Proving F=-constant*(delta y) for delta y < d

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The discussion centers on proving the equation F = -constant * (delta y) for delta y < d, involving forces acting on a white ball in a stable system. Key forces include weight (mg), tension (T), and electric force (F = k * q² / d²). Participants suggest using Newton's second law to derive relationships between these forces and the angle theta, ultimately leading to a substitution involving delta y and the radius r. The goal is to demonstrate that the negative charge behaves like a spring system, oscillating vertically.

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I need to prove that F=-constant*(delta y) if delta y < d

I have no idea how to go about this. Please help!
 

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Mango12 said:
I need to prove that F=-constant*(delta y) if delta y < d

I have no idea how to go about this. Please help!
Look at one of the white balls. There are three forces acting on it: 1) the weight, 2) the tension, and last but not least, 3) the electric force. If the system is stable then the sum of these forces is 0 N.

Can you finish from there or do you need more detail?

-Dan
 
topsquark said:
Look at one of the white balls. There are three forces acting on it: 1) the weight, 2) the tension, and last but not least, 3) the electric force. If the system is stable then the sum of these forces is 0 N.

Can you finish from there or do you need more detail?

-Dan

well I think weight is mgsin(theta) and tension is Fesin(theta) + mgcos(theta)

But I don't know theta and I don't get what delta y has to do with it
 
Mango12 said:
well I think weight is mgsin(theta) and tension is Fesin(theta) + mgcos(theta)

But I don't know theta and I don't get what delta y has to do with it
I didn't notice this last night...What force are you trying to find?

Either way you are likely to have to consider at least some of the following...

Consider the charge on the right. The electric force between two point charges is [math]F = \frac{k q^2}{d^2}[/math]. The weight of the ball is mg, and you have a tension force,T, acting up and to the left at an angle [math]\theta[/math]. Using the usual xy coordinate system we have Newton's 2nd:
[math]\sum F_x = -T~cos( \theta ) + \frac{kq^2}{d^2} = 0 [/math]

[math]\sum F_y = T~sin( \theta ) - mg = 0[/math]

Use the y equation to get an equation for T. Put that T value into the x equation. That will get you an equation for d in terms of [math]\theta[/math]. Beyond that I don't know what force you are trying to calculate.

-Dan
 
topsquark said:
I didn't notice this last night...What force are you trying to find?

Either way you are likely to have to consider at least some of the following...

Consider the charge on the right. The electric force between two point charges is [math]F = \frac{k q^2}{d^2}[/math]. The weight of the ball is mg, and you have a tension force,T, acting up and to the left at an angle [math]\theta[/math]. Using the usual xy coordinate system we have Newton's 2nd:
[math]\sum F_x = -T~cos( \theta ) + \frac{kq^2}{d^2} = 0 [/math]

[math]\sum F_y = T~sin( \theta ) - mg = 0[/math]

Use the y equation to get an equation for T. Put that T value into the x equation. That will get you an equation for d in terms of [math]\theta[/math]. Beyond that I don't know what force you are trying to calculate.

-Dan

I'm not sure. I do know I need to show that the negative charge would bounce up and down as if on a spring.
 
Mango12 said:
I'm not sure. I do know I need to show that the negative charge would bounce up and down as if on a spring.
Show that F = -constant * delta y

This is what I have so far that we did in class. He said we needed to make a substitution somewhere but I don’t understand where

2F=[Kq^2/r^2]2

Fy = [2Kq^2/r^2] sin theta

Sin theta = delta y/r
Sin theta = delta y/(d/2)
 

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