Question

In a lecture demonstration, a professor pulls apart two hemispherical steel shells (diameter \(D\)) with ease using their attached handles. She then places them together, pumps out the air to an absolute pressure of \(p\), and hands them to a bodybuilder in the back row to pull apart. (a) If atmospheric pressure is \({p_0}\) , how much force must the bodybuilder exert on each shell? (b) Evaluate your answer for the case \(p\) \(=\) 0.025 atm, \(D\) \(=\) 10.0 cm.

Short answer

The bodybuilder must exert approximately 780 N on each shell.

Step-by-step solution

Understanding the Problem

We need to find the force that the bodybuilder must exert to separate the two hemispheres after the air has been pumped out. Once the air inside is pumped out, the pressure difference between the inside and the atmospheric pressure will create a force. This force is what the bodybuilder needs to overcome to pull the hemispheres apart.

Calculate the Area of the Hemisphere

The force exerted by the pressure difference acts over the area of the hemisphere. The effective area over which the pressure acts is the area of the circular face of the hemisphere. Since the diameter is given as \( D = 10.0 \) cm, the radius \( r \) is \( \frac{D}{2} = 5.0 \) cm. The area \( A \) of the circular face is \( \pi r^2 \). So, \( A = \pi (5\, \text{cm})^2 \).

Convert Diameter and Calculate Area

First, convert the radius into meters: \( r = 0.05 \) m. Then calculate the area: \( A = \pi (0.05)^2 = 7.85 \times 10^{-3} \) m².

Calculate the Force Required

The force needed can be determined by the formula \( F = (p_0 - p) \cdot A \). Here \( p_0 \) is atmospheric pressure and \( p \) is the internal pressure. Convert pressures into the same units. Atmospheric pressure \( \approx 1 \) atm, which equals \( 101325 \) Pa (Pascals). Let \( p = 0.025 \) atm \( = 0.025 \times 101325 \) Pa. Calculate \( F \) using these values.

Substitute and Solve for Force

Substitute the known values into the equation: Pressure difference \( (p_0 - p) = 101325 - (0.025 \times 101325) \). Calculate this to find the difference in Pascals. Then, \( F = (p_0 - p) \times A = 99391.875 \times 7.85 \times 10^{-3} \) which gives \( F = 780 \) N approximately.

Key concepts

Atmospheric Pressure

Atmospheric pressure is the force exerted by the weight of the air above us. It is a key concept in understanding how forces work on objects in the air. - Imagine a column of air above your head; that's the weight of the atmosphere pressing down. - Typically, at sea level, atmospheric pressure is about 101,325 Pascals (Pa) or 1 atmosphere (atm). When you pump air out from between the hemispheres, the atmospheric pressure outside becomes significant. It starts pushing the hemispheres together. This pressure difference is what the bodybuilder needs to overcome. To wrap your mind around it, think of it like a suction cup: once you remove the air inside, the outside pressure holds it tight to the surface. This is why it becomes hard to separate the hemispheres after air removal.

Force Calculation

Force calculation in this context revolves around understanding how pressure differences generate force. - The atmosphere presses down on the hemispheres with a pressure of about 101,325 Pa, while the pressure inside is significantly lower due to the air being pumped out. - Force can be defined by the formula: \[F = (p_0 - p) \cdot A\]where \( p_0 \) is the atmospheric pressure, \( p \) the internal pressure, and \( A \) the area over which the pressure acts.This force equation tells us the amount of force the bodybuilder must apply to overcome the holding power of the pressure difference. You first find the pressure difference between inside and outside, then multiply it by the area of contact to get the force necessary for separation. It highlights how small changes in pressure can result in significant forces.

Hemisphere Area Calculation

Before calculating the force, it's vital to determine the area over which the pressure acts, which, in this case, is the circular face of the hemisphere. - Given a hemisphere of diameter 10 cm, the radius \( r \) is half of that, \( r = 5 \) cm or \( 0.05 \) m when converted to meters.- The area \( A \) of this circular face can be calculated using the formula for the area of a circle: \[A = \pi r^2\]- Substituting \( r = 0.05 \) m gives: \[A = \pi (0.05)^2 \approx 7.85 \times 10^{-3} \text{ m}^2\]This area calculation is crucial because it determines the region over which the external atmospheric pressure is active, enabling the calculation of the required force to separate the hemispheres after the air is pumped out.