A sixlegged water bug with a mass 11 g is suspended on the s

A six-legged water bug with a mass 1.1 g is suspended on the surface of a pond by surface tension. If each leg is in contact with the water over the same length, what is that length? Assume water does not wet the legs. What additional assumptions did you have to make to answer that question?

Solution

Assumptions :

The atmosphere of air and water in contact with each other are at standard temperature and pressure (STP).

At STP conditions, the surface tension of water in contact with air is S = 0.073 N/m.

The total force on the water bug due to surface tension is

Fs = S * 6 legs * L

Where L is the contact length of each leg with water and

S is the surface tension = 0.073 N/m

This force acts upwards on the water bug

The other force on the water bug is due to gravity. It is given by

Weight of water bug, Fg = Mass of bug * acceleration due to gravity.

Fg = 1.1 *10^(-3) kg * 9.81 m/s^2

===> Fg = 0.010791 N

This force acts downwards.

As the water bug just floats on the water surface, these forces balance each other.

Thus we have Fg = Fs

===> 0.010791 N = 0.073 N/m * 6 legs * L m ===> L = 0.02464 m or 24.64 mm

Thus the contact length of each leg of the water bug with water surface is L = 24.64 mm