Inez is putting up decorations for her sister\'s quinceanera (fifteenth birthday
ID: 1566842 • Letter: I
Question
Inez is putting up decorations for her sister's quinceanera (fifteenth birthday party). She ties three light silk ribbons together to the top of a gateway and hangs a rubber balloon from each ribbon (see figure below). To include the effects of the gravitational and buoyant forces on it, each balloon can be modeled as a particle of mass 2.20 g, with its center 48.5 cm from the point of support. Inez rubs the whole surface of each balloon with her woolen scarf, making the balloons hang separately with gaps between them. Looking directly upward from below the balloons, Inez notices that the centers of the hanging balloons form a horizontal equilateral triangle with sides 27.3 cm long. What is the common charge each balloon carries?
µC
Explanation / Answer
Let Fe be the electrostatic force between one balloon and another. That force will be horizontal. Looking down from above, the horizontal force on any one balloon is the vector sum of the forces from the adjacent balloons. An equilateral triangle's angle is 60º, so the angle between the forces is 60º. The resultant force will bisect that angle, and the magnitude will be 2*Fe*cos30º. That force must be balanced by the horizontal component of the weight of the balloon, W = m*g. That component will depend on the angle the ribbon makes with the horizontal. The tie point of the ribbon is directly above the geometric center of the triangle. The distance from a vertex to the center of an equilateral triangle is L/ (2*cos30º) if L is the side of the triangle. The ribbon is the hypotenuse and the vertex distance the adjacent side to the angle the ribbon makes with the horizontal. If the ribbon length is R, then that angle is
ø = arccos(L/(2*cos30º) / R)
ø = arccos L/ (2*cos30º * R).
For L = 27.3cm and R =48.5 cm,
Then ø = 71.04º
The vertical component of tension in the ribbon must support the vertical weight of the balloon,
So Tr *sinø= m*g
Tr = m*g/sinø.
The horizontal component of Tr is Tr*cosø,
So Fh = m*g*cotø
Now equate Fh with the resultant electrostatic force .
m*g*cotø = 2*Fe*cos30º
0.0022*9.8*0.344 = 2*Fe*0.866
Fe = 0.0043 N
To produce that force between two charges of value q:
Fe = K*q²/L² (L is the distance between the charges, the length of the side of the triangle).
q = [Fe*L²/K]
q = L* [Fe/K]
We know K = 9*109 N*m²/C²
q = 0.273*[0.0043/9*10^9]
Therefore common charge on each balloon is q = 0.189 µC