Hanging pith ball problem with unequal masses and charges

  • Thread starter furth721
  • Start date
  • #1
6
0

Homework Statement



Suppose that two balls are suspended by identical threads of length .10m anchored to the same point. The pith balls have different masses and charges. The charge of one is 2.0 X 10^-7 C and the other one is 6.0 X 10^-8 C. Both threads make a 25 degree angle at equilibrium with the vertical. What is the mass of each ball?



Homework Equations



culombs law = kq1q2/r^2

weight of ball 1 = m1g

weight of ball 2 = m2g

Tension in x direction = Tsin(25) = kq1q2/r^2

Tension in y direction of ball 1 = Tcos(25) = m1g

Tension in y direction of ball 2 = Tcos(25) = m2g

r = distance between charges = .10sin(25)

The Attempt at a Solution



I drew a free body diagram of the first ball and set the x and y components of the tension equal to the weight and the force of the like charges applied on ball one. Since both strings make a 25 degree angle with the vertical the pith balls would be the same distance away from the vertical and have equal tensions and masses. This is where I think I am making a mistake. I solve for the mass of the first ball by solving for T in one equation and subbing that into the other equation. I get that the mass of both balls are something like 3.3 x 10 ^ -3 kg. The answer in the back of the book says that the masses of both balls is 1.2 x 10 ^ -4 kg. I would really appreciate if someone could give me their input on the problem.
 

Answers and Replies

  • #2
The force due to gravity (its weight), the force due to the tension and the electrostatic repulsive force should all cancel out to give a net force of zero since the balls are no longer moving.

Also, the distance between charges is incorrect I think, that's half the distance. It should be (0.10*sin (25))*2
 
  • #3
6
0
Oh yeah, the distance is .20sin(25) that was a typo on my part. I understand that because I've done problems like this before where the masses and charges were equal. To solve it you have to set the weight equal to Tcos(25) which is the y component of the tension and the force of the charge equal of the x component then solve for T and substitute into the other equation. The thing I am having trouble with is, the balls are supposed to be two different masses but they make the same angle with the vertical which means that the tension acting on both of the balls is equal, and the Force of charge on the balls is equal shouldn't that make the mass equal as well?
 
  • #4
Hmm, that's interesting. I actually didn't notice it stated different masses until you said it now. I would've thought that being two different masses the angles should not be the same provided all other things remain unchanged for both balls.

I'm stumped, maybe someone else can provide some insight in to this, because I'm interested to know as well.
 

Related Threads on Hanging pith ball problem with unequal masses and charges

  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
12K
Replies
2
Views
1K
  • Last Post
Replies
12
Views
28K
  • Last Post
Replies
20
Views
3K
Replies
3
Views
666
  • Last Post
Replies
2
Views
6K
Replies
4
Views
2K
Top