The air pressure in the top of the 2 -liter pop bottle shown in Video \(\mathrm{V} 2.6\) and Fig. \(\mathrm{P} 2.118\) is \(40 \mathrm{psi}\), and the pop depth is 10 in. The bottom of the bottle has an irregular shape with a diameter of 4.3 in (a) If the bottle cap has a diameter of 1 in. what is the magnitude of the axial force required to hold the cap in place? (b) Determine the force needed to secure the bottom 2 in. of the bottle to its cylindrical sides. For this calculation assume the effect of the weight of the pop is negligible. (c) By how much does the weight of the pop increase the pressure 2 in. above the bottom? Assume the pop has the same specific weight as that of water.

Step by step solution

01

Calculation of Axial Force on Cap

Use the formula for pressure: Pressure = Force / Area. Here, Area of the cap = \( \pi (Diameter/2)^2 \) = \( \pi (1/2)^2 \) = \( 0.7854\, in^2 \). We have been given pressure = 40 psi. We can rearrange the formula and find the force: Force = Pressure * Area = 40 psi * 0.7854 \( in^2 \) = 31.416 lb.

02

Find the Force Securing the Bottom Part of the Bottle

The area can be calculated using the given diameter which will be \( \pi (4.3/2)^2 = 14.522 in^2 \). The pressure will again be 40 psi. Thus the required force will be = Pressure * Area = 40 psi * 14.526 \( in^2 \) = 580.96 lb.

03

Calculate the Increase in Pressure due to the Weight of the Pop

The pressure due to the weight of the pop can be determined with the formula Pressure = density * g * h. Here, density of water = 62.43 \( lb/ft^3 \), h (2 inches) = 0.167 ft and specific weight of pop = specific weight of water (assumed). Pressure increase caused by the weight of the pop = 62.43 \( lb/ft^3 \) * 32.2 \( ft/s^2 \) * 0.167 ft = 336.59 Pa. Note that 1 psi = 6894.76 Pa. Hence, the increase in pressure will be 336.59 / 6894.76 psi = 0.0488 psi.

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