Question

I would like a step by step solution please and clear to read.

Surface Tension and Bubbles Surface tension is a force caused by the attraction of liquid molecules on a boundary surface for each other. It acts tangential to the surface and is modeled as a single number, the surface tension Consider a square film on a wire frame. The bottom edge is movable and, when the frame is vertical, as shown, the surface tension of the film pulls upward and gravity pulls downward. The boundary of interest here is the bottom edge of the film on the moving bar, and the film is a double layer, so the force due to surface tension is This must balance the force of gravity, so we have Now, we can apply this idea to a bubble, either a gas bubble in a liquid or a drop of liquid. In either case, there is a single gas- liquid boundary and the surface tension will be that of the liquid. We will assume a spherical shape (which is what you see for small bubbles and drops) at equilibrium with the pressure. Consider such a small spherical drop of liquid with external pressure To analyze the surface tension pressure equilibrium, we will focus on one hemisphere and the pressure difference across the boundary [1] If we considered the entire drop, what would the total pressure difference in the x direction be, including direction? The pressure difference across the surface of the hemisphere and the surface tension along the surface must balance to be in equilibrium. We wil first do this in a straightforward way using calculus to get the sum of forces on the parts of the hemisphere, then we will analyze the result and come up with a smarter way to do it which will be very helpful when we study electromagnetic fields. [2] First, consider symmetry. With the x axis along the axis of the hemisphere, what happens to any force components in the y and z directions, once the forces on all the patches on the hemisphere are added? [3] Next, focus on the strip of width in the figure. The pressure difference gives an outward force, normal to the surface. Keeping in mind that the width of the strip is extremely small, write the expression for the area of the strip. [4]Give the force due to the pressure difference acting on the strip, using the area expression you gave above

Answer 1

DF = (P_in - P_out)dA: dA = 2 pi R sin theta . Rd theta dF = (P_in - P_out)2pi R sin theta Rd theta F_y = integral dF cos theta = integral (P_in - P_out)2pi R sin theta Rd theta. Cos theta = (P_in - P_out)pi R^2 integral^pi/2_0 sin 2 theta d theta { 2 sin theta cos theta = sin 2 theta } = (P_in - P_out)pi R^2 [-cos 2 theta/2]^pi/2_0 { integral sin (ax + b)dx = -cos (ax + b/a + c) = (P_in - P_out)pi R^2 [-cos pi/2 - (- cos theta/2] F_y = (P_in - P_out)pi R^2 F_3 = integral dF sin theta = 0 cancels out due to symmetry. \r\n dF_y = -[gamma 2 pi R sin (theta + d theta/2) sin (theta + d theta/2) - gamma 2 pi R sin (theta - d theta/2)sin (theta - d theta/2) dF_y = -2 pi R gamma [sin^2 (theta + d theta/2) - sin^2 (theta - d theta/2)] dF_y = -2pi R gamma [sin (theta