Problem 3 - College Physics

Question

Four acrobats of mass \(75.0 \mathrm{~kg}, 68.0 \mathrm{~kg}, 62.0 \mathrm{~kg}\), and \(55.0 \mathrm{~kg}\) form a human tower, with each acrobat standing on the shoulders of another acrobat. The \(75.0-\mathrm{kg}\) acrobat is at the bottom of the tower. (a) What is the normal force acting on the \(75-\mathrm{kg}\) acrobat? (b) If the area of each of the 75.0-kg acrobat's shoes is \(425 \mathrm{~cm}^{2}\), what average pressure (not including atmospheric pressure) does the column of acrobats exert on the floor? (c) Will the pressure be the same if a different acrobat is on the bottom?

Step-by-Step Solution

Answer

The normal force acting on the 75.0-kg acrobat is 2548 N. The pressure exerted by the acrobats on the floor is 29976.47 Pa. If a different acrobat is at the bottom, the pressure would remain the same because the total weight of the acrobats and the area of the shoes do not change.

Step 1: Calculate the total weight

The total weight of the acrobats is calculated by multiplying their total mass by the acceleration due to gravity (let's assume it to be 9.8 m/s²). The total mass is \(75.0 kg + 68.0 kg + 62.0 kg + 55.0 kg = 260.0 kg\). Hence, the total weight is \(260.0 kg * 9.8 m/s² = 2548 N\).

Step 2: Find the normal force on the 75-kg Acrobat

The normal force on the 75.0-kg acrobat is equal to the total weight of the acrobats, which we have found to be 2548 N.

Step 3: Calculate pressure exerted

The pressure exerted by the acrobats is the force divided by the area, where the force is the normal force and the area is the combined area of the acrobat's shoes. Convert area from cm² to m² by multiplying by \(1 \times 10^{-4}\). Therefore, the area in m² is \(425 cm^{2} * 2 * 1 \times 10^{-4} m²/cm² = 0.085 m²\). Therefore, the pressure is \(2548 N / 0.085 m² = 29976.47 Pa\).

Step 4: Evaluate if changing the acrobat would alter pressure

If we consider a different acrobat at the bottom, the total weight of the acrobats (which is the force exerted) and the area of the shoes would remain constant, hence, the pressure exerted on the floor would not change.

Key Concepts

Normal Force Calculation

When we talk about the normal force, we're referring to the support force exerted by a surface when an object is in contact with it. Think of it as the floor pushing upwards against your feet. The normal force is always perpendicular to the surface contact. In our exercise example, the acrobat at the base of the human tower is experiencing a normal force from the ground, which equates to supporting the total weight of the tower.
This force can be calculated by summing the weights of all the acrobats above. Each acrobat's weight is their mass multiplied by the acceleration due to gravity. The total is the normal force exerted on the bottom acrobat. Understanding this force is crucial as it relates to other concepts like friction, and it also informs us about the structural load capacities that must be considered when designing objects like furniture or bridges.

Pressure Calculation

Pressure is a measure of how force is distributed over an area. In physics, it's the amount of force applied per unit area and is usually measured in Pascals (Pa). When we calculate pressure, we divide the total force exerted on a surface by the area of that surface.

Application in the Exercise

Here, the force involved is the total weight of the acrobats, and the area is the contact area of the bottom acrobat's shoes against the ground. By dividing the normal force by the shoe area, converts the area from square centimeters to square meters to match the standard units for pressure calculation. Understanding pressure is vitally important in everything from designing comfortable shoes to understanding atmospheric science, as it helps us comprehend how forces are applied and experienced.

Newton's Laws of Motion

Newton's Laws of Motion are the cornerstones of classical mechanics and are deeply involved in solving the type of physics problem presented here. The first law, also known as the law of inertia, states that an object at rest stays at rest, and an object in motion stays in motion unless acted upon by an external force. The second law provides the calculation for force (\( F = ma \), where F is force, m is mass, and a is acceleration). This is directly used to find the weight of the acrobats. The third law, known as action and reaction, implies that for every action, there is an equal and opposite reaction; in our case, the ground exerts an upward normal force equal to the downward force due to the acrobats' weight.

Force and Area Relationship

Force and area are inseparably linked when it comes to calculating pressure. The relationship between them is inversely proportional; as the area over which a force is spread increases, the pressure decreases, assuming the force remains constant. Conversely, if the area decreases and the force remains the same, the pressure increases.

Exploring the Relationship

This concept shows why it doesn't hurt when someone steps on your foot with a flat shoe but might be painful with a high heel — even though the force (the person's weight) is the same, the area of contact is much smaller with the heel, leading to higher pressure. In our acrobat example, the pressure exerted on the ground is constant regardless of which acrobat is at the bottom because the force (the total weight of the tower) and the contact area (the bottom acrobat's shoe size) do not change.