taki Posted January 26, 2007 Report Share Posted January 26, 2007 A student performs Coulomb's classic experiment. Shes adds Y coulomb of positive charge to a small metal sphere that is suspended by a thread. Shes adds X coulomb of positive charge to a small metal sphere that is held on a glass rod. She positions the X coulomb sphere as indicated in the diagram and because like charges repel eachother the Y coulomb sphere is pushed out to the right. Calculate the mass of the Y coulomb spere. hint:It is bigger than you might guess. angle=30 degreesX = 3Y = 6 l l 30 cm l l O O X Y Quote Link to comment Share on other sites More sharing options...
taki Posted January 26, 2007 Author Report Share Posted January 26, 2007 Ok maybe someone know how distance, coulomb force, and mass is related? Quote Link to comment Share on other sites More sharing options...
Pyrotex Posted January 26, 2007 Report Share Posted January 26, 2007 Ok maybe someone know how distance, coulomb force, and mass is related?Let me dust off my favorite Physics text book. What fun!! Quote Link to comment Share on other sites More sharing options...
taki Posted January 26, 2007 Author Report Share Posted January 26, 2007 Can someone tell me if this is right? 30 cm = 0.3 m(0.3m)(tan 30) = 0.17 m (distance) F= (9.0x10^9N.m^2C^-2)(3.00 C)(6.00 C)/ (0.17m^2) = 5.6e12 N (coulomb's law)^^ 1N = 1 kg*m/s^2 F=mgm = F/gm = (5.6e12 kg*m/s^2)/(9.8 m/s^2) = 5.7e11 kg Quote Link to comment Share on other sites More sharing options...
taki Posted January 26, 2007 Author Report Share Posted January 26, 2007 Anyone? Maybe someone can tell me if I am to determine the mass, do I only consider the forces in the x-direction. Or the sum of the forces in both directions. I would think F=mg would be enough to relate the coulmb force sphere Y "feels" and mass of the sphere. However I think maybe the tension should be considered. Someone please help. Quote Link to comment Share on other sites More sharing options...
taki Posted January 28, 2007 Author Report Share Posted January 28, 2007 Anyone know about Coulomb's law, mass, distance? Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted January 28, 2007 Report Share Posted January 28, 2007 Ok buddy, slow down, your questions will be answered, you just need to give us some time :cup: From what I can surmise of your diagram, the Y sphere has been pushed out and now subtends a 30 degree angle with the vertical - is this correct? Yes 30cm is .3m :shrug: Quote Link to comment Share on other sites More sharing options...
taki Posted January 28, 2007 Author Report Share Posted January 28, 2007 Ok buddy, slow down, your questions will be answered, you just need to give us some time :cup: From what I can surmise of your diagram, the Y sphere has been pushed out and now subtends a 30 degree angle with the vertical - is this correct? Yes 30cm is .3m :shrug: Yes, sorry the diagram is not so good. Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted January 28, 2007 Report Share Posted January 28, 2007 Assuming a few things along the way, this is what I have got to: Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted January 28, 2007 Report Share Posted January 28, 2007 LOL :doh: accidently did the wrong one.. [math]cos{30}=\frac{mg}{T}[/math][math]m=\frac{T}{g}cos{30}[/math] Quote Link to comment Share on other sites More sharing options...
taki Posted January 28, 2007 Author Report Share Posted January 28, 2007 Assuming a few things along the way, this is what I have got to: YES, that is exactly right. First mistake I made was I used tan instaed of sin (thanks). Now since I can calulate the Coulomb's force on Y, can I relate it to F=mg? I am concerned that maybe I didn't consider all the forces in the X and Y direction, like tension. Then again when I think about "mg" it seems it only has do with the Y direction. Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted January 28, 2007 Report Share Posted January 28, 2007 sorry see post #10 :shrug: I was un-needingly complicating the matter! Quote Link to comment Share on other sites More sharing options...
taki Posted January 28, 2007 Author Report Share Posted January 28, 2007 sorry see post #10 :shrug: I was un-needingly complicating the matter! What do you put in for tension "T" (for this particular problem)? From what I can see in your diagram you used the length of the string (.3m) for T. Is that correct? Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted January 28, 2007 Report Share Posted January 28, 2007 No, this is the part where I am wrong.. I have been trying to work it out, when I get home I will have a look in my physics book! It is definitely not .3m, for mass to be correct you must know the T in newtons.. :shrug: Ill be back J Quote Link to comment Share on other sites More sharing options...
ronthepon Posted January 29, 2007 Report Share Posted January 29, 2007 For problems involving three forces in equilibrium (like these) you can use the wonderful Lami's theorem. I'll use JQ's picture as a reference. The forces [math]T[/math], [math]c[/math] and [math]mg[/math] are in equilibrium. By the usage of elementary geometry, you'll figure that there's an angle of 30+90 = 120 degrees between [math]c[/math] and [math]T[/math]. Ofcourse, that's when you assume that [math]c[/math] and [math]mg[/math] are at 90 degrees to each other. What remains is the angle between [math]T[/math] and [math]mg[/math], and it's 360 - (120 + 90) = 150 degrees. Now the only thing remaining for you to do is use Lami's theorem (for three forces in equilibrium) It's going to be written out as: [math] \frac{T}{Sin90} = \frac{mg}{Sin120} = \frac{c}{Sin150} [/math] Use the values you know, and find all the unknowns. For more info on Lami's theorem, do see the wiki link I've given. Quote Link to comment Share on other sites More sharing options...
ronthepon Posted January 29, 2007 Report Share Posted January 29, 2007 And, er, for general problems like these (eg: wit more or less than just three forces) You can perform a vector resolution of all the forces present in any two mutually perpendicular directions. Have you studied vectors, taki? Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted January 29, 2007 Report Share Posted January 29, 2007 I am not so sure our approach will work in this case ron.. well not yet at any rate, we first need to find what C or T is Quote Link to comment Share on other sites More sharing options...
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