# Two electrically charged pith balls

Two electrically charged pith balls, each having a mass of 0.015 kg, are suspended from light threads of equal length. Determine the resultant horizontal force of repulsion, F, acting on each ball if the measured distance between them is r = 200 mm.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

We will focus on one pith ball. You can pick either, but we focused on the right pith ball. Let us draw the free body diagram for the right pith ball.

To figure out the angle (shown in the free body diagram), we will use the given dimensions.

The angle can be calculated using the inverse of cosine (adjacent over hypotenuse).

$\text{cos}^{-1}(\frac{90}{150})= 53.1^0$

We will assume forces going $\rightarrow^+$ to be positive and $\uparrow+$ to be positive.

Let us now write equations of equilibrium for the y-axis forces first, as this will give us a value for T.

$T\text{sin}\,(53.1^0)-0.14715=0$
(solve for T)

$T=0.184$ N

Show me the FBD again

Now, we will write equations of equilibrium for the x-axis forces.

$F-0.184\text{cos}\,(53.1^0)=0$
(Substituted the value of T we found earlier. Solve for F)

$F=0.11$ N

Show me the FBD again