taki Posted January 29, 2007 Author Report 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 I am not sure about T, but if c is the force on the Y sphere, can't we use coulomb's law? F = {Kc* lq1l *lq2l}/r^2 (Kc times absolute value of test charge times absolute value of source charge divided by r squared) constant: Kc = 9.0x10^9 Nm^2C^-2 BTW...thanks for the replies so far Quote
Jay-qu Posted January 29, 2007 Report Posted January 29, 2007 well then I think we would have to have more info about the charges, clearly they must be opposite since they repel, but are they equal in magnitude? Quote
ronthepon Posted January 29, 2007 Report Posted January 29, 2007 Well, you do know [math]c[/math]. All you need to do now is to take the terms two-at-a-time and find the respective forces. However you'll need to know the values of the 'X' and 'Y' charges for this to be solvable for numerical values. Quote
taki Posted January 30, 2007 Author Report Posted January 30, 2007 Well, you do know [math]c[/math]. All you need to do now is to take the terms two-at-a-time and find the respective forces. However you'll need to know the values of the 'X' and 'Y' charges for this to be solvable for numerical values. Thank you! That is exactly right. After finding "c," all I had to do was sum the forces in the x diretion to find "T" (tension) . Then plug tension into the sum the forces in the y direction, and then solve for mass. F(x) = c + (-Tsin30) = 0 F(y) = -mg + (T cos30) = 0m = (Tcos30)/g It was simple, but the fundamental part I previously overlooked was the fact that the objects were not moving therfore, sum of the forces in the x and y direction equal zero. Thanks to all who helped! Quote
InfiniteNow Posted January 30, 2007 Report Posted January 30, 2007 Taki, I love your enthusiasm. Keep up your hard work and engagement with learning. It's the good kind of contagious. :lol: Quote
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