# Finding the charge on two beads?

by janetxo
 P: 5 1. The problem statement, all variables and given/known data Two small insulating spherical bead are placed on a taut(insulating, isolated) string which has a knot at its lower end. The string is stretched at an angle of 60degress to the horizontal. The beads can move frictionlessly on the string. The beads are given identical positive charges (which are distributed evenly over the tiny volumes of the beads). The lower bead rests against the knot. The other bead comes to rest 5cm away from the knot on the string. The mass of each bead is 0.1g. What is the charge on the beads? 2. Relevant equations 3. The attempt at a solution Would I use work? Or something else? Not quite sure Just an idea of even what equation to start with would be nice
 PF Patron HW Helper Thanks P: 3,666 Think about how it is possible that the bead can remain at rest. What forces act on the bead?
 P: 5 well there would be the force that pushes the two beads apart, and gravity?
P: 5

## Finding the charge on two beads?

 Quote by TSny Think about how it is possible that the bead can remain at rest. What forces act on the bead?
Well there would be the force that pushed the two beads away, and gravity?
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 Quote by janetxo Well there would be the force that pushed the two beads away, and gravity?
Yes. But there is a third force. Hint: The bead is in contact with the string.

What is the condition that must be met by these 3 forces for the bead to remain at rest?
P: 5
 Quote by TSny Yes. But there is a third force. Hint: The bead is in contact with the string. What is the condition that must be met by these 3 forces for the bead to remain at rest?
So, a normal force between the bed and the string? (as there is no friction). and the net forces would have to be 0 in order for it to be at rest. Fnet=ma=0.
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 Quote by janetxo So, a normal force between the bed and the string? (as there is no friction). and the net forces would have to be 0 in order for it to be at rest. Fnet=ma=0.
Yes, that's right.
P: 5
 Quote by TSny Yes, that's right.
thanks for walking me through that bit! Wasnt quite sure how to start, :) really apperiate it
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