Final answer:
The question involves finding the maximum diameter of a steel needle that can float on water due to surface tension. The floating condition is achieved when the force due to surface tension equals the weight of the needle. To calculate dmax, the surface tension, the density of the needle, and gravitational force are taken into account.
Step-by-step explanation:
The scenario involves a steel needle floating on the surface of a liquid due to surface tension. To find the maximum diameter dmax that a needle can have to float on water, we can use the concept of surface tension and balancing forces. The weight of the needle is balanced by the surface tension force acting along the perimeter of the needle in contact with water.
For a cylindrical needle with diameter d, length L, and density rho_n (the symbol rho represents density), the weight of the needle is W = rho_n * g * Volume, where g is the acceleration due to gravity and Volume is pi * (d/2)^2 * L. Since we are assuming a contact angle of 0 degrees and neglecting buoyancy, the maximum supporting force due to surface tension is Fmax = 2 * pi * (d/2) * L * Y, where Y is the surface tension of the liquid.
By setting W = Fmax and solving for d, we arrive at dmax which is the maximum diameter of the needle that can float. For a steel needle (SG = 7.84) in water at 20 degrees Celsius (Y = 0.0728 N/m), we substitute rho_n as SG * rho_water, and using the equation, we can calculate dmax.
To calculate dmax:
- Find the volume of the needle: Volume = pi * (d/2)^2 * L
- Calculate the weight of the needle: W = rho_n * g * Volume
- Calculate the maximum force by surface tension: Fmax = 2 * pi * (d/2) * L * Y
- Balance the weight and surface tension force: W = Fmax
- Solve for d to find the dmax