Electrostatic Force: Finding mA:mB Ratio

[peterpang1994]

I have some problems withe this question:

Two positively charged particles A and B of masses of m_A and m_B 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 3^1/2 : 1)

 

[tiny-tim]

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!

 

[peterpang1994]

thanks a lot tim,
First I let the electrostatic force be F, T_A and T_B 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,
T_Asin30°=Fsinθ=T_Bsin60°
T_A=(T_Bsin60°)√3
T_A=(T_B)√3

Vertically,
T_Acos30°=Fcosθ + m_Ag
T_Bcos60°=Fcosθ - m_Bg

(T_Acos30° + T_Bcos60°)/g = m_A + m_B
m_A + m_B = 2T_B/g

I am stuck here.

 

[tiny-tim]

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? ​

 

[peterpang1994]

Yes θ is 75° , the balls are consider as point masses

 

[tiny-tim]

yes, but where is the centre of mass of the pair?

 

[peterpang1994]

that is unknown

 

[tiny-tim]

think!

 

[peterpang1994]

the distance between the centre of mass and the ball A (R) and d be the distance between ball A and B (r) :

R = m_Br/(m_A+m_B)

 

[peterpang1994]

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

 

[tiny-tim]

take moments about the pivot

 

[peterpang1994]

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

rT_Bsin45°=rm_Bsn75°
m_B=T_Bsin45°/sin75°

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

rT_Asin45°=rm_Asn75°
m_A=T_Asin45°/sin75°

m_A/m_B = T_A/T_B = √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]

… 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?

 

[peterpang1994]

MC is multiple choices question . But this MC is inside the chapter about electrostatic force. I did try to find the tension T.T