Electrostatic force

  • #1
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)
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi peterpang1994! welcome to pf! :smile:

(have a square-root: √ and a degree: ° and try using the X2 icon just above the Reply box :wink:)

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
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.
 
  • #4
tiny-tim
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hi peterpang1994! :wink:

looks fine so far :smile:

(btw, you can find θ easily from the fact that there's an isoceles right-angled triangle)

now how about the centre of mass? :wink:
 
  • #5
Yes θ is 75° , the balls are consider as point masses
 
  • #6
tiny-tim
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yes, but where is the centre of mass of the pair?
 
  • #8
tiny-tim
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think! :rolleyes:
 
  • #9
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)
 
  • #10
I still have no idea with this question, I just keep on asking how come this mc so difficult
 
  • #11
tiny-tim
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take moments about the pivot :wink:
 
  • #12
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??
 
  • #13
tiny-tim
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… 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" ? :confused:)

it is simple! :smile:

you didn't need to use the tensions, did you? :wink:
 
  • #14
MC is multiple choices question . But this MC is inside the chapter about electrostatic force. I did try to find the tension T.T
 

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