Electrodynamics problem - force equilibrium

[y_r123]

Homework Statement

Hello! I have a question about the following problem:

Two point masses m1 and m2 are attached to isolating wires to point P. They are both positively charges (charge Q1 and Q2) and in the picture you can see the situation at equilibrium.

What is the proportion of the masses (m1 / m2)?[/B]

Homework Equations

The Attempt at a Solution

So this is how I would do it:

We have the weight force for m₁ which is W₁ = m₁ * g and for m₂ we have W₂ = m₂ * g

The forces W are is in y-direction

For other forces of the y-direction we have to break down the tension force into its x and y components.

For m₁ we can say that the tension force in y direction is T_1y = T₁ * cos (60°) and for m₂ the tension force in y direction is T_2y = T₂ * cos(30°)

this means that T₁ * cos (60°) - m₁ * g = 0
and T₂ * cos (30°) - m₂ g )= 0

or T₁ * cos (60°) = m₁ * g
and T₂ * cos (30°) = m₂ * g

If we divide the both equations we get
T1 / T2 * cos(60°) / cos(30°) = m1 / m2

for cos(60) = 1/2
for cos (30) = sqrt (3) /2

cos(60)/cos(30) = 1/sqrt(3)

which leads us to T1 / T2 * 1/sqrt(3) = m1/m2

how can I get rid of T1/T2 and what is m1/m2?

Thanks for your help!

 

[BvU]

Hi Y,

we have to break down the tension force into its x and y components

So a this point you need another equation involving these two ratios. Any idea ?

 

[y_r123]

well the x components are T₂ * sin(60°) - T₁ * sin(30°) = 0

which is T₂ * sqrt (3)/2 = T₁ * 1/2

so T1 / T2 = sqrt (3)

if we but that in our first equation then we get m₁ / m₂ = sqrt (3) * 1/sqrt(3) which should be 1

and the possible given solutions are:

A) 3 B) 1/3 C) 1/sqrt(3)

so I don't know where I did something wrong

Y

 

[gneill]

well the x components are T2 * sin(60°) - T1 * sin(30°) = 0

Are you sure you've chosen the right trig function?

 

[y_r123]

ok I see the mistake and if I turn it around it should be 1/3

 

[gneill]

Right. You might have also taken a slightly different approach by considering the torque about point P. Since the assembly is in equilibrium that torque should be zero. You should know the relative lengths of the sides of the common 30-60-90 triangle.