^2):
weight = (0.730283 kg) * (9.8 m/s^2)
Once you have both values, compare them and see how they relate to each other.