Question

Calculate the position of the centre of mass of \(2 \mathrm{~kg}\) placed at \(x=1,3 \mathrm{~kg}\) placed at \(x=4,1 \mathrm{~kg}\) placed at \(x=6\) and \(6 \mathrm{~kg}\) placed at \(x=-5\).

Short answer

Answer: The position of the center of mass is at \(x_{cm} = -\frac{5}{6}\).

Step-by-step solution

Identify the masses and positions

We are given four masses and their positions: 1. \(m_1 = 2\,kg\) at \(x_1 = 1\) 2. \(m_2 = 3\,kg\) at \(x_2 = 4\) 3. \(m_3 = 1\,kg\) at \(x_3 = 6\) 4. \(m_4 = 6\,kg\) at \(x_4 = -5\)

Calculate the numerator of the equation

Find the sum of the product of each mass and its position: $$\sum{m_i x_i} = m_1 x_1 + m_2 x_2 + m_3 x_3 + m_4 x_4$$ $$\sum{m_i x_i} = (2\times1) + (3\times4) + (1\times6) + (6\times-5)$$ $$\sum{m_i x_i} = 2 + 12 + 6 - 30 = -10$$

Calculate the denominator of the equation

Find the sum of the masses: $$\sum{m_i} = m_1 + m_2 + m_3 + m_4$$ $$\sum{m_i} = 2 + 3 + 1 + 6 = 12$$

Calculate the position of the centre of mass

Divide the sum of the products found in Step 2 by the sum of the masses found in Step 3: $$x_{cm} = \frac{\sum{m_i x_i}}{\sum{m_i}}$$ $$x_{cm} = \frac{-10}{12} = -\frac{5}{6}$$ So, the position of the centre of mass is at \(x_{cm} = -\frac{5}{6}\).

Key concepts

Mass Distribution

Understanding the concept of mass distribution is essential when calculating the center of mass. Mass distribution refers to how mass is spread out in a space. In our example, we have discrete masses placed at different positions along a line, which creates a certain distribution pattern. When performing center of mass calculations, you're essentially finding the balance point, or the average location, of the entire mass distribution. If you were to balance this system on a fulcrum, the center of mass would be the point where the system remains in balance without tipping over. This has huge implications in engineering designs such as in the manufacturing of vehicles where the mass distribution must ensure stability.

Mechanical Equilibrium

Mechanical equilibrium is a state where the sum of forces and moments about any point is zero; there's no net movement or rotation. While calculating the center of mass, we implicitly assume that the system is in mechanical equilibrium. This means all the masses in the system are at rest relative to one another. If the calculated center of mass in our example were physically supported, the objects would remain stationary. This concept is pivotal as the stability of structures often depends on maintaining mechanical equilibrium. For instance, understanding this helps engineers ensure that buildings don't topple by keeping the center of mass within the base of support.

Physics for Engineers

The principles of physics are the foundation of engineering. Mechanics, particularly the study of forces and equilibrium, is a branch of physics that is heavily relied upon in engineering disciplines. These principles enable engineers to design structures and systems that can withstand various loads and stresses. Our example of center of mass calculation is a basic application of physics for engineers. It demonstrates how understanding the mechanics of objects can contribute to more effective and safe designs. Moreover, routinely working through physics problems equips engineers with the analytical skills necessary to solve real-world challenges.

Statics

Statics is a branch of mechanics concerned with the analysis of loads (force, moment) on physical systems in static equilibrium—that is, in a state where the relative positions of subsystems do not vary over time. In the context of our exercise, statics involves the calculation of the center of mass in a system that is not moving or rotating. By learning how to perform these static calculations, like finding the center of mass for various objects, engineering students gather essential skills for solving practical problems, such as determining where to place the supports for a bridge or figuring out the load distribution on a structural beam.