# Calculation of the weight of an insect floating by surface tension

• srm

## Homework Statement

The surface of a liquid is just able to support the weight of a six-legged insect. The leg ends can be assumed to be spheres each of radius 3.2 × 10−5 m and the weight of the insect is distributed equally over the six legs. The coefficient of surface tension in this case is 0.1 N/m and the angle of the footfall with respect to the vertical is θ = π/3 radians (see figure). The mass of the insect is close to_

## Homework Equations

For calculating the surface tension F= T×L where T is the surface tension in F/m and L is the length over which the surface tension works.

## The Attempt at a Solution

I assumed that the sphere is floating halfway, so the surface tensions acts on the length 2πR. I calculated 2πR which is 20.1×10-5m. Then I multiplied it with T, and the force turns out to be 2.01×10-5 N. Multiplied it by cosπ/3 and obtained 1×10-5. This must be the force on one leg. So, I multiplied it with 6 and got 6×10-5N. Now if you calculate the mass, it turns out to be about 6.122×10-6. But the answer is 5×10-6.
Where am I going wrong?

Hello srm, What a nice exercise !

I assumed that the sphere is floating halfway
Does that correspond to what you (not me, I don't even see a figure ) see in the figure ?