# Electrostatic force

I have some problems withe this question:

Two positively charged particles A and B of masses of mA and mB respectively are suspended by two insulating threads of the same length from O. Due to the electrostatic repulsive force between A and B, threads from AO and BO make 30 degrees and 60 degrees respectively with the vertical as shown. Find the ratio mA:mB

I can find mA+mB but I can't find mA:mB

(the correct answer is 31/2 : 1)

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## Answers and Replies

tiny-tim
Science Advisor
Homework Helper
welcome to pf!

hi peterpang1994! welcome to pf! (have a square-root: √ and a degree: ° and try using the X2 icon just above the Reply box )

Show us what you've tried, and where you're stuck, and then we'll know how to help! thanks a lot tim,
First I let the electrostatic force be F, TA and TB be the tension of the threads linking the ball A and B respectively and θ be the angle between the vertical and the electrostatic force . And the balls are steady.
Horizontally,
TAsin30°=Fsinθ=TBsin60°
TA=(TBsin60°)√3
TA=(TB)√3

Vertically,
TAcos30°=Fcosθ + mAg
TBcos60°=Fcosθ - mBg

(TAcos30° + TBcos60°)/g = mA + mB
mA + mB = 2TB/g

I am stuck here.

tiny-tim
Science Advisor
Homework Helper
hi peterpang1994! looks fine so far (btw, you can find θ easily from the fact that there's an isoceles right-angled triangle)

now how about the centre of mass? Yes θ is 75° , the balls are consider as point masses

tiny-tim
Science Advisor
Homework Helper
yes, but where is the centre of mass of the pair?

that is unknown

tiny-tim
Science Advisor
Homework Helper
think! the distance between the centre of mass and the ball A (R) and d be the distance between ball A and B (r) :

R = mBr/(mA+mB)

I still have no idea with this question, I just keep on asking how come this mc so difficult

tiny-tim
Science Advisor
Homework Helper
take moments about the pivot If I take the ball A as the center of rotation, the net moment = 0 ,

rTBsin45°=rmBsn75°
mB=TBsin45°/sin75°

If I take the ball B as the center of rotation, the net moment = 0 ,

rTAsin45°=rmAsn75°
mA=TAsin45°/sin75°

mA/mB = TA/TB = √3

thanks a lot!!!! But is there any other simple way to solve this MC? Since that is just a MC in my book??

tiny-tim
Science Advisor
Homework Helper
… thanks a lot!!!! But is there any other simple way to solve this MC? Since that is just a MC in my book??

(what's an "MC" ? )

it is simple! you didn't need to use the tensions, did you? MC is multiple choices question . But this MC is inside the chapter about electrostatic force. I did try to find the tension T.T